**Authors:** Darrin Taylor

In base 3, the presence of leading 1s during division has a one to one correlation with the 3n+1 operation. This is because dividing a leading 1 in base 3 is the only way to lose a digit and 3n+1 shifts are the only way to gain a digit. Total digit length doesn't change around a loop so they must equal each other. Because the leading 1 pattern among a series of divides only has 2 segments either 1->2 or 1->2->1 there are a limited number of patterns that can make up a loop. Naming 1->2->(next segments leading 1) as segment A Naming 1->2->1->(next segments leading 1) as segment B We can see that A is 2 divides and 1 non localized shift while B is 3 divides and 2 non localized shifts. The pattern ABB...ABB descends because 8 divides and 5 shifts descends for numbers larger than 1000 and lower then 1000 have been numerically disqualified previously. So the sequence BBB must exist at least once in every loop. BBB implies ABBB or BBBB if BBBB then "expel" a B which ascends and keep searching for the segment before the sequence. Once ABBB is found this implies AABBB or BABBB and AABBB is disproven as not possible. Once BABBB is known this implies ABABBB or BBABBB and ABABBB is disproven. Once BBABBB is known this implies ABBABBB or BBBABBB and ABBABBB is disproven. Once BBBABBB is found we can "expel" ABBB which ascends and BBB(ABBB) becomes BBB and we are back where we started. Once the entire loop has been traversed this way the sequence has expelled only (B) or (ABBB) and the remaining sequence is BBB and all of these ascend. Loops must have ascending and descending segments for a total non ascending and non descending but this loop always ascends. Thus it cannot be a loop and no loops of As and Bs can exist as those with fewer Bs than ABBABBABB…..always descend and adding a single B makes it always ascend. And As and Bs are the only possible segments to add.

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[v1] 2019-07-09 01:25:02

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