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Sorry for the repost and for my "bad" English. I made a lot of errors in the previous one, so here's my actual question:

Let's take a look at this sequence:

(1)

$[a_1,a_2,a_3,a_4,...,a_x]$

where

$$a_1=\frac{3 \cdot 2^{n_1}+2^{n_2}}{2^{n_3}-9}$$

$$a_2=\frac{9 \cdot 2^{n_1}+3 \cdot 2^{n_2}+2^{n_3}}{2^{n_4}-27}$$

$$a_x=\frac{3^{x}\cdot 2^{n_1}+3^{x-1}\cdot 2^{n_2}+3^{x-2}\cdot 2^{n_3}+ ... +3 \cdot 2^{n_{x}}+2^{n_{x+1}}}{2^{n_{x+2}}-3^{x+1}}$$

$n_1,n_2,n_3,...,n_x ∈ \mathbb N$

$n_1≤n_2≤n_3≤...≤n_x$

Is there a way to prove that every term of (1) is not a natural number?

Thanks.

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    $\begingroup$ something to do with the collatz conjecture? $\endgroup$ – snulty Jul 28 '15 at 18:27
  • $\begingroup$ yes, solving this would prove there's no other cycle than 4,2,1, but only this. $\endgroup$ – Dea5 Jul 28 '15 at 18:28

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