<s>[INST] Numbers of the form $5 \cdot 2^{n}-1$ divisible by $3^k$ for large values of $k$ Let $n_k$ be the smallest integer such that $5 \cdot 2^{n_k}-1$ is divisble by $3^k$ where $k$ is a positive integer. Can one say something about the growth of $n_k$ with respect to $k$ ? Is it exponential ?
The first values are $n_k=1,1,13,31,139,\ldots$ for $k=1,2,3,4,5,\ldots$.
 [/INST] Here is some progress I made ...
Looking closely at the values given by Lucian, we remark that $$n_k - n_{k-1} = 2 \cdot a_{k} \cdot 3^{k-2} \text{ where } a_{k}=0,1,2 \text{ for } k>1.$$
Example: $n_6 - n_5= 324 = 4 \cdot 3^4$.
My sketch of proof below:
First, by induction, we have
$$ 2^{2 \cdot 3^{k-2}} \equiv 1 + 3^{k-1} \pmod{3^k}.$$
Then, assuming that $$5 \cdot 2^{n_{k-1}} \not\equiv 1 \pmod{3^k}$$ we can write
$$5 \cdot 2^{n_{k-1}} \equiv 1 + i \cdot 3^{k-1} \pmod{3^k} \text{ where } i=1,2,$$ and we obtain 
$$5 \cdot 2^{n_{k-1} + 2(3-i)3^{k-2}} \equiv (1 + i \cdot 3^{k-1})(1 + 3^{k-1})^{3-i} \equiv 1 \pmod{3^k}.$$
So we can take $a_{k} = 3-i$.
Finally we take $a_k = 0$ in the simple case $5 \cdot 2^{n_{k-1}} \equiv 1 \pmod{3^k}$.
$\square$
Thus we have the general formula
$$n_k = 1 + 2\sum_{j=2}^k a_{j} \cdot 3^{j-2}.$$
The sequence of $a_k$ values ($k\geq2$) is $$0, 2, 1, 2, 2, 0, 1, 2, 2, 0, 1, 1, 0, 0, 2, 0, 1, 0, 2, \ldots$$
and $n_{17} = n_{16} = n_{15} + 4 \cdot 3^{14} = 19641181$.
An interesting result is:
$$\text{if }  a_k \neq 0 \text{ then } 2 \cdot 3^{k-2} < n_k < 2 \cdot 3^{k-1}.$$
But the problem remains if $a_k=0$ for many consecutive $k$ integers. And it is difficult to say much more, since $a_k$ sequence appears to be pseudorandom. Any idea ?
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