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Re-viewing some older fiddlings on more elementary aspects of the existence of "1-cycles" in the Collatz-problem I came at the following problem.

Let $N \gt 0$ denote the number of odd steps of an assumed "1-cycle", $\gamma = \log_2(3)$, then $S_N=\lceil N \cdot \gamma \rceil$,$B_N=\lceil N \cdot (\gamma-1) \rceil$.
Then from a well known formula for the existence of "1-cycles" (adapted to my notation) with length $N$ asking for (positive) integer $k$ in
$$k = {2^{B_N}-1\over 2^{S_N}-3^N} \tag 1$$ Now I write $3^N$ modular with base $2^{B_N}$ as $3^N = m \cdot 2^{B_N} - r_N$ where $r_N \in \{0 \ldots (2^{B_N}-1)\}$.
This gives $$k(2^{S_N}-m \cdot 2^{B_N} + r_N) = 2^{B_N}-1 \\ k\cdot 2^{B_N}(2^N-m) +k\cdot r_N = 2^{B_N}-1 \tag {2a}$$ Obviously in the lhs the term $2^N-m$ must be zero for a solution at all; (of course this asks for the analysis of the number of leading ones in the binary expansion of $3^N$ but this is another route). What I found more impressing - and perhaps easier to prove - is the observation by the subsequent more explicite requirement, $$k\cdot r_N = 2^{B_N}-1 \tag {2b}$$ that the residue $r_N$ in $3^N = m \cdot 2^{B_N} - r_N $ must be a divisor of $2^{B_n}-1$ and that this reduces the set of possible solutions -seemingly- for $N$ to a nice small set - a property, which might have been found already elsewhere.

One can as well write $r_N= (-3^N) \bmod 2^{B_N}$, so the formula in the subject line came to existence.

For this I found the amazing heuristic that for $N$ up to some $10000$ the residue $r_N \;\mid \; 2^{B_N} - 1 $ or in the notation of the subject line $$ (2^{B_N} - 1) \bmod \left( -3^N \bmod 2^{B_N}\right) =0 \implies N \in \{1,2,3, 4,8,12\} \tag 3$$ or, what caught my amazement, a bit sloppy: $ N \in \{1,2,3\} \times \{1,4\}$ .

Thus my question: is this known, that the set of solutions of eq (3) is a) finite and b) is just the set of the found small solutions?

P.S.: I'd propose, in following comments or answers, to use simply $S$ and $B$ -notation instead of the full $S_N$ and $B_N$ - to simplify reading & writing arguments. I used that explicite notation here only to avoid any initial misunderstanding

Note: this problem is "weaker" than the question of existence of the "1"-cycle, and so might not need the tools of transcendental number theory which is employed in the Steiner/Simons-proofs

The following picture shows the primefactorization of $2^B-1$ in contrast to the values of $3^N \pmod {2^B}$, where for each $2^B$ the sequence of $3^k \; k=1..20$ is documented. The yellow marked entries show the solutions $N\in\{1,2,3\} \cup \{4,8,12\}$. Interesting that the periodicity for each $2^B$ over consecutive $3^k$ seems to be just $2^{B-2}$. Is there some hint in this?
picture

Gottfried Helms
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    if $r=2^{S_N}-3^N$, shouldn't you have $r=(-3^N) % 2^{S_N}$ – Collag3n Jan 27 '21 at 16:05
  • @Collag3n - this is well true. And, by the complete rhs, of course smaller than $2^B$, because the numerator is divided by a value which is larger than $1$ in all cases except $N=1$. So by construction, $r$ must also be in the interval $0 \ldots (2^B-1)$. I didn't check, whether taking the residue modulus $2^S$ would give the same restrictive set of solutions, but possibly it does. I took my version because I guess - if at all - then the result would possibly be known in that form. – Gottfried Helms Jan 27 '21 at 16:30
  • If for b) it is true that these are the only solutions then isn't that considerably stronger than the 1-cycle result ? One would then only need an easy check for these values. –  Jan 27 '21 at 17:18
  • @S.Dolan - I'd say, it is "weaker" because this condition alone does not guarantee that no "1"-cycle exists: by this criterion alone we could have "1-cycles" with $6$ different $N$ but still have only one (the trivial one, plus another one with $N=2$ in the negative integers). The other, let's call it, complementary or supplementary condition is, that $m=2^N$ and this is empirically untrue which makes the complete relation to the "1-cycle". I've looked at this -in terms of leading ones in the binary representation of $3^N$; (...) – Gottfried Helms Jan 27 '21 at 17:24
  • (...) there is some little literature on this, for instance by M. Bennet (from the 1990ies), but I didn't try to combine my given condition to the bounds in his article so far. Just the specific and small set of solutions for my stated problem amazed me, so I thought, this might be known in a wider/different context. – Gottfried Helms Jan 27 '21 at 17:30
  • Ehmm, Michael Bennet "An ideal Waring problem with restricted summands" (1994) in "Acta Arithmetica" discusses $|| {(N+1)^k\over N^k} ||$ (the double "||" meaning smallest distance to integers) with a more general focus - downloadable. (unfortunately I didn't save the link... :-( ) – Gottfried Helms Jan 27 '21 at 17:38
  • I have to dig deeper. Either I end up on this: https://math.stackexchange.com/questions/3862457/is-this-argument-valid-for-showing-given-integers-n1-and-2m-3n-2m/3866590#3866590 (where $k=b$) or on this:https://math.stackexchange.com/questions/3338466/showing-3iq-2i-p-neq2pq-1-with-p-lceil-i-log-23-rceil-q-left (where $k=a$). In the later case, $k=2^B\left{\frac{2^N\cdot3^{2^{B-2}}}{2^S\cdot3^N}\right}$ – Collag3n Jan 28 '21 at 16:15
  • @Collag3n - yes, I knew that questions. Well, I think excursions from here into the known handling of the "1-cycle" should not give new insights - this specific little problem is surely worth to be a stand-alone one, and my hope was/is, it might be solvable by simpler means than the transcendental number theory approach behind the Simons/ Steiner solution(s). I'd liked to see, what Steven Finch would have to say about this, but his book "Mathematical constants" has its price... At your formula: this formulation surprises me! I'll look what I can take from this. – Gottfried Helms Jan 28 '21 at 19:03
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    perhaps with $k=(3^{2^{B-2}-N}); % ; 2^B=(a\mod c)$ and $r= (-3^N) ; % ; 2^B=(b\mod c)$ and using $((a\mod c)(b\mod c))\mod c=(ab)\mod c$ ? The LHS is $kr=2^B-1$ or $(-1);%;2^B$ and the RHS is $(-3^{2^{B-2}}); % ; 2^B$ – Collag3n Jan 28 '21 at 19:54
  • I don't think it will help, $(-3^{2^{B-2}})\equiv (-1)\mod 2^B$ anyway – Collag3n Jan 28 '21 at 21:54
  • ...yes @Collag3n, because it ($2^{B-2}$) is just the length-of-the-periode... or not? (I'm back tomorrow) – Gottfried Helms Jan 28 '21 at 21:56
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    yes, the order of $3 \mod 2^B$ where $\lambda(2^B)=2^{B-2}$ (Carmichael) – Collag3n Jan 28 '21 at 22:04
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    also note that it would simplify what I wrote here https://math.stackexchange.com/questions/3338466/showing-3iq-2i-p-neq2pq-1-with-p-lceil-i-log-23-rceil-q-left to $(3^N(3^{-N} % 2^B)-1)2^{-B}\leq 2^N(3^{-N} % 2^B)-1$ with equality in case of a cycle ($N=i$) – Collag3n Jan 29 '21 at 22:43

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