In cycles you can also use 2^k = \prod\limits_{i=0}^{x-1} (3+ \cfrac{1}{a_i})\leqslant (3+ \cfrac{1}{a_{min}})^x\leqslant 4^x=2^{2x} (for a_i\geqslant a_{min}\geqslant1). You can even skip the a_{min} part and just use a_i\geqslant1
From there you also have 2^k \leqslant (3+ \cfrac{1}{a_{min}})^x \rightarrow a_{min}\leqslant \cfrac{1}{2^{k/x}-3}