**Previous months:**

2007 - 0703(3) - 0706(2)

2008 - 0807(1) - 0809(1) - 0810(1) - 0812(2)

2009 - 0901(2) - 0904(2) - 0907(2) - 0908(4) - 0909(1) - 0910(2) - 0911(1) - 0912(1)

2010 - 1001(3) - 1002(1) - 1003(55) - 1004(50) - 1005(36) - 1006(7) - 1007(11) - 1008(16) - 1009(21) - 1010(8) - 1011(7) - 1012(13)

2011 - 1101(14) - 1102(7) - 1103(13) - 1104(3) - 1105(1) - 1106(2) - 1107(1) - 1108(2) - 1109(2) - 1110(5) - 1111(4) - 1112(4)

2012 - 1201(2) - 1202(7) - 1203(6) - 1204(6) - 1205(7) - 1206(6) - 1207(5) - 1208(5) - 1209(11) - 1210(14) - 1211(10) - 1212(4)

2013 - 1301(5) - 1302(9) - 1303(16) - 1304(15) - 1305(12) - 1306(12) - 1307(25) - 1308(11) - 1309(8) - 1310(13) - 1311(15) - 1312(21)

2014 - 1401(20) - 1402(10) - 1403(26) - 1404(10) - 1405(17) - 1406(20) - 1407(33) - 1408(50) - 1409(47) - 1410(16) - 1411(16) - 1412(18)

2015 - 1501(14) - 1502(14) - 1503(33) - 1504(23) - 1505(18) - 1506(12) - 1507(15) - 1508(14) - 1509(13) - 1510(11) - 1511(9) - 1512(25)

2016 - 1601(14) - 1602(17) - 1603(77) - 1604(53) - 1605(28) - 1606(17) - 1607(17) - 1608(15) - 1609(22) - 1610(22) - 1611(12) - 1612(19)

2017 - 1701(19) - 1702(23) - 1703(25) - 1704(32) - 1705(25) - 1706(25) - 1707(21) - 1708(26) - 1709(17) - 1710(26) - 1711(23) - 1712(34)

2018 - 1801(32) - 1802(21) - 1803(22) - 1804(27) - 1805(31) - 1806(16) - 1807(18) - 1808(14) - 1809(22) - 1810(17) - 1811(26) - 1812(33)

2019 - 1901(13) - 1902(13) - 1903(15)

Any replacements are listed farther down

[1952] **viXra:1903.0390 [pdf]**
*submitted on 2019-03-21 22:53:24*

**Authors:** Soerivhe Iriene, J. Oquibo Ihwaiuwaue

**Comments:** 6 Pages.

The paper "Proof of the Polignac Prime Conjecture and other Conjectures", (although listed under the title "Elementary Proof of the Goldbach Conjecture") first published in 2017 claimed to have proven Polignac's conjecture, and in doing so also the twin prime conjecture. The said paper had several problems, not least of which was a catastrophic basic error that completely invalidated the proof. Polignac's conjecture remains unproven, as does the twin primes conjecture.

**Category:** Number Theory

[1951] **viXra:1903.0387 [pdf]**
*submitted on 2019-03-22 04:31:14*

**Authors:** Juan Moreno Borrallo

**Comments:** 6 Pages. Spanish language

En este breve artículo se propone y demuestra una curiosa identidad de la función zeta, equivalente a la suma de las progresiones geométricas de los recíprocos de todos los enteros positivos que no son potencias, con numeradores cuyo valor es la función divisor del exponente de cada término de la progresión.

**Category:** Number Theory

[1950] **viXra:1903.0353 [pdf]**
*submitted on 2019-03-19 14:19:09*

**Authors:** Sally Myers Moite

**Comments:** 9 Pages.

Numbers of form 6N – 1 and 6N + 1 factor into numbers of the same form. This observation provides elimination sieves for numbers N that lead to primes and prime pairs. The sieves do not explicitly reference primes.

**Category:** Number Theory

[1949] **viXra:1903.0333 [pdf]**
*submitted on 2019-03-18 18:07:39*

**Authors:** Toshiro Takami

**Comments:** 180 Pages.

I also found a zero point which seems to deviate from 0.5.
I thought that the zero point outside 0.5 can not be found very easily in the area
which can not be shown in the figure, but this area can not be represented in the
figure but can be found one after another.
It is completely unknown whether this axis is distorted in the 0.5 axis or just by
coincidence.
The number of zero points in the area that can not be shown in the figure is now 43.
No matter how you looked it was not found in other areas.
It seemed that there is no other way to interpret this axis as 0.5 axis is distorted in
this area.
Somewhere on the net there is a memory that reads the mathematician's view that
"there are countless zero points in the vicinity of 0.5 on high area".
We are reporting that the zero point search of the high-value area of the imaginary
part which was giving up as it is no longer possible with the supercomputer is no
longer possible, is reported.
43 zero-point searches in the high-value area of the imaginary part are thus
successful.
This means that the zero point search in the high-value area of the imaginary part
has succeeded in the 43.
We will also write 43 zero point searches of the successful high-value area of the
imaginary part.
There are many counterexamples far beyond 0.5, which is far beyond the limit, but
the computer can not calculate it.
Moreover, I believe that it can only be confirmed on supercomputer whether this is
really counterexample. In addition, it is necessary to make corrections in the
supercomputer.

**Category:** Number Theory

[1948] **viXra:1903.0296 [pdf]**
*submitted on 2019-03-15 19:04:45*

**Authors:** Masashi Furuta

**Comments:** 21 Pages.

We define the "Div sequence" that sets up the number of times divided by 2 in the Collatz operation.
Using this and the "infinite descent", we prove the Collatz conjecture.

**Category:** Number Theory

[1947] **viXra:1903.0295 [pdf]**
*submitted on 2019-03-15 22:14:20*

**Authors:** Aaron Chau

**Comments:** 2 Pages.

也因为多与少，即填得满与填不满的视觉凭证是零点空格，所以，零点空格证明黎猜不成立。

**Category:** Number Theory

[1946] **viXra:1903.0209 [pdf]**
*submitted on 2019-03-11 18:46:22*

**Authors:** Bambore Dawit Geinamo

**Comments:** 2 Pages. For more improvement comments and corrections are expected

This paper magically shows very interesting and simple proof of Fermata Last Theorem. The proof describes sufficient conditions of that the equation holds and contradictions on them to satisfy the theorem. If Fermat had proof most probably his proof may be similar with this one.

**Category:** Number Theory

[1945] **viXra:1903.0200 [pdf]**
*submitted on 2019-03-12 06:40:54*

**Authors:** Maik Becker-Sievert

**Comments:** 1 Page.

Which Cube is sum of six cubes?

**Category:** Number Theory

[1944] **viXra:1903.0167 [pdf]**
*submitted on 2019-03-09 10:51:21*

**Authors:** Zeolla Gabriel Martín

**Comments:** 11 Pages. Idioma Español

Este documento desarrolla y demuestra el descubrimiento de un nuevo algoritmo de multiplicación que funciona absolutamente con todos los números.

**Category:** Number Theory

[1943] **viXra:1903.0157 [pdf]**
*submitted on 2019-03-10 00:49:01*

**Authors:** Toshiro Takami

**Comments:** 20 Pages.

I also found a zero point which seems to deviate from 0.5.
I thought that the zero point outside 0.5 can not be found very easily in the area which can not be shown in the figure, but this area can not be represented in the figure but can be found one after another.
It is completely unknown whether this axis is distorted in the 0.5 axis or just by coincidence.
The number of zero points in the area that can not be shown in the figure is now 43.
No matter how you looked it was not found in other areas.
It seemed that there is no other way to interpret this axis as 0.5 axis is distorted in this area.
Somewhere on the net there is a memory that reads the mathematician's view that "there are countless zero points in the vicinity of 0.5 on high area".
We are reporting that the zero point search of the high-value area of the imaginary part which was giving up as it is no longer possible with the supercomputer is no longer possible, is reported.
43 zero-point searches in the high-value area of the imaginary part are thus successful.
This means that the zero point search in the high-value area of the imaginary part has succeeded in the 43.
We will also write 43 zero point searches of the successful high-value area of the imaginary part.
There are many counterexamples far beyond 0.5, which is far beyond the limit, but the computer can not calculate it.
Moreover, I believe that it can only be confirmed on supercomputer whether this is really counterexample. In addition, it is necessary to make corrections in the supercomputer.

**Category:** Number Theory

[1942] **viXra:1903.0144 [pdf]**
*submitted on 2019-03-08 12:57:37*

**Authors:** Emmanuil Manousos

**Comments:** 10 Pages.

In this article, we define a pair of sequences (α, β). By using the properties of the pair (α, β), we establish a method for determining large prime numbers.

**Category:** Number Theory

[1941] **viXra:1903.0104 [pdf]**
*submitted on 2019-03-07 01:59:40*

**Authors:** Kouji Takaki

**Comments:** 22 Pages.

We have obtained the conclusion that there are no odd perfect numbers when n≧5.

**Category:** Number Theory

[1940] **viXra:1903.0059 [pdf]**
*submitted on 2019-03-05 05:22:37*

**Authors:** Espen Gaarder Haug

**Comments:** 4 Pages.

In this paper, we point out an interesting asymmetry in the rules of fundamental mathematics between positive and negative numbers. Further, we show that there exists an alternative numerical system that is basically identical to today’s system, but where positive numbers dominate over negative numbers. This is like a mirror symmetry of the existing number system. The asymmetry in both of these systems leads to imaginary and complex numbers.
We suggest an alternative number system with perfectly symmetric rules – that is, where there is no dominance of negative numbers over positive numbers, or vice versa, and where imaginary and complex numbers are no longer needed. This number system seems to be superior to other number systems, as it brings simplicity and logic back to areas that have been dominated by complex rules for much of the history of mathematics. We also briefly discuss how the Riemann hypothesis may be linked to the asymmetry in the current number system.

**Category:** Number Theory

[1939] **viXra:1903.0031 [pdf]**
*submitted on 2019-03-02 16:28:58*

**Authors:** Ahmad Telfah

**Comments:** 10 pages

this paper carrying a method to introduce the distribution of the densities of the prime numbers and the composite numbers along in natural numbers, the method basically depends on the direct deduction of the composite numbers in a specified intervals that also with using some corrections and modifications to reach maximum and minimum values of the composites and primes densities, this allowed us to detect some special conjectures related to the primes density.

**Category:** Number Theory

[1938] **viXra:1903.0030 [pdf]**
*submitted on 2019-03-02 16:40:03*

**Authors:** Ahmad Telfah

**Comments:** 5 pages

this paper carrying a method to calculate an approximation to the number of the prime numbers in the natural numbers interval I={1,2,3,4,……,P_n,P_n+1,P_n+2,……,P_n^2 } by using the primes (2,3,5,…,P_n ) to specify the primes density in the sub intervals I(P_n ) as
I(P_n )={P_n^2 〖,P〗_n^2+1,P_n^2+2,P_n^2+3,……,P_(n+1)^2-1} has primes density of (d(P_n ))= ( ∏_(i=1)^(i=n)▒〖( 1- 1/P_i 〗 ).

**Category:** Number Theory

[1937] **viXra:1902.0406 [pdf]**
*submitted on 2019-02-25 03:49:20*

**Authors:** Dariusz Gołofit

**Comments:** 8 Pages.

If we elimate an ordered subset from the ordered set, we will receive a subset od orderly character.

**Category:** Number Theory

[1936] **viXra:1902.0405 [pdf]**
*submitted on 2019-02-25 03:54:25*

**Authors:** Dariusz Gołofit

**Comments:** 12 Pages.

If we elimate an ordered subset from the ordered set, we will receive a subset od orderly character.

**Category:** Number Theory

[1935] **viXra:1902.0395 [pdf]**
*submitted on 2019-02-23 20:22:29*

**Authors:** Nicholas R. Wright

**Comments:** 7 Pages.

This proof identifies the three solutions to the three ABC-conjecture formulations. Given that the ABC-conjecture’s relevance to a slew of unsolved problems, other equations will be proven by inspection. Aside from the ABC conjecture, this proof will solve for a hypothetical Moore graph of diameter 2, girth 5, degree 57 and order 3250 (degree-diameter problem); the Collatz conjecture; and the Beal conjecture. Discussion and conclusion will review a unifying solution by spectral graph theory.

**Category:** Number Theory

[1934] **viXra:1902.0390 [pdf]**
*submitted on 2019-02-24 03:18:22*

**Authors:** ZhangAik, Leet_Noob

**Comments:** 4 Pages.

The elementary proof to the twin conjecture.

**Category:** Number Theory

[1933] **viXra:1902.0235 [pdf]**
*submitted on 2019-02-13 05:23:19*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 6 Pages. Submitted to the The Ramanujan Journal. Comments welcome.

In this paper, we consider the abc conjecture in the case c=a+1. Firstly, we give the proof of the first conjecture that c

**Category:** Number Theory

[1932] **viXra:1902.0200 [pdf]**
*submitted on 2019-02-11 06:24:07*

**Authors:** Faisal Amin Yassein Abdelmohssin

**Comments:** 2 Pages.

I claim that the sum of following distinct proper fractions [(1/2),(1/3),(1/6)] is the only triple of distinct proper fraction that sum to 1 {i.e. [(1/2)+(1/3)+(1/6)]=1}.

**Category:** Number Theory

[1931] **viXra:1902.0147 [pdf]**
*submitted on 2019-02-08 09:11:21*

**Authors:** Kenneth A. Watanabe

**Comments:** 13 Pages.

The Near-Square Prime conjecture, states that there are an infinite number of prime numbers of the form x^2 + 1. In this paper, a function was derived that determines the number of prime numbers of the form x^2 + 1 that are less than n^2 + 1 for large values of n. Then by mathematical induction, it is proven that as the value of n goes to infinity, the function goes to infinity, thus proving the Near-Square Prime conjecture.

**Category:** Number Theory

[1930] **viXra:1902.0106 [pdf]**
*submitted on 2019-02-06 07:50:18*

**Authors:** Algirdas Anatano Maknickas

**Comments:** 2 Pages.

This remark gives analytical solution of Last Fermat's Theorem

**Category:** Number Theory

[1929] **viXra:1902.0040 [pdf]**
*submitted on 2019-02-02 16:34:38*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 9 Pages. A Complete proof of the abc conjecture using elementary calculus with numerical examples. Submitted to the Ramanujan Journal. Your comments are welcome.

In this paper, we consider the abc conjecture. Firstly, we give a proof of a first conjecture that c

**Category:** Number Theory

[1928] **viXra:1902.0036 [pdf]**
*submitted on 2019-02-03 00:15:01*

**Authors:** Simon Plouffe

**Comments:** 10 Pages.

Une famille de formules permettant d'obtenir une suite de longueur arbitraire de nombres premiers. Ces formules sont nettement plus efficaces que celles de Mills ou Wright. Le procédé permet de produire par exemple une suite dont la croissance est double exponentielle mais l'exposant = 101/100.

**Category:** Number Theory

[1927] **viXra:1902.0005 [pdf]**
*submitted on 2019-02-01 10:53:12*

**Authors:** James Edwin Rock

**Comments:** 2 Pages of exposition. 5 Tables. Copyright 2019 James Edwin Rock Creative Commons Attribution-ShareAlike 4.0 International License

Let P_n be the n_th prime. For twin primes P_n – P_(n-1) = 2. Let X be the number of (6j –1, 6j+1) pairs in the interval [P_n, P_n^2]. The number of twin primes (TPAn) in [P_n, P_n^2] can be approximated by the formula
(a_3 /5)(a_4 /7)(a_5 /11)…(a_n /P_n)(X) for 3 ≤ m ≤ n, a_m = P_m –2 .
We establish a lower bound for TPAn (3/5)(5/7)(7/9)…(P_n–2)/P_n)(X) = 3X/P_n < TPAn.
We exhibit a formula showing as P_n increases, the number of twin primes in the interval [P_n, P_n^2] also increases. Let P_n – P_(n-1) = c. For all n (TPAn-1)(1+(2c –2)/2P_(n-1)+(c^2–2c)/2P_(n-1)^2) < TPAn

**Category:** Number Theory

[1926] **viXra:1901.0436 [pdf]**
*submitted on 2019-01-29 10:25:44*

**Authors:** Kenneth A. Watanabe

**Comments:** 10 Pages.

Legendre's conjecture, states that there is a prime number between n^2 and (n + 1)^2 for every positive integer n. In this paper, an equation was derived that determines the number of prime numbers less than n for large values of n. Then by mathematical induction, it is proven that there is at least 1 prime number between n^2 and (n + 1)^2 for all positive integers n thus proving Legendre’s conjecture.

**Category:** Number Theory

[1925] **viXra:1901.0430 [pdf]**
*submitted on 2019-01-29 01:24:31*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, we consider the $ABC$ conjecture,then we give a proof that $C<rad^2(ABC)$ that it will be the key of the proof of the $ABC$ conjecture.

**Category:** Number Theory

[1924] **viXra:1901.0300 [pdf]**
*submitted on 2019-01-19 14:41:09*

**Authors:** Toshiro Takami

**Comments:** 4 Pages.

On calculation by Euler 's formula(3), at least notice that the zero point of the real part is not on x = 0.5.
Using Euler 's formula, we found that at least the real part' s zero point is not x = 0.5 but about x = 0.115444. Moreover, the imaginary point is around i14.524.
And s=0.8355 +i39.
And s=0.1645 +i39.
And s=0.884556 +i14.524.
And s=0.115444 +i14.524
Also, replacing sin with cos, the imaginary part becomes zero.
I do not know at all whether the collapse of Riemann hypothesis or not?
In addition, books are printed as cos instead of sin.
Also, I have collected ζ on the left side.
In addition, (8) is Euler's formula found overseas, which is also a singular point in this. Also, I have collected ζ on the left side.
In this case, sin is printed instead of cos, but if sin and cos are exchanged, the zero point moves only from the real part to the imaginary part.

**Category:** Number Theory

[1923] **viXra:1901.0297 [pdf]**
*submitted on 2019-01-19 22:35:49*

**Authors:** Michael Grützmann

**Comments:** 2 Pages.

every prime number can be a sum of p=3+...+3+2 or
q=3+...+3+4, the number of '3's in both equatons always being odd.
there are the for two primes p+q the combinations
'p+q', 'p+p' and 'q+q'.
we consider case 1: p+q=3k+2+3l+4
3k must be odd, as the product of two odd numbers,
3l must be odd, for the same reason.
but the sum of two odd numbers is an even number always. also, if you add more even numbers, like 2 and 4, the result will always be even also.
So this results in an even number.
cases 'p+p' and 'q+q' analogue.

**Category:** Number Theory

[1922] **viXra:1901.0227 [pdf]**
*submitted on 2019-01-16 12:07:43*

**Authors:** James Edwin Rock

**Comments:** 5 Pages. Copyright 2018 James Edwin Rock Create Commons Attribution-ShareAlike 4.0 International License

Collatz sequences are formed by applying the Collatz algorithm to any positive integer. If it is even repeatedly divide by two until it is odd, then multiply by three and add one to get an even number and vice versa. If the Collatz conjecture is true eventually you always get back to one. A connected Collatz Structure is created, which contains all positive integers exactly once. The terms of the Collatz Structure are joined together via the Collatz algorithm. Thus, every positive integer forms a Collatz sequence with unique terms terminating in the number one.

**Category:** Number Theory

[1921] **viXra:1901.0155 [pdf]**
*submitted on 2019-01-11 06:22:54*

**Authors:** Edgar Valdebenito

**Comments:** 1 Page.

This note presents three trigonometric identities.

**Category:** Number Theory

[1920] **viXra:1901.0116 [pdf]**
*submitted on 2019-01-10 02:36:15*

**Authors:** Quang Nguyen Van

**Comments:** 2 Pages.

The equation a^5 + b^5 = c^2 has no solution in integer. However, related to Fermat- Catalan conjecture, the equation a^5 + b^5 = 2c^2 has a solution in integer. In this article, we give a parametric equation of the equation a^5 + b^5 = 2c^2.

**Category:** Number Theory

[1919] **viXra:1901.0108 [pdf]**
*submitted on 2019-01-08 11:13:12*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. A Proof of ABC conjecture is submitted to the Journal of Number Theory (2019). Comments Welcome.

In this paper, we assume that the ABC conjecture is true, then we give a proof that Beal conjecture is true. We consider that Beal conjecture is false then we arrive to a contradiction. We deduce that the Beal conjecture is true.

**Category:** Number Theory

[1918] **viXra:1901.0101 [pdf]**
*submitted on 2019-01-09 00:16:39*

**Authors:** Johnny E. Magee

**Comments:** 7 Pages.

We identify equivalent restatements of the Brocard-Ramanujan diophantine equation, $(n! + 1) = m^2$; and employing the properties and implications of these equivalencies, prove that for all $n > 7$, there are no values of $n$ for which $(n! + 1)$ can be a perfect square.

**Category:** Number Theory

[1917] **viXra:1901.0046 [pdf]**
*submitted on 2019-01-04 11:35:20*

**Authors:** Nazihkhelifa

**Comments:** 4 Pages.

Relation between
The Euler Totient,
the counting prime formula
and the prime generating Functions
The theory of numbers is an area of mathematis hiih eals ith
the propertes of hole an ratonal numbers... In this paper I ill
intro uie relaton bet een Euler phi funiton an prime iountnn
an neneratnn formula, as ell as a ioniept of the possible
operatons e ian use ith them. There are four propositons hiih
are mentone in this paper an I have use the efnitons of these
arithmetial funitons an some Lemmas hiih refeit their
propertes, in or er to prove them

**Category:** Number Theory

[1916] **viXra:1901.0030 [pdf]**
*submitted on 2019-01-03 17:17:16*

**Authors:** Robert C. Hall

**Comments:** 24 Pages.

The Benford's Law Summation test consists of adding all numbers that begin with a particular first or first two digits and determining its distribution with respect to these first or first two digits numbers. Most people familiar with this test believe that the distribution is a uniform distribution for any distribution that conforms to Benford's law i.e. the distribution of the mantissas of the logarithm of the data set is uniform U[0,1). The summation test that results in a uniform distribution is true for an exponential function (geometric progression) but not true for a data set that conforms to a Log Normal distribution even when the Log Normal distribution itself closely approximates a Benford's Law distribution.

**Category:** Number Theory

[1915] **viXra:1901.0007 [pdf]**
*submitted on 2019-01-01 16:19:56*

**Authors:** Rachid Marsli

**Comments:** 11 Pages.

In this work, we show a sufficient and necessary condition for an integer of the form
(z^n-y^n)/(z-y)to be divisible by some perfect nth power p^n, where p is an odd prime. We also show how to construct such integers. A link between
the main result and Fermat’s last theorem is discussed. Other related ideas, examples and applications are provided.

**Category:** Number Theory

[1914] **viXra:1812.0497 [pdf]**
*submitted on 2018-12-31 12:45:35*

**Authors:** Pedro Hugo García Peláez

**Comments:** 5 Pages.

Series De Números Cuyos Factores Son La Lista De Los Números Primos Desde El Principio y Enumerando Todos Sin excepción

**Category:** Number Theory

[1913] **viXra:1812.0496 [pdf]**
*submitted on 2018-12-31 13:08:20*

**Authors:** Pedro Hugo García Peláez

**Comments:** 5 Pages.

Series of numbers whose factors are the list of prime numbers from the beginning and listing all without exception

**Category:** Number Theory

[1912] **viXra:1812.0495 [pdf]**
*submitted on 2018-12-31 14:29:57*

**Authors:** James Edwin Rock

**Comments:** Pages.

Let P_n be the n_th prime. For twin primes P_n – P_n-1 = 2. We show that in the interval (P_n, P_n^2) as P_n gets larger, there is an increasing number of twin primes.

**Category:** Number Theory

[1911] **viXra:1812.0494 [pdf]**
*submitted on 2018-12-31 09:51:52*

**Authors:** Simon Plouffe

**Comments:** 5 Pages.

We show here a new set of formulas for producing primes with a growth rate much smaller than the ones of Mills and Wright. Several examples of formulas are given. All results are empirical.

**Category:** Number Theory

[1910] **viXra:1812.0488 [pdf]**
*submitted on 2018-12-30 10:59:58*

**Authors:** Zeolla Gabriel Martín

**Comments:** 8 Pages.

This paper develops the analysis of Simple composite numbers by golden patterns. Examine how the Simple composite numbers are distributed in different combinations of multiples.

**Category:** Number Theory

[1909] **viXra:1812.0439 [pdf]**
*submitted on 2018-12-27 18:34:37*

**Authors:** Robert C. Hall

**Comments:** 5 Pages.

While it is fairly easy to prove that the Log Normal distribution becomes a Benford distribution as the standard deviation approaches infinity (see appendix A), it is a bit more difficult to prove that as the standard deviation approaches zero that the distribution becomes a Normal distribution with a mean of e^u where u is the mean of the natural logarithms of the data set values.

**Category:** Number Theory

[1908] **viXra:1812.0422 [pdf]**
*submitted on 2018-12-25 18:07:31*

**Authors:** A. A. Frempong

**Comments:** 9 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. The proof would be complete after showing that if A and B have a common prime factor, and C^z can be produced from the sum A^x + B^y. In the proof, one begins with A^x + B^y and change this sum to the single power, C^z as was done in the preliminaries. It was determined that if A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[1907] **viXra:1812.0418 [pdf]**
*submitted on 2018-12-24 06:01:30*

**Authors:** Nicolò Rigamonti

**Comments:** 15 Pages.

This paper shows the importance of two properties, which are at the base of the Riemann hypothesis. The key point of all the reasoning about the validity of the Riemann hypothesis is in the fact that only if the Riemann hypothesis is true, these two properties, which are satisfied by the non-trivial zeros, are both true. In fact, only if these two properties are both true , all non-trivial zeros lie on the critical line. Leave a comment, a reflection or an opinion about the paper. You shouldn’t keep your doubts and questions in yourself: if you think that it’s right or can’t attack it,share it, otherwise “ Why don’t you start a discussion?.........”.

**Category:** Number Theory

[1906] **viXra:1812.0400 [pdf]**
*submitted on 2018-12-24 05:12:29*

**Authors:** Jesús Sánchez

**Comments:** Pages.

b=0;
p=input('Input a number : ');
m=fix((p+1)/2);
for k=2:m+1;
fun=@(w) exp(p.*1j.*w).*exp(-2.*k.*1j.*w).*(1-exp(-m.*k.*1j.*w))./(1-exp(-k.*1j.*w));
a=integral(fun,-pi,pi);
b=b+a;
end;
b=b/(2*pi);
disp(b);
Above simple MATLAB/Octave program, can detect if a number is prime or not. If the result is zero (considering zero being less than 1e-5), the number introduced is a prime.
If the result is an integer, this result will tell us how many permutations of two divisors, the input number has. As you can check, no recurrent division by odd or prime numbers is done. Just this strange integral:
1/2π ∫_(-π)^π▒〖e^pjω (∑_(k=2)^(k=m+1)▒〖e^(-2kjω) ((1-e^(-mkjω))/(1-e^(-kjω) )) 〗)dω 〗
Being p the number that we want to check if it is a prime or not. And being m whatever integer number higher than (p+1)/2(the lowest, the most efficient the operation). As k and ω are independent variables, the sum can be taken outside the integral (as it is in the above program).
To get to this point, we will do the following. First, we will create a domain with all the composite numbers. This is easy, as you can just multiply one by one all the integers (greater or equal than 2) in that domain. So, you will get all the composite numbers (not getting any prime) in that domain.
Then, we will use the Fourier transform to change from this original domain (called discrete time domain in this regards) to the frequency domain. There, we can “ask”, using Parseval’s theorem, if a certain number is there or not. The use of Parseval’s theorem leads to the above integral. If the number p that we want to check is not in the domain, the result of the integral is zero and the number is a prime. If instead, the result is an integer, this integer will tell us how many permutations of two divisors the number p has. And, in consequence information how many factors, the number p has.
So, for any number p lower than 2m-1, you can check if it is prime or not, just making the numerical definite integration (but even this integral, if no further developments are done, the numerical integration is inefficient computing-wise compared with brute-force checking for example). To be added, is the question regarding the level of accuracy needed (number of decimals and number of steps in the numerical integration) to have a reliable result for large numbers. This will be commented on the paper, but a separate study will be needed to have detailed conclusions.
Of course, the best would be that in the future, an analytical result (or at least an approximation) for the summation or for the integration is achieved.

**Category:** Number Theory

[1905] **viXra:1812.0362 [pdf]**
*submitted on 2018-12-20 21:07:15*

**Authors:** Ho Soo Shin

**Comments:** 2 Pages. comments welcome!

As is well known, the Collatz sequence, which is also named as the hailstone sequence, follows the rule of Collatz conjecture. The rule requires us to divide any positive even integer by 2. We must multiply every positive odd number by 3 and then add 1 according to the rule. By investigating residues modulo 3, I will prove any integer multiple of 3 cannot appear more than one time in a Collatz sequence, which implies any multiple of 3 cannot be included in a possible cycle of the Collatz sequence.

**Category:** Number Theory

[1904] **viXra:1812.0340 [pdf]**
*submitted on 2018-12-19 17:50:09*

**Authors:** Wu ShengPing

**Comments:** 3 Pages.

The main idea of this article is simply calculating integer
functions in module. The algebraic in the integer modules is studied in
completely new style. By a careful construction the result that
two finite numbers is with unequal logarithms in a corresponding module is proven, which result is applied to solving
a kind of high degree diophantine equation.

**Category:** Number Theory

[1903] **viXra:1812.0322 [pdf]**
*submitted on 2018-12-18 09:16:24*

**Authors:** James Edwin Rock

**Comments:** 2 Pages. Copyright 2018 James Edwin Rock Create Commons Attribution-ShareAlike 4.0 International License2

Let Pn be the n_th prime. For twin primes Pn – Pn-1 = 2. We give a heuristic argument that in the interval (Pn, Pn2) as Pn gets larger, there is an increasing number of twin primes.

**Category:** Number Theory

[1902] **viXra:1812.0312 [pdf]**
*submitted on 2018-12-19 05:32:58*

**Authors:** Julian Beauchamp

**Comments:** 3 Pages.

In 2002 Preda Mihailescu used the theory of cyclotomic fields and Galois modules to prove Catalan's Conjecture. In this short paper, we give a very simple proof. We first prove that no solutions exist for a^x-b^y=1 for a,b>1 and x,y>2. Then we prove that when x=2 the only solution for y is y=3.

**Category:** Number Theory

[1901] **viXra:1812.0300 [pdf]**
*submitted on 2018-12-17 23:40:30*

**Authors:** Aaron chau

**Comments:** 1 Page.

无论是在历史的任何时间，如不谈黎曼猜想则已；如谈，我们大家当然先要彻底弄明白：
黎曼临界线上从小到大的一组零点及其（零点空格），它们在数论上究竟是什么意思呢？

**Category:** Number Theory

[1900] **viXra:1812.0299 [pdf]**
*submitted on 2018-12-17 23:47:57*

**Authors:** Aaron chau

**Comments:** 2 Pages.

本文强调：质数与孪生质数分别都是无限的依据是算术中的（加减乘除）。
比如在古希腊，欧几里德证明质数无限，他所应用的是（乘除法）来表述反证法。
又比如，现时在伦敦来证明孪生质数无限，即无限存在二个质数的距离＝2；本文应用的是（加减法）来表述：就在（单数空格）里，单数与奇合数的个数分别在每一数段里的多与少。

**Category:** Number Theory

[1899] **viXra:1812.0296 [pdf]**
*submitted on 2018-12-18 01:33:20*

**Authors:** Joseph Olloh

**Comments:** 13 Pages. The content of the paper provides a fundamentally new way of looking at numbers.

We differentiate even and odd numbers into various groups and subgroups. We provide the properties of the forms of numbers which fall into each groups and subgroups. We expound on the relationship of a special group of even numbers and the collatz conjecture, we also derive an accurate formula to calculate the steps involved when an even number of the group is the initial value of the collatz operation.
For each group and subgroup of odd and even numbers, we discuss the observed pattern of their sequences and also derive accurate formulas for each sequence. Throughout, b, d, k, N, n, x, m, and z all denote positive integers, with d, and N denoting odd numbers, x and z denoting even numbers, and b denoting special even-even numbers
The order of priority of the properties of each group is key in the differentiation of the numbers into their various groups and subgroups.

**Category:** Number Theory

[1898] **viXra:1812.0287 [pdf]**
*submitted on 2018-12-16 09:20:54*

**Authors:** Zhang Tianshu

**Comments:** 14 Pages.

If regard positive integers which have a common prime factor as a kind, then the positive half line of the number axis consists of infinite many recurring segments which have same permutations of c kinds of integer’s points, where c≥1. In this article, the author shall prove Grimm’s conjecture by the method that changes orderly symbols of each kind of composite number’s points at the original number axis, so as to form consecutive composite number’s points within limits that proven Bertrand's postulate restricts.

**Category:** Number Theory

[1897] **viXra:1812.0242 [pdf]**
*submitted on 2018-12-13 08:45:16*

**Authors:** Jean Pierre Morvan

**Comments:** 6 Pages.

L'hypothèse de RIEMANN est fausse parce que la fonction zéta n'est pas nulle avec s = 1/2 +ib

**Category:** Number Theory

[1896] **viXra:1812.0208 [pdf]**
*submitted on 2018-12-11 16:27:17*

**Authors:** Kenneth A. Watanabe

**Comments:** 9 Pages.

A twin prime is defined as a pair of prime numbers (p1,p2) such that p1 + 2 = p2. The Twin Prime Conjecture states that there are an infinite number of twin primes. The first mention of the Twin Prime Conjecture was in 1849, when de Polignac made the more general conjecture that for every natural number k, there are infinitely many primes p such that p + 2k is also prime. The case where k = 1 is the Twin Prime Conjecture. In this document, I derive a function that corresponds to the number of twin primes less than n for large values of n. This equation is identical to that used to prove the Goldbach Conjecture. Then by proof by induction, it is shown that as n increases indefinitely, the function also increases indefinitely thus proving the Twin Prime Conjecture. Using the same methodology, de Polignac’s conjecture is also shown to be true.

**Category:** Number Theory

[1895] **viXra:1812.0182 [pdf]**
*submitted on 2018-12-10 10:58:25*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 4 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, we use the Fermat's Last Theorem (FLT) to give a proof of the ABC conjecture. We suppose that FLT is false ====> we arrive that the ABC conjecture is false. Then taking the negation of the last statement, we obtain: ABC conjecture is true ====> FLT is true. But, as FLT is true, then we deduce that the ABC conjecture is true.

**Category:** Number Theory

[1894] **viXra:1812.0154 [pdf]**
*submitted on 2018-12-08 16:12:41*

**Authors:** M. Sghiar

**Comments:** 7 Pages. french version © Copyright 2018 by M. Sghiar. All rights reserved. Respond to the author by email at: msghiar21@gmail.com

I show here that if $ x \in \mathbb{N}^*$ then $1 \in \mathcal{O}_S (x)= \{ S^n(x), n \in \mathbb{N}^* \} $ where $ \mathcal{O}_S (x)$ is the orbit of the function S defined on $\mathbb{R}^+$ by $S(x)= \frac{x}{2} + (x+\frac{1}{2}) sin^2(x\frac{\pi}{2})$, and I deduce the proof of the Syracuse conjecture.

**Category:** Number Theory

[1893] **viXra:1812.0150 [pdf]**
*submitted on 2018-12-08 21:32:29*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT simply for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not by way of contradiction) of FLT is based on our algebraic identity, denoted, {for convenience}, as $(r)^n+(s)^n=(t)^n$ with $r,s,t>0$ as functions of variables. We infer that $\{(r,s,t)|r,s,t\in\mathbb{N},(r)^n+(s)^n+(t)^n\}=\{(x,y,z)|x,y,z\in\mathbb{N},(x)^n+y^n=z^n\}$ for $n\in\mathbb{N}, n>2$. In addition, we show, for integral values of $n>2$, that $\{(r,s,t)|r,s,t\in\mathbb{N},(r)^n+(s)^n=t^n\}=\varnothing$. Hence, for $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[1892] **viXra:1812.0133 [pdf]**
*submitted on 2018-12-07 13:32:39*

**Authors:** James Edwin Rock

**Comments:** 5 Pages. Copyright 2018 James Edwin Rock Create Commons Attribution-ShareAlike 4.0 International License

Collatz sequences are formed by applying the Collatz algorithm to a positive integer. If it is even repeatedly divide by two until it is odd, then multiply by three and add one to get an even number and vice versa. Eventually you get back to one. The Collatz Structure is created, which contains all positive integers exactly once. The terms of the Collatz Structure are joined together via the Collatz algorithm. Thus, every positive integer forms a Collatz sequence with unique terms terminating in the number one.

**Category:** Number Theory

[1891] **viXra:1812.0131 [pdf]**
*submitted on 2018-12-07 18:37:16*

**Authors:** Toshiro Takami

**Comments:** 5 Pages.

Calculationally, \sqrt(24a+1)=t is excellent as a prime number production formula.
But, \sqrt(24a+1)=t is a sequence that increases only by 2, 4, 2, 4 alternately.
The prime number may only be 2, 4, 2, 4 and the increasing sequence of complexity is increasing.
Occasionally, it may be prime that 2 and 4 combine to become 6, 8, 10, and so on.
\sqrt(12a+1)=t, \sqrt(8a+1)=t, \sqrt(6a+1)=t (a is positive integer)
as an examples.
\sqrt(24a+1)=t are exactly the same as these.
\sqrt(20a+1)=t, \sqrt(18a+1)=t, \sqrt(16a+1)=t, \sqrt(10a+1)=t, etc, tried, but thease did not become a prime number production formula at all.

**Category:** Number Theory

[1890] **viXra:1812.0130 [pdf]**
*submitted on 2018-12-07 21:04:48*

**Authors:** Zhang Tianshu

**Comments:** 12 Pages.

Since there are infinitely many consecutive satisfactory values of ε to enable A+B=C satisfying C>(rad(A, B, C))1+ε, thus the author uses a representative equality, namely 1+2N(2N-2)=(2N-1)2 satisfying (2N-1)2>[rad(1, 2N(2N-2), (2N-1)2)]1+ε, and that first let ε equal a value near the greater end of the infinitely many consecutive satisfactory values to prove the ABC conjecture; again let ε equal a value near the smaller end to negate the ABC conjecture. This shows that the ABC conjecture is in the ambiguity in which case of ε>0.

**Category:** Number Theory

[1889] **viXra:1812.0107 [pdf]**
*submitted on 2018-12-06 14:23:59*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 8 Pages. Comments welcome.

In this paper, we give the elliptic curve (E) given by the equation:
y^2=x^3+px+q
with $p,q \in Z$ not null simultaneous. We study a part of the conditions verified by $(p,q)$ so that it exists (x,y) \in Z^2 the coordinates of a point of the elliptic curve (E) given by the equation above.

**Category:** Number Theory

[1888] **viXra:1812.0074 [pdf]**
*submitted on 2018-12-04 16:03:11*

**Authors:** Stephen Marshall

**Comments:** 7 Pages.

The Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1/2
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000).
It was proposed by Bernhard Riemann (1859), after whom it is named. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields.
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000). The Riemann zeta function is defined for complex s with real part greater than 1 by the absolutely convergent infinite series:
ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ...
The Riemann hypothesis asserts that all interesting solutions of the equation:
ζ(s) = 0
lie on a certain vertical straight line.
In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers:
Hn = 1 + 1/2+1/3+1/4+⋯+ 1/n = ∑_(n=1)^n▒1/n
Harmonic numbers have been studied since early times and are important in various branches of number theory. They are sometimes loosely termed harmonic series, are closely related to the Riemann zeta function.
The harmonic numbers roughly approximate the natural logarithm function and thus the associated harmonic series grows without limit, albeit slowly. In 1737, Leonhard Euler used the divergence of the harmonic series to provide a new proof of the infinity of prime numbers. His work was extended into the complex plane by Bernhard Riemann in 1859, leading directly to the celebrated Riemann hypothesis about the distribution of prime numbers

**Category:** Number Theory

[1887] **viXra:1812.0071 [pdf]**
*submitted on 2018-12-04 22:22:02*

**Authors:** Colin James III

**Comments:** 2 Pages. © Copyright 2018 by Colin James III All rights reserved. Respond to the author by email at: info@ersatz-systems dot com.

Using the standard wiki definition of the Collatz conjecture, we map a positive number to imply that a divisor of two implies either an even numbered result (unchanged) or an odd numbered result (changed to the number multiplied by three plus one) to imply the final result of one. This is the shortest known confirmation of the conjecture, and in mathematical logic.

**Category:** Number Theory

[1886] **viXra:1812.0020 [pdf]**
*submitted on 2018-12-01 15:11:29*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 6 Pages. Comments welcome.

In this paper, we assume that Beal conjecture is true, we give a complete proof of the ABC conjecture. We consider that Beal conjecture is false $\Longrightarrow$ we arrive that the ABC conjecture is false. Then taking the negation of the last statement, we obtain: ABC conjecture is true $\Longrightarrow$ Beal conjecture is true. But, if the Beal conjecture is true, then we deduce that the ABC conjecture is true

**Category:** Number Theory

[1885] **viXra:1812.0019 [pdf]**
*submitted on 2018-12-01 16:17:59*

**Authors:** Kamal Barghout

**Comments:** 10 Pages. This work is copyrighted material. No part of this work is to be copied without prior permission from the author

: A probabilistic proof of the Collatz conjecture is described via identifying a sequential permutation of even natural numbers by divisions by 2 that follows a recurrent pattern of the form x,1,x,1…, where x represents divisions by 2 more than once. The sequence presents a probability of 50:50 of divisions by 2 once as opposed to divisions by 2 more than once over the even natural numbers. The sequence also gives the same 50:50 probability of consecutive Collatz even elements when counted for division by 2 more than once as opposed to division by 2 once ratio of 2.8. Considering Collatz function producing random numbers, over sufficient iterations this probability distribution produces a descending order of its elements that leads to convergence of the Collatz function to the cycle 1-2-4-1.

**Category:** Number Theory

[1884] **viXra:1812.0018 [pdf]**
*submitted on 2018-12-01 17:00:45*

**Authors:** Khalid Ibrahim

**Comments:** 91 Pages.

In this paper, not only did we disprove the Riemann Hypothesis (RH) but also we showed that zeros of the Riemann zeta function $\zeta (s)$ can be found arbitrary close to the line $\Re (s) =1$. Our method to reach this conclusion is based on analyzing the fine behavior of the partial sum of the Dirichlet series with the Mobius function $M (s) = \sum_n \mu (n) /n^s$ defined over $p_r$ rough numbers (i.e. numbers that have only prime factors greater than or equal to $p_r$). Two methods to analyze the partial sum fine behavior are presented and compared. The first one is based on establishing a connection between the Dirichlet series with the Mobius function $M (s) $ and a functional representation of the zeta function $\zeta (s)$ in terms of its partial Euler product. Complex analysis methods (specifically, Fourier and Laplace transforms) were then used to analyze the fine behavior of partial sum of the Dirichlet series. The second method to estimate the fine behavior of partial sum was based on integration methods to add the different co-prime partial sum terms with prime numbers greater than or equal to $p_r$. Comparing the results of these two methods leads to a contradiction when we assume that $\zeta (s)$ has no zeros for $\Re (s) > c$ and $c <1$.

**Category:** Number Theory

[1883] **viXra:1812.0002 [pdf]**
*submitted on 2018-12-01 05:01:55*

**Authors:** Zhang Tianshu

**Comments:** 21 Pages.

In this article, the author first classify A, B and C according to their respective odevity, and thereby get rid of two kinds which belong not to AX+BY=CZ. Then, affirm the existence of AX+BY=CZ in which case A, B and C have at least a common prime factor by several concrete equalities. After that, prove AX+BY≠CZ in which case A, B and C have not any common prime factor by the mathematical induction with the aid of the distinct odd-even relation on the premise whereby even number 2W-1HZ as symmetric center of positive odd numbers concerned after divide the inequality in four. Finally, reach a conclusion that the Beal’s conjecture holds water via the comparison between AX+BY=CZ and AX+BY≠CZ under the given requirements.

**Category:** Number Theory

[988] **viXra:1903.0387 [pdf]**
*replaced on 2019-03-22 08:40:24*

**Authors:** Juan Moreno Borrallo

**Comments:** 7 Pages. Spanish Language

At this brief paper, it is proposed and demonstrated a curious identity of Zeta Function, equivalent to the sum of the geometric progression of reciprocals of all the positive integers which are not perfect powers, having as numerators the number of divisors of the exponent of each term of the progression.

**Category:** Number Theory

[987] **viXra:1903.0200 [pdf]**
*replaced on 2019-03-13 09:22:44*

**Authors:** Maik Becker-Sievert

**Comments:** 1 Page.

Two cubes are a sum of nine cubes

**Category:** Number Theory

[986] **viXra:1903.0200 [pdf]**
*replaced on 2019-03-13 04:09:18*

**Authors:** Maik Becker-Sievert

**Comments:** 1 Page.

Which Cube is sum of six cubes?

**Category:** Number Theory

[985] **viXra:1903.0157 [pdf]**
*replaced on 2019-03-11 16:26:03*

**Authors:** Toshiro Takami

**Comments:** 10 Pages.

I also found a zero point which seems to deviate from 0.5.
I thought that the zero point outside 0.5 can not be found very easily in the area which can not be shown in the figure, but this area can not be represented in the figure but can be found one after another.
It is completely unknown whether this axis is distorted in the 0.5 axis or just by coincidence.
The number of zero points in the area that can not be shown in the figure is now 43.
No matter how you looked it was not found in other areas.
It seemed that there is no other way to interpret this axis as 0.5 axis is distorted in this area.
Somewhere on the net there is a memory that reads the mathematician's view that "there are countless zero points in the vicinity of 0.5 on high area".
We are reporting that the zero point search of the high-value area of the imaginary part which was giving up as it is no longer possible with the supercomputer is no longer possible, is reported.
43 zero-point searches in the high-value area of the imaginary part are thus successful.
This means that the zero point search in the high-value area of the imaginary part has succeeded in the 43.
We will also write 43 zero point searches of the successful high-value area of the imaginary part.
There are many counterexamples far beyond 0.5, which is far beyond the limit, but the computer can not calculate it.
Moreover, I believe that it can only be confirmed on supercomputer whether this is really counterexample. In addition, it is necessary to make corrections in the supercomputer.

**Category:** Number Theory

[984] **viXra:1903.0104 [pdf]**
*replaced on 2019-03-19 10:13:33*

**Authors:** Kouji Takaki

**Comments:** 23 Pages.

We have obtained the conclusion that there are no odd perfect numbers.

**Category:** Number Theory

[983] **viXra:1903.0104 [pdf]**
*replaced on 2019-03-14 07:06:07*

**Authors:** Kouji Takaki

**Comments:** 22 Pages.

We have obtained the conclusion that there are no odd perfect numbers when n≧5.

**Category:** Number Theory

[982] **viXra:1903.0104 [pdf]**
*replaced on 2019-03-12 16:02:18*

**Authors:** Kouji Takaki

**Comments:** 23 Pages.

We have obtained the conclusion that there are no odd perfect numbers.

**Category:** Number Theory

[981] **viXra:1902.0468 [pdf]**
*replaced on 2019-03-13 17:36:31*

**Authors:** Toshiro Takami

**Comments:** 9 Pages. serios mistake consist, re-load-up

Let a be real number of 0< a <1.
(1)= cos[x*ln1]/1^a – cos[x*ln2]/2^a + cos[x*ln3]/3^a – cos[x*ln4]/4^a + cos[x*ln5]/5^a............
(2)= sin[x*ln1]/1^a – sin[x*ln2]/2^a + sin[x*ln3]/3^a – sin[x*ln4]/4^a + sin[x*ln5]/5^a.............
Then, at this time,
The imaginary solution of the equation (1)^2+(2)^2=0 exists only when a = 0.5.
x is an infinite non-trivial zero.
At the same time satisfying (1) and (2) is x, that is, an infinitely present non-tribial zero.
(1) is
\sum_{n=1}^{infty} [cos(x*ln(2n-1))/(2n-1)^0.5)- cos(x*ln(2n)/(2n)^0.5)] =0
(2) is
\sum_{n=1}^{infty} [sin(x*ln(2n-1))/(2n-1)^0.5) - sin(x*ln(2n)/(2n)^0.5)] =0

**Category:** Number Theory

[980] **viXra:1902.0468 [pdf]**
*replaced on 2019-03-12 19:14:08*

**Authors:** Toshiro Takami

**Comments:** 8 Pages.

Let a be real number of 0< a <1.
(1)= cos[x*ln1]/1^a – cos[x*ln2]/2^a + cos[x*ln3]/3^a – cos[x*ln4]/4^a + cos[x*ln5]/5^a............
(2)= sin[x*ln1]/1^a – sin[x*ln2]/2^a + sin[x*ln3]/3^a – sin[x*ln4]/4^a + sin[x*ln5]/5^a.............
Then, at this time,
The imaginary solution of the equation (1)^2+(2)^2=0 exists only when a = 0.5.
x is an infinite non-trivial zero.
At the same time satisfying (1) and (2) is x, that is, an infinitely present non-tribial zero.
(1) is
\sum_{n=1}^{infty} [cos(x*ln(2n-1))/(2n-1)^0.5)- cos(x*ln(2n)/(2n)^0.5)] =0
(2) is
\sum_{n=1}^{infty} [sin(x*ln(2n-1))/(2n-1)^0.5) - sin(x*ln(2n)/(2n)^0.5)] =0

**Category:** Number Theory

[979] **viXra:1902.0390 [pdf]**
*replaced on 2019-03-16 21:21:15*

**Authors:** ZhangAik, Leet_Noob

**Comments:** 2 Pages.

The elementary proof to the twin conjecture.

**Category:** Number Theory

[978] **viXra:1902.0390 [pdf]**
*replaced on 2019-03-15 08:50:15*

**Authors:** ZhangAik, Leet_Noob

**Comments:** 2 Pages.

The elementary proof to the twin conjecture.

**Category:** Number Theory

[977] **viXra:1902.0390 [pdf]**
*replaced on 2019-03-09 15:54:30*

**Authors:** ZhangAik, Leet_Noob

**Comments:** 3 Pages.

The elementary proof to the twin conjecture.

**Category:** Number Theory

[976] **viXra:1902.0390 [pdf]**
*replaced on 2019-03-08 17:57:51*

**Authors:** ZhangAik, Leet_Noob

**Comments:** 3 Pages.

The elementary proof to the twin conjecture.

**Category:** Number Theory

[975] **viXra:1902.0117 [pdf]**
*replaced on 2019-02-12 01:24:03*

**Authors:** Toshiro Takami

**Comments:** 24 Pages.

2^s/(2^-1)*3^s/(3^-1)*5^s/(5^s-1)*7^s/(7^s-1)………
Whether the above equation converges to 0 was verified.
Convergence is extremely slow, and divergence tendency was rather rather abundant when the prime number was 1000 or more.
It was thought that the above equation could possibly be an expression that can be composed only of real numbers.

**Category:** Number Theory

[974] **viXra:1902.0106 [pdf]**
*replaced on 2019-02-10 10:53:42*

**Authors:** Algirdas Antano Maknickas

**Comments:** 2 Pages.

This remark gives analytical solution of Last Fermat's Theorem

**Category:** Number Theory

[973] **viXra:1902.0106 [pdf]**
*replaced on 2019-02-09 02:11:25*

**Authors:** Algirdas Antano Maknickas

**Comments:** 2 Pages. In previous version was mistake in middle name.

This remark gives analytical solution of Last Fermat's Theorem

**Category:** Number Theory

[972] **viXra:1902.0005 [pdf]**
*replaced on 2019-03-04 08:20:36*

**Authors:** James Edwin Rock

**Comments:** 7 Pages.

Let P_n be the n_th prime. For twin primes P_n – P_(n-1) = 2. Let X be the number of (6j –1, 6j+1) pairs in the interval [P_n, P_n^2]. The number of twin primes (TPAn) in [P_n, P_n^2] can be approximated by the formula
(a_3 /5)(a_4 /7)(a_5 /11)…(a_n /P_n)(X) for 3 ≤ m ≤ n, a_m = P_m –2 .
We establish a lower bound for TPAn (3/5)(5/7)(7/9)…(P_n–2)/P_n)(X) = 3X/P_n < TPAn.
We exhibit a formula showing as P_n increases, the number of twin primes in the interval [P_n, P_n^2] also increases. Let P_n – P_(n-1) = c. For all n (TPAn-1)(1+(2c –2)/2P_(n-1)+(c^2–2c)/2P_(n-1)^2) < TPAn

**Category:** Number Theory

[971] **viXra:1902.0005 [pdf]**
*replaced on 2019-02-19 08:32:39*

**Authors:** James Edwin Rock

**Comments:** 2 pages of exposition and 7 pages with supporting tables. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Let P_n be the n_th prime. For twin primes P_n – P_(n-1) = 2. Let X be the number of (6j –1, 6j+1) pairs in the interval [P_n, P_n^2]. The number of twin primes (TPAn) in [P_n, P_n^2] can be approximated by the formula
(a_3 /5)(a_4 /7)(a_5 /11)…(a_n /P_n)(X) for 3 ≤ m ≤ n, a_m = P_m –2 .
We establish a lower bound for TPAn (3/5)(5/7)(7/9)…(P_n–2)/P_n)(X) = 3X/P_n < TPAn.
We exhibit a formula showing as P_n increases, the number of twin primes in the interval [P_n, P_n^2] also increases. Let P_n – P_(n-1) = c. For all n (TPAn-1)(1+(2c –2)/2P_(n-1)+(c^2–2c)/2P_(n-1)^2) < TPAn

**Category:** Number Theory

[970] **viXra:1902.0005 [pdf]**
*replaced on 2019-02-16 12:01:31*

**Authors:** James Edwin Rock

**Comments:** 2 Pages of exposition and 7 pages of graphs

**Category:** Number Theory

[969] **viXra:1902.0005 [pdf]**
*replaced on 2019-02-13 08:29:18*

**Authors:** James Edwin Rock

**Comments:** 2 pages of exposition and 6 pages with supporting tables. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

**Category:** Number Theory

[968] **viXra:1901.0436 [pdf]**
*replaced on 2019-02-26 16:28:14*

**Authors:** Kenneth A. Watanabe

**Comments:** 13 Pages.

Legendre's conjecture, states that there is a prime number between n^2 and(n + 1)^2 for every positive integer n. In this paper, an equation was derived that accurately determines the number of prime numbers less than n for large values of n. Then it is proven by induction that there is at least one prime number between n^2 and (n + 1)^2 for all positive integers n thus proving Legendre’s conjecture.

**Category:** Number Theory

[967] **viXra:1901.0436 [pdf]**
*replaced on 2019-02-13 08:35:02*

**Authors:** Kenneth A. Watanabe

**Comments:** 8 Pages.

Legendre's conjecture, states that there is a prime number between n^2 and (n + 1)^2 for every positive integer n. In this paper, an equation was derived that determines the number of prime numbers less than n for large values of n. Then it is proven by mathematical induction that there is at least one prime number between n^2 and (n + 1)^2 for all positive integers n thus proving Legendre’s conjecture.

**Category:** Number Theory

[966] **viXra:1901.0436 [pdf]**
*replaced on 2019-02-11 08:33:18*

**Authors:** Kenneth A. Watanabe

**Comments:** 7 Pages.

Legendre's conjecture, states that there is a prime number between n^$ and (n + 1)^2 for every positive integer n. In this paper, an equation was derived that determines the number of prime numbers less than n for large values of n. Then it is proven by mathematical induction that there is at least 1 prime number between n^2 and (n + 1)^2 for all positive integers n thus proving Legendre’s conjecture.

**Category:** Number Theory

[965] **viXra:1901.0430 [pdf]**
*replaced on 2019-01-31 12:44:22*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Paper corrected of a mistake in the last version. Comments welcome.

In this paper, we consider the abc conjecture, then we give a proof of the conjecture c<rad^2(abc) that it will be the key to the proof of the abc conjecture.

**Category:** Number Theory

[964] **viXra:1901.0430 [pdf]**
*replaced on 2019-01-29 08:20:57*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, we consider the ABC conjecture, then we give a proof that C<rad*2(ABC) that it will be the key of the proof of the ABC conjecture.

**Category:** Number Theory

[963] **viXra:1901.0227 [pdf]**
*replaced on 2019-03-18 08:58:13*

**Authors:** James Edwin Rock

**Comments:** 11 Pages.

Collatz sequences are formed by applying the Collatz algorithm to any positive integer. If it is even repeatedly divide by two until it is odd, then multiply by three and add one to get an even number and vice versa. If the Collatz conjecture is true eventually you always get back to one. A connected Collatz Structure is created, which contains all positive integers exactly once. The terms of the Collatz Structure are joined together via the Collatz algorithm. Thus, every positive integer forms a Collatz sequence with unique terms terminating in the number one.

**Category:** Number Theory

[962] **viXra:1901.0227 [pdf]**
*replaced on 2019-02-25 10:42:27*

**Authors:** James Edwin Rock

**Comments:** 12 pages. Section 5 rewritten This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Collatz sequences are formed by applying the Collatz algorithm to any positive integer. If it is even repeatedly divide by two until it is odd, then multiply by three and add one to get an even number and vice versa. If the Collatz conjecture is true eventually you always get back to one. A connected Collatz Structure is created, which contains all positive integers exactly once. The terms of the Collatz Structure are joined together via the Collatz algorithm. Thus, every positive integer forms a Collatz sequence with unique terms terminating in the number one.

**Category:** Number Theory

[961] **viXra:1901.0227 [pdf]**
*replaced on 2019-02-21 08:01:40*

**Authors:** James Edwin Rock

**Comments:** 12 pages. Section 5 is rewritten.This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

**Category:** Number Theory

[960] **viXra:1901.0227 [pdf]**
*replaced on 2019-01-29 11:21:03*

**Authors:** James Edwin Rock

**Comments:** 7 pages. This paper should be much easier to understand than the paper it is replacing

**Category:** Number Theory

[959] **viXra:1901.0227 [pdf]**
*replaced on 2019-01-17 08:32:12*

**Authors:** James Edwin Rock

**Comments:** 5 pages. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

**Category:** Number Theory

[958] **viXra:1901.0104 [pdf]**
*replaced on 2019-03-05 14:25:19*

**Authors:** Toshiro Takami

**Comments:** 30 Pages.

I tried to prove that(Riemann hypothesis), but I realized that I can not prove how I did it.
When we calculate by the sum method of (1) we found that the nontrivial zero point will never converge to zero.
Calculating ζ(2), ζ(3), ζ(4), ζ(5) etc. by the method of the sum of (1) gives the correct calculation result.
This can be considered because convergence is extremely slow in the case of complex numbers, but there is no tendency to converge at all. Rather, it tends to diffuse.
In other words, it is inevitable to conclude that Riemann's hypothesis is a mistake.
We will fundamentally completely erroneous ones, For 150 years, We were trying to prove it.

**Category:** Number Theory

[957] **viXra:1901.0104 [pdf]**
*replaced on 2019-02-15 21:57:44*

**Authors:** Toshiro Takami

**Comments:** 8 Pages.

I tried to prove that(Riemann hypothesis), but I realized that I can not prove how I did it.
When we calculate by the sum method of (1) we found that the nontrivial zero point will never converge to zero.
Calculating ζ(2), ζ(3), ζ(4), ζ(5) etc. by the method of the sum of (1) gives the correct calculation result.
This can be considered because convergence is extremely slow in the case of complex numbers, but there is no tendency to converge at all. Rather, it tends to diffuse.
In other words, it is inevitable to conclude that Riemann's hypothesis is a mistake.
We will fundamentally completely erroneous ones, For 150 years, We were trying to prove it.

**Category:** Number Theory

[956] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-30 16:38:19*

**Authors:** James Edwin Rock

**Comments:** 5 Pages. This replacement paper adds a ratio TPA / 2.06 column to Table 2 and Table 1 contains a comparison of the twin prime pairs formula for [743,743^2] and [19993,19993^2]. It gives a better explanation for the twin prime pairs formula.

Let P_n be the n_th prime. For twin primes P_n – P_n-1 = 2. We exhibit formula for calculating the number of twin primes in the closed interval [P_n, P_n^2]. We prove that as P_n increases the number of twin primes in the interval [P_n, Pn^2] also increases.

**Category:** Number Theory

[955] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-25 15:03:02*

**Authors:** James Edwin Rock

**Comments:** 2 pages of exposition and 3 pages with supporting tables. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Let P_n be the n_th prime. For twin primes P_n – P_n-1 = 2. We exhibit formula for calculating the number of twin primes in the closed interval [P_n, P_n^2]. We prove that as P_n increases the number of twin primes in the interval [P_n, Pn^2] also increases.

**Category:** Number Theory

[954] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-22 09:05:33*

**Authors:** James Edwin Rock

**Comments:** 5 Pages.

Let Pn be the n_th prime. For twin primes Pn – Pn-1 = 2. We exhibit two formulas for calculating the number of twin primes in the closed interval [Pn, Pn^2]. We show there is a lower limit for the number of twin primes in the closed interval [Pn, Pn^2].

**Category:** Number Theory

[953] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-14 10:52:26*

**Authors:** James Edwin Rock

**Comments:** 2 pages of exposition and 3 pages with supporting tables. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Let Pn be the n_th prime. For twin primes Pn – Pn-1 = 2. We exhibit two formulas for calculating the number of twin primes in the closed interval [Pn, Pn^2]. We show there is a lower limit for the number of twin primes in the closed interval [Pn, Pn^2].

**Category:** Number Theory

[952] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-07 12:49:24*

**Authors:** James Edwin Rock

**Comments:** 1 page of exposition and 3 pages with supporting tables. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Let P_n be the n_th prime. For twin primes P_n – P_n-1 = 2. We exhibit two formulas for calculating the number of twin primes in the closed interval [P_n, P_n^2]. We show there is a lower limit for the number of twin primes in the closed interval [P_n, Pn^2].

**Category:** Number Theory

[951] **viXra:1812.0494 [pdf]**
*replaced on 2019-01-18 17:04:43*

**Authors:** Simon Plouffe

**Comments:** 10 Pages.

A set of new formulas for primes are given. These formulas have a growth rate much smaller than the ones of Mills and Wright.

**Category:** Number Theory

[950] **viXra:1812.0494 [pdf]**
*replaced on 2019-01-07 23:07:56*

**Authors:** Simon Plouffe

**Comments:** 7 Pages.

In 1947, W. H. Mills published a paper describing a formula that gives primes : if A = 1.3063778838630806904686144926… then ⌊A^(3^n ) ⌋ is always prime, here ⌊x⌋ is the integral part of x. Later in 1951, E. M. Wright published another formula, if g_0=α = 1.9287800… and g_(n+1)=2^(g_n ) then
⌊g_n ⌋= ⌊2^(…2^(2^α ) ) ⌋ is always prime.
The primes are uniquely determined by α , the prime sequence is 3, 13, 16381, …
The growth rate of these functions is very high since the fourth term of Wright formula is a 4932 digit prime and the 8’th prime of Mills formula is a 762 digit prime.
A new set of formulas is presented here, giving an arbitrary number of primes minimizing the growth rate. The first one is : if a_0=43.8046877158…and a_(n+1)= 〖〖〖(a〗_n〗^(5/4))〗^n , then if S(n) is the rounded values of a_n, S(n) = 113,367,102217,1827697,67201679,6084503671, …. Other exponents can also give primes like 11/10, or 101/100. If a_0 is well chosen then it is conjectured that any exponent > 1 can also give an arbitrary series of primes. The method for obtaining the formulas is explained. All results are empirical.

**Category:** Number Theory

[949] **viXra:1812.0494 [pdf]**
*replaced on 2018-12-31 12:31:57*

**Authors:** Simon Plouffe

**Comments:** 6 Pages.

A new set of formulas for primes is presented, they are better than the ones of Mills and Wright.

**Category:** Number Theory

[948] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-15 03:19:29*

**Authors:** A. A. Frempong

**Comments:** 11 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. In the numerical equations, two approaches have been used to change the sum, A^x + B^y, of two powers to a single power, C^z. In one approach, the application of factorization is the main principle, while in the other approach, a formula derived from A^x + B^y was applied. The two approaches changed the sum A^x + B^y to a single power, C^z, perfectly. The derived formula confirmed the validity of the assumption that it is necessary that the sum A^x + B^y has a common prime factor before C^z can be derived. It was concluded that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor.

**Category:** Number Theory

[947] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-08 13:30:23*

**Authors:** A. A. Frempong

**Comments:** 4 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor By applying numerical examples, it is shown that one can begin with the sum A^x + B^y and change this sum to a product and to the single power, C^z. It is concluded that it is necessary that the sum
A^x + B^y has a common prime factor before C^z can be derived. It was determined that if
A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[946] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-05 16:32:17*

**Authors:** A. A. Frempong

**Comments:** 4 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor By applying numerical examples, it is shown that one can begin with the sum A^x + B^y and change this sum to a product and to the single power, C^z. It is concluded that it is necessary that the sum
A^x + B^y has a common prime factor before C^z can be derived. It was determined that if
A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[945] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-04 11:56:04*

**Authors:** A. A. Frempong

**Comments:** 4 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor By applying numerical examples, it is shown that one can begin with the sum A^x + B^y and change this sum to a product and to the single power, C^z. It is concluded that it is necessary that the sum
A^x + B^y has a common prime factor before C^z can be derived. It was determined that if
A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[944] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-01 21:14:39*

**Authors:** A. A. Frempong

**Comments:** 4 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. By applying numerical examples, it is shown that one can begin with the sum A^x + B^y and change this sum to a product and to the single power, C^z. It is concluded that it is necessary that the sum
A^x + B^y has a common prime factor before C^z can be derived. It was determined that if
A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[943] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-01 03:48:04*

**Authors:** A. A. Frempong

**Comments:** 4 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. Using concrete examples, the proof would be complete after showing that if A and B have a common prime factor, the sum A^x + B^y can be changed to a product and to a single power, C^z. In the proof, one begins with A^x + B^y and changes this sum to the single power, C^z . It was determined that if A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[942] **viXra:1812.0422 [pdf]**
*replaced on 2018-12-28 14:18:23*

**Authors:** A. A. Frempong

**Comments:** 8 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. The proof would be complete after showing that if A and B have a common prime factor, and C^z can be produced from the sum A^x + B^y. In the proof, one begins with A^x + B^y and changes this sum to the single power, C^z as was done in the preliminaries. It was determined that if A^x + B^y = C^z , then A, B and C have a common prime factor.

**Category:** Number Theory

[941] **viXra:1812.0400 [pdf]**
*replaced on 2018-12-25 07:45:30*

**Authors:** Jesús Sánchez

**Comments:** 21 Pages.

b=0;
p=input('Input a number : ');
m=fix((p+1)/2);
for k=2:m+1;
fun=@(w) exp(p.*1j.*w).*exp(-2.*k.*1j.*w).*(1-exp(-m.*k.*1j.*w))./(1-exp(-k.*1j.*w));
a=integral(fun,-pi,pi);
b=b+a;
end;
b=b/(2*pi);
disp(b);
Above simple MATLAB/Octave program, can detect if a number is prime or not. If the result is zero (considering zero being less than 1e-5), the number introduced is a prime.
If the result is an integer, this result will tell us how many permutations of two divisors, the input number has. As you can check, no recurrent division by odd or prime numbers is done. Just this strange integral:
1/2π ∫_(-π)^π▒〖e^pjω (∑_(k=2)^(k=m+1)▒〖e^(-2kjω) ((1-e^(-mkjω))/(1-e^(-kjω) )) 〗)dω 〗
Being p the number that we want to check if it is a prime or not. And being m whatever integer number higher than (p+1)/2(the lowest, the most efficient the operation). As k and ω are independent variables, the sum can be taken outside the integral (as it is in the above program).
To get to this point, we will do the following. First, we will create a domain with all the composite numbers. This is easy, as you can just multiply one by one all the integers (greater or equal than 2) in that domain. So, you will get all the composite numbers (not getting any prime) in that domain.
Then, we will use the Fourier transform to change from this original domain (called discrete time domain in this regards) to the frequency domain. There, we can “ask”, using Parseval’s theorem, if a certain number is there or not. The use of Parseval’s theorem leads to the above integral. If the number p that we want to check is not in the domain, the result of the integral is zero and the number is a prime. If instead, the result is an integer, this integer will tell us how many permutations of two divisors the number p has. And, in consequence information how many factors, the number p has.
So, for any number p lower than 2m-1, you can check if it is prime or not, just making the numerical definite integration (but even this integral, if no further developments are done, the numerical integration is inefficient computing-wise compared with brute-force checking for example). To be added, is the question regarding the level of accuracy needed (number of decimals and number of steps in the numerical integration) to have a reliable result for large numbers. This will be commented on the paper, but a separate study will be needed to have detailed conclusions.
Of course, the best would be that in the future, an analytical result (or at least an approximation) for the summation or for the integration is achieved.

**Category:** Number Theory

[940] **viXra:1812.0340 [pdf]**
*replaced on 2018-12-21 15:57:08*

**Authors:** Wu ShengPing

**Comments:** 4 Pages.

The main idea of this article is simply calculating integer
functions in module. The algebraic in the integer modules is studied in
completely new style. By a careful construction the result that
two finite numbers is with unequal logarithms in a corresponding module is proven, which result is applied to solving
a kind of high degree diophantine equation.

**Category:** Number Theory

[939] **viXra:1812.0312 [pdf]**
*replaced on 2018-12-19 06:26:14*

**Authors:** Julian Beauchamp

**Comments:** 3 Pages.

In 2002 Preda Mihailescu used the theory of cyclotomic fields and Galois modules to prove Catalan's Conjecture. In this short paper, we give a very simple proof. We first prove that no solutions exist for a^x-b^y=1 for a,b>0 and x,y>2. Then we prove that when x=2 the only solution for a is a=3 and the only solution for y is y=3.

**Category:** Number Theory

[938] **viXra:1812.0305 [pdf]**
*replaced on 2018-12-25 04:29:58*

**Authors:** Juan Moreno Borrallo

**Comments:** 24 Pages.

In this paper it is proposed and proved an exact formula for the prime-counting function, finding an expression of Legendre's formula. As corollaries, they are proved some important conjectures regarding prime numbers distribution.

**Category:** Number Theory

[937] **viXra:1812.0242 [pdf]**
*replaced on 2018-12-14 12:37:47*

**Authors:** Jean Pierre Morvan

**Comments:** 6 Pages.

L'hypothèse de RIEMANN est fausse parce que la fonction zéta n'est pas nulle avec s = 1/2 + ib

**Category:** Number Theory

[936] **viXra:1812.0208 [pdf]**
*replaced on 2019-01-23 08:21:08*

**Authors:** Kenneth A. Watanabe

**Comments:** 14 Pages. Acknowledgement section was added and proof of delta pi was added

A twin prime is defined as a pair of prime numbers (p1,p2) such that p1 + 2 = p2. The Twin Prime Conjecture states that there are an infinite number of twin primes. A more general conjecture by de Polignac states that for every natural number k, there are infinitely many primes p such that p + 2k is also prime. The case where k = 1 is the Twin Prime Conjecture. In this document, a function is derived that corresponds to the number of twin primes less than n for large values of n. Then by proof by induction, it is shown that as n increases indefinitely, the function also increases indefinitely thus proving the Twin Prime Conjecture. Using this same methodology, the de Polignac Conjecture is also shown to be true.

**Category:** Number Theory

[935] **viXra:1812.0208 [pdf]**
*replaced on 2019-01-09 08:19:00*

**Authors:** Kenneth A. Watanabe

**Comments:** 13 Pages. Paper was modified to include future directions and additional references were added

A twin prime is defined as a pair of prime numbers (p1,p2) such that p1 + 2 = p2. The Twin Prime Conjecture states that there are an infinite number of twin primes. A more general conjecture by de Polignac states that for every natural number k, there are infinitely many primes p such that p + 2k is also prime. The case where k = 1 is the Twin Prime Conjecture. In this document, a function is derived that corresponds to the number of twin primes less than n for large values of n. Then by proof by induction, it is shown that as n increases indefinitely, the function also increases indefinitely thus proving the Twin Prime Conjecture. Using this same methodology, the de Polignac Conjecture is also shown to be true.

**Category:** Number Theory

[934] **viXra:1812.0208 [pdf]**
*replaced on 2018-12-27 09:47:28*

**Authors:** Kenneth A. Watanabe

**Comments:** 12 Pages.

A twin prime is defined as a pair of prime numbers (p1,p2) such that p1 + 2 = p2. The Twin Prime Conjecture states that there are an infinite number of twin primes. The first mention of the Twin Prime Conjecture was in 1849, when de Polignac made the more general conjecture that for every natural number k, there are infinitely many primes p such that p + 2k is also prime. The case where k = 1 is the Twin Prime Conjecture. In this document, I derive a function that corresponds to the number of twin primes less than n for large values of n. This equation is identical to that used to prove the Goldbach Conjecture. Then by proof by induction, it is shown that as n increases indefinitely, the function also increases indefinitely thus proving the Twin Prime Conjecture. Using the same methodology, de Polignac’s conjecture is also shown to be true.

**Category:** Number Theory

[933] **viXra:1812.0182 [pdf]**
*replaced on 2018-12-30 15:25:04*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, from a,b,c positive integers relatively prime with c=a+b, we consider a bounded of c depending of a,b. Then we do a choice of K(\epsilon) and finally we obtain that the ABC conjecture is true. Four numerical examples confirm our proof.

**Category:** Number Theory

[932] **viXra:1812.0182 [pdf]**
*replaced on 2018-12-16 05:58:38*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, we use the Fermat's Last Theorem (FLT) to give a proof of the ABC conjecture. We suppose that FLT is false ===> we arrive that the ABC conjecture is false. Then taking the negation of the last statement, we obtain: ABC conjecture is true ===> FLT is true. But, as FLT is true, then we deduce that the ABC conjecture is true.

**Category:** Number Theory

[931] **viXra:1812.0154 [pdf]**
*replaced on 2018-12-13 06:58:30*

**Authors:** M. Sghiar

**Comments:** 7 Pages. Accepted & french version © Copyright 2018 by M. Sghiar. All rights reserved. Respond to the author by email at: msghiar21@gmail.com

I show here that if $ x \in \mathbb{N}^*$ then $1 \in \mathcal{O}_S (x)= \{ S^n(x), n \in \mathbb{N}^* \} $ where $ \mathcal{O}_S (x)$ is the orbit of the function S defined on $\mathbb{R}^+$ by $S(x)= \frac{x}{2} + (x+\frac{1}{2}) sin^2(x\frac{\pi}{2})$, and I deduce the proof of the Syracuse conjecture.

**Category:** Number Theory

[930] **viXra:1812.0150 [pdf]**
*replaced on 2019-03-14 09:19:34*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-(2^\frac{2}{n}q)^n=((r-2q^n)^\frac{1}{n})^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $2^\frac{2}{n}q$ by $t$; and, we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $u$. For any given $n>2$ : Since the term $t$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null set $\{(s,t,u)|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null set $\{z,y,x|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$: We show it is true, for $n>0$, that $\{t|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}=\{y|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[929] **viXra:1812.0150 [pdf]**
*replaced on 2019-03-07 18:34:05*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-(2^\frac{2}{n}q)^n=((r-2q^n)^\frac{1}{n})^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $2^\frac{2}{n}q$ by $t$; and, we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $u$. For any given $n>2$ : Since the term $t$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null set $\{(s,t,u)|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null set $\{z,y,x|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$: We show it is true, for $n>0$, that $\{t|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}=\{y|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[928] **viXra:1812.0150 [pdf]**
*replaced on 2019-03-02 20:18:33*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-(2^\frac{2}{n}q)^n=((r-2q^n)^\frac{1}{n})^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $2^\frac{2}{n}q$ by $t$; and, we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $u$. For any given $n>2$ : Since the term $t$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null set $\{(s,t,u)|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null set $\{z,y,x|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$: We show it is true, for $n>0$, that $\{t|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}=\{y|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[927] **viXra:1812.0150 [pdf]**
*replaced on 2019-02-23 19:59:18*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

**Category:** Number Theory

[926] **viXra:1812.0150 [pdf]**
*replaced on 2019-02-18 09:09:13*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

**Category:** Number Theory

[925] **viXra:1812.0150 [pdf]**
*replaced on 2019-02-13 19:33:18*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

**Category:** Number Theory

[924] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-31 20:27:12*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-(2^\frac{2}{n}q)^n=((r-2q^n)^\frac{1}{n})^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $2^\frac{2}{n}q$ by $t$; and, we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $u$. For any given $n>2$ : Since the term $t$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null sets $\{(s,t,u)|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null sets $\{z,y,x|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$: We show it is true, for $n>0$, that $\{t|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}=\{y|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[923] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-29 20:38:53*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-((r-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $t$; and, we \emph{denote} $2^\frac{2}{n}q$ by $u$. For any given $n>2$ : Since the term $u$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null sets $\{(s,t,u)|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null sets $\{z,y,x|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$. We show it is true, for $n>0$, that $\{u|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}=\{x|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[922] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-22 19:55:05*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-((r-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $t$; and, we \emph{denote} $2^\frac{2}{n}q$ by $u$. For any given $n>2$ : Since the term $u$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null sets $\{(s,t,u)|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}$ with subsequently proven null sets $\{z,y,x|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$. We show it is true, for $n>0$, that $\{u|s,t,u\in\mathbb{N},s,t,u>0,s^n-t^n=u^n\}=\{x|z,y,x\in\mathbb{N},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[921] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-19 20:14:58*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-((r-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ for which $n$ is any given positive natural number, $r$ is positive real and $q$ is positive rational such that the set of triples $\{((r+2q^n)^\frac{1}{n},(r-2q^n)^\frac{1}{n},2^\frac{2}{n}q)\}$ is not empty with $(r+2q^n)^\frac{1}{n},(r-2q^n)^\frac{1}{n},(2^\frac{2}{n}q)\in\mathbb{N}$. We relate this set of triples to $\{z,y,x|z,y,x\in\mathbb{N}$ for which the transposed \emph{Fermat equation} $z^n-y^n=x^n$ holds. We demonstrate, for any given value of $n$, that $2^\frac{2}{n}q=x$. Clearly, for $n>2$, the term $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational. Consequently, for values of $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[920] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-14 21:02:12*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-((r-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ for which $n$ is any given positive natural number, $r$ is positive real and $q$ is positive rational such that the set of triples $\{((r+2q^n)^\frac{1}{n},(r-2q^n)^\frac{1}{n},2^\frac{2}{n}q)\}$ is not empty with $(r+2q^n)^\frac{1}{n},(r-2q^n)^\frac{1}{n},(2^\frac{2}{n}q)\in\mathbb{N}$. We relate this set of triples to $\{z,y,x|z,y,x\in\mathbb{N}$ for which the transposed \emph{Fermat equation} $z^n-y^n=x^n$ holds. We demonstrate, for any given value of $n$, that $2^\frac{2}{n}q=x$. Clearly, for $n>2$, the term $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational. Consequently, for values of $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[919] **viXra:1812.0150 [pdf]**
*replaced on 2019-01-09 21:44:36*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT simply for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not by way of contradiction) of FLT is based on our algebraic identity $((r^n+2q^n)^\frac{1}{n})^n-((r^n-2q^n)^\frac{1}{n})^n=(2^\frac{2}{n}q)^n$ such that $n$ is any given positive natural number, $r$ is unrestricted positive real and $q$ are all positive rationals, for which the set of triples $\{((r^n+2q^n)^\frac{1}{n},(r^n-2q^n)^\frac{1}{n},2^\frac{2}{n}q\}$ is not empty with $(r^n+2q^n)^\frac{1}{n},(r^n-2q^n)^\frac{1}{n},(2^\frac{2}{n}q)\in\mathbb{N}$. We relate this identity to the transposed \emph{Fermat equation} $z^n-y^n=x^n$ for which $z,y,x$ are such natural numbers. We demonstrate, for any given value of $n$, that $2^\frac{2}{n}q=x$. Clearly, for $n>2$, the term $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational. Consequently, for values of $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[918] **viXra:1812.0150 [pdf]**
*replaced on 2018-12-11 10:03:58*

**Authors:** Phil Aaron Bloom

**Comments:** 2 Pages.

An open problem is proving FLT simply for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not by way of contradiction) of FLT is based on our algebraic identity, denoted, {for convenience}, as $(r)^n+(s)^n=(t)^n$ with $r,s,t>0$ as functions of variables. We infer that $\{(r,s,t)|r,s,t\in\mathbb{N},(r)^n+(s)^n+(t)^n\}=\{(x,y,z)|x,y,z\in\mathbb{N},(x)^n+y^n=z^n\}$ for $n\in\mathbb{N}, n>2$. In addition, we show, for integral values of $n>2$, that $\{(r,s,t)|r,s,t\in\mathbb{N},(r)^n+(s)^n=t^n\}=\varnothing$. Hence, for $n\in\mathbb{N},n>2$, it is true that $\{(x,y,z)|x,y,z\in\mathbb{N},x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[917] **viXra:1812.0074 [pdf]**
*replaced on 2018-12-10 15:13:28*

**Authors:** Stephen Marshall

**Comments:** 8 Pages.

The Riemann hypothesis is a conjecture that the Riemann zeta function has its zero’s only at the negative even integers and complex numbers with real part 1/2
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000).
It was proposed by Bernhard Riemann (1859), after whom it is named. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields.
The Riemann hypothesis implies significant results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000). The Riemann zeta function is defined for complex s with real part greater than 1 by the absolutely convergent infinite series:
ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ...
The Riemann hypothesis asserts that all interesting solutions of the equation:
ζ(s) = 0
lie on a certain vertical straight line.
In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers:
Hn = 1 + 1/2+1/3+1/4+⋯+ 1/n = ∑_(n=1)^n▒1/n
Harmonic numbers have been studied since early times and are important in various branches of number theory. They are sometimes loosely termed harmonic series, are closely related to the Riemann zeta function.
The harmonic numbers roughly approximate the natural logarithm function and thus the associated harmonic series grows without limit, albeit slowly. In 1737, Leonhard Euler used the divergence of the harmonic series to provide a new proof of the infinity of prime numbers. His work was extended into the complex plane by Bernhard Riemann in 1859, leading directly to the celebrated Riemann hypothesis about the distribution of prime numbers.

**Category:** Number Theory

[916] **viXra:1812.0074 [pdf]**
*replaced on 2018-12-06 08:38:16*

**Authors:** stephen Marshall

**Comments:** 7 Pages.

The Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1/2
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000).
It was proposed by Bernhard Riemann (1859), after whom it is named. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields.
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000). The Riemann zeta function is defined for complex s with real part greater than 1 by the absolutely convergent infinite series:
ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ...
The Riemann hypothesis asserts that all interesting solutions of the equation:
ζ(s) = 0
lie on a certain vertical straight line.
In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers:
Hn = 1 + 1/2+1/3+1/4+⋯+ 1/n = ∑_(n=1)^n▒1/n
Harmonic numbers have been studied since early times and are important in various branches of number theory. They are sometimes loosely termed harmonic series, are closely related to the Riemann zeta function.
The harmonic numbers roughly approximate the natural logarithm function and thus the associated harmonic series grows without limit, albeit slowly. In 1737, Leonhard Euler used the divergence of the harmonic series to provide a new proof of the infinity of prime numbers. His work was extended into the complex plane by Bernhard Riemann in 1859, leading directly to the celebrated Riemann hypothesis about the distribution of prime numbers.

**Category:** Number Theory

[915] **viXra:1812.0022 [pdf]**
*replaced on 2018-12-25 01:58:45*

**Authors:** Pankaj Mani

**Comments:** 12 Pages.

Riemann Hypothesis is TRUE if we look at the Functional Equation satisfied by the Riemann Zeta function upon analytical continuation in Game Perspective way as visualized by David Hilbert. It uses technical game theoretical concepts e.g. Nash Equilibrium to assert that Riemann Hypothesis has to be True. Needs to be looked at the Foundational Principles underlying Mathematics. In other words,it’s the game of arranging Zeros in the complex plane using the functional equation.

**Category:** Number Theory

[914] **viXra:1812.0019 [pdf]**
*replaced on 2018-12-13 09:03:08*

**Authors:** Kamal Barghout

**Comments:** 16 Pages. The material in this article is copyrighted. Please obtain authorization from the author before the use of any part of the manuscript

A probabilistic proof of the Collatz conjecture is described via identifying a sequential permutation of even natural numbers by divisions by 2 that follows a recurrent pattern of the form x,1,x,1…, where x represents divisions by 2 more than once. The sequence presents a probability of 50:50 of division by 2 more than once as opposed to divisions by 2 once over the even natural numbers. The sequence also gives the same 50:50 probability of consecutive Collatz even elements when counted for division by 2 more than once as opposed to division by 2 once and a ratio of 3:1 of high counts. Considering Collatz function producing random numbers and over sufficient iterations, this probability distribution produces numbers in descending order that lead to the convergence of the Collatz function to 1, assuming the only cycle of the function is 1-4-2-1.

**Category:** Number Theory

[913] **viXra:1812.0019 [pdf]**
*replaced on 2018-12-08 04:52:18*

**Authors:** Kamal Barghout

**Comments:** 12 Pages. The material in this article is copyrighted. Please obtain authorization from the author before the use of any part of the manuscript

A probabilistic proof of the Collatz conjecture is described via identifying a sequential permutation of even natural numbers by divisions by 2 that follows a recurrent pattern of the form x,1,x,1…, where x represents divisions by 2 more than once. The sequence presents a probability of 50:50 of division by 2 more than once as opposed to divisions by 2 once over the even natural numbers. The sequence also gives the same 50:50 probability of consecutive Collatz even elements when counted for division by 2 more than once as opposed to division by 2 once and a ratio of 3:1 of high counts. Considering Collatz function producing random numbers and over sufficient iterations, this probability distribution produces numbers in descending order that lead to the convergence of the Collatz function to 1, assuming the only cycle of the function is 1-4-2-1.

**Category:** Number Theory

[912] **viXra:1812.0018 [pdf]**
*replaced on 2019-01-08 21:11:56*

**Authors:** Khalid Ibrahim

**Comments:** 96 Pages.

In this paper, not only did we disprove the Riemann Hypothesis (RH) but we also showed that zeros of the Riemann zeta function $\zeta (s)$ can be found arbitrary close to the line $\Re (s) =1$. Our method to reach this conclusion is based on analyzing the fine behavior of the partial sum of the Dirichlet series with the Mobius function $M (s) = \sum_n \mu (n) /n^s$ defined over $p_r$ rough numbers (i.e. numbers that have only prime factors greater than or equal to $p_r$). Two methods to analyze the partial sum fine behavior are presented and compared. The first one is based on establishing a connection between the Dirichlet series with the Mobius function $M (s) $ and a functional representation of the zeta function $\zeta (s)$ in terms of its partial Euler product. Complex analysis methods (specifically, Fourier and Laplace transforms) were then used to analyze the fine behavior of partial sum of the Dirichlet series. The second method to estimate the fine behavior of partial sum was based on integration methods to add the different co-prime partial sum terms with prime numbers greater than or equal to $p_r$. Comparing the results of these two methods leads to a contradiction when we assume that $\zeta (s)$ has no zeros for $\Re (s) > c$ and $c <1$.

**Category:** Number Theory

[911] **viXra:1812.0018 [pdf]**
*replaced on 2018-12-22 01:07:51*

**Authors:** Khalid Ibrahim

**Comments:** 95 Pages.

In this paper, not only did we disprove the Riemann Hypothesis (RH) but we also showed that zeros of the Riemann zeta function $\zeta (s)$ can be found arbitrary close to the line $\Re (s) =1$. Our method to reach this conclusion is based on analyzing the fine behavior of the partial sum of the Dirichlet series with the Mobius function $M (s) = \sum_n \mu (n) /n^s$ defined over $p_r$ rough numbers (i.e. numbers that have only prime factors greater than or equal to $p_r$). Two methods to analyze the partial sum fine behavior are presented and compared. The first one is based on establishing a connection between the Dirichlet series with the Mobius function $M (s) $ and a functional representation of the zeta function $\zeta (s)$ in terms of its partial Euler product. Complex analysis methods (specifically, Fourier and Laplace transforms) were then used to analyze the fine behavior of partial sum of the Dirichlet series. The second method to estimate the fine behavior of partial sum was based on integration methods to add the different co-prime partial sum terms with prime numbers greater than or equal to $p_r$. Comparing the results of these two methods leads to a contradiction when we assume that $\zeta (s)$ has no zeros for $\Re (s) > c$ and $c <1$.

**Category:** Number Theory