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Let $$ a_{n}=\dfrac{3n-1}{2n-1}. $$

Show that any product of at least $2$ terms of this sequence is not a power of $2$.

In other words:

The equation $$ a_{n_1} a_{n_2} \cdots a_{n_k} = 2^a $$ has no solutions where $a, n_1, n_2, \dots, n_k$ are positive natural numbers and $k \geq 2$.

I know $$ \dfrac{3n-1}{2n-1}\equiv \dfrac{n-1}{-1}\equiv n-1\pmod 2 . $$

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closed as unclear what you're asking by Ivan Neretin, Shailesh, Namaste, Parcly Taxel, user223391 Oct 10 '16 at 22:31

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  • $\begingroup$ What does "show that this sequence any product of a first term not powers of $2$" mean? $\endgroup$ – lulu Oct 8 '16 at 14:55
  • $\begingroup$ @lulu,Now can you understand? $\endgroup$ – function sug Oct 8 '16 at 15:02
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    $\begingroup$ I think so, though $a_1=2$ is a counterexample. Are you just excluding that one? $\endgroup$ – lulu Oct 8 '16 at 15:05
  • $\begingroup$ As a quick remark, no subscript $n$ with $n\equiv 2 \pmod 3$ can occur. That's because, for such $n$, $2n-1\equiv 0\pmod 3$ so the denominator is divisible by $3$, and it is clearly not possible for the numerator to ever cancel that. $\endgroup$ – lulu Oct 8 '16 at 15:10