If $\sum\limits_{i=0}^{k-1} 2^{a_i} 3^i$ is divisible by $2^m - 3^k$, then the quotient is $1$ or $2$ (Existence of cycles in the Collatz-problem) Given any two positive integers $m,k$ and a sequence of numbers $a_0, a_1,...,a_{k-1}$ such that $m>a_0>a_1...>a_{k-1}\ge0$,
then $$S=\frac{\sum_{i=0}^{k-1}{2^{a_i}3^i}}{2^m-3^k}\in \mathbb{N} \implies  S=1,2$$
Any idea how to solve this? I quickly wrote up a python script and its true for $m\le20$. Pretty sure this isn't within my level of math but any help or points of attack could be useful
I think this is loosely related to the Collatz conjecture because it involves powers of 3 and 2
 A: This is exactly the question of existence of cycles in the Collatz-problem, and is still open. There are partial solutions, for instance if the sequence of the $a_k$ is $[0,1,2,3,...,N-1$ then we have the "1-cycle" problem for which R. Steiner showed 1976 that there is only one positive integer soultion for the quotient, namely $1$.               
The formula occurs in the following way; I use different letters here: $N$ for your $k$ and $S$ for your $m$.          
Write the "Syracuse"-notation of the Collatz-problem in odd positive numbers $a$ and $b$ $$a_{k+1} = {3a_k+1 \over 2^{A_k}} \qquad \qquad \text{where }  A_k =v_2(3a_k+1) \tag 1$$
Put this as functional notation $$ a_2=T(a_1;[A_1]) \tag 2$$ and the $N$-fold iteration with the sum $S$ of the $A_k$ then a general transformation can be written as
$$ a_{N+1} = T(a_1;[A_1,A_2,A_3,...,A_N]) \tag 3$$ with the restriction that all $A_k>0$ then the full transformation is expanded
$$ a_{N+1} = {3^N \over 2^S} a_1 + { 3^{N-1} + 3^{N-2}2^{A_1}+ 3^{N-3}2^{A_1+A_2} + ... + 2^{A_1+A_2+...+A_{N-1}} \over 2^S} \tag {4.1}$$
or, introducing a short notation for the long numerator
$$a_{N+1}= {3^N \over 2^S} a_1 + {Q([A_1,A_2,...,A_N])\over 2^S} \tag {4.2}$$
The the question of the existence of a cycle $a_{N+1}=a_1$ occurs as diophantine quation
$$a_1= {3^N \over 2^S} a_1 + {Q([A_1,A_2,...,A_N])\over 2^S} \\
 a_1 (2^S-3^N) = Q([A_1,A_2,...,A_N]) \\
 a_1  = {Q([A_1,A_2,...,A_N]) \over 2^S-3^N} \\  \tag 5
$$
for which a solution is sought for a positive odd integer $a_1$ (which is the "quotient" in your question) given a vector of exponents $[A_1,...,A_N]$. Current knowledge is that there is only one such solution which is $1 = T(1;[2])= T(1;[2,2])=...= T(1;[2,2,...,2])$ which is called the "trivial cycle".
To have a positive solution in $a_1$ we must have $2^S>3^N$ ; for the case $2^S<3^N$ there are $3$ further solution for negative odd $a_1$ (three cycles exist in the negative numbers).  See more on this for instnace at this link 
But that question in full generality is still open.       

The non-excistence of further positive quotients ($a_1$ in the odd positive numbers) is only known for a certain structure of the exponents-vector, namely of the form $E_{N,s}=[1,1,1,...,1,A_N]$ . Here  $A_N>1$ is such that $S=N-1+A_N$ and $2^S>3^N$ . With this restricion on the exponents there is no positive integer solution for $a_1$ (proof of R. Steiner of non-existence of 1-cycles) and subsequently no solution for $m$-cycles up to $m=72$ . $m$-cycles are defined by $m$ concatenations of that $E$-type vectors of different lengthes and/or trailing $A_N$. The proof were done by J. Simons/B.d.Weger in the years from 2000 to 2008 .

Update: A concise and nice remark about your quotient can be seen at Terence Tao's blog , look there for the paragraph containing the name "Sontacchi" and conjecture 3. 
