Collatz Cycles with double the even numbers than odd This question is with reference to the collatz conjecture. It is known that a number $A$ in a collatz cycle is of the form: 
$$A = \sum_{i=1}^k \frac{ 3^{k-i} \cdot 2^{k_i}}{2^n - 3^k},$$
$0 = k_1< k_2 < k_2 ...k_k < n$.
Here, $n$ represents the number of divisions by $2$ in the cycle and $k$ represents the number of odd numbers in the cycle. For the trivial cycle $n = 2k$.
My question is this: has it been shown that there are no non-trivial cycles with $n = 2k$? If so, could i get a reference? 
 A: A cycle of $a_i$ for $k$ odd steps and $n$ even steps must satisfy a product formula
$$   2^n  = (3+{1\over  a_1})(3+{1\over a_2})\cdots (3+{1\over a_k})
$$
Because the rhs is larger than $3^k$ we must have that $n \ge \lceil k \cdot \ln_2(3) \rceil \approx \lceil 1.5 k \rceil $
If we want $n=2k$ then we have
$$   2^{2k} = (3+{1\over  a_1})(3+{1\over  a_2})\cdots (3+{1\over a_k}) \\
  4^{k} = \underset{ k \text{ - parentheses}}{\underbrace{(3+{1\over  a_1})(3+{1\over  a_2})\cdots (3+{1\over a_k})}} \\$$
and from this, since no parenthese on the rhs can be larger than $4$, no other one can be smaller than $4$ and thus all of them must equal $4$ and finally all $a_i=1$ is required.        
So this all defines the "trivial cycle" $1 \to 1 \to \cdots $ of length $k$ and no other solutions are possible.
 
I have a bit of discussion in my text on my homepage which might interest you although this is no (serious) "reference". (Unfortunately I do not know any reference which discusses this detail explicitely but of course the product formula and some consequences of it are well known at least since R.Crandall1 in the 70'ies)
1 Crandall, R. E., On the ”3x+1” problem, Math. Comput. 32, 1281-1292 (1978). ZBL0395.10013.
