Timeline for Proof that $3^ns + \sum_{k=0}^{n-1} 3^{n-k-1}2^{a_k}=2^m.$

5 events

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Nov 2, 2019 at 20:53	history	edited	Gottfried Helms	CC BY-SA 4.0	added 50 characters in body
Nov 2, 2019 at 17:06	comment	added	Gottfried Helms		@ReverseFlow - yes, that is much harder. I've left this route when I was more engaged in the analysis of the Collatz-problem, and looked into the question of cycles instead, and to have some analyzable material discussed few-step cycles and also the so-called "1-cycle". There is much untractable there ... sigh... :)
Nov 2, 2019 at 15:08	comment	added	ReverseFlowControl		So...finding orbits with 2 odd integers is trivial. Just take $b_n=\frac{4^n-1}{3}$, which is exactly what you have listed there. It is also easy to prove that this is the only form there is for 2 odd integers in an orbit containing s and 1.The interesting part is taking it beyond that using explicit formulas without using recursive constructs. That is much harder.
Nov 2, 2019 at 14:59	history	edited	Gottfried Helms	CC BY-SA 4.0	added picture
Nov 2, 2019 at 13:04	history	answered	Gottfried Helms	CC BY-SA 4.0