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Suppose that I have a number of the form

$$ x = \frac{1}{3^m}(2^{h} - \sum\limits_{k=1}^{m}3^{m-k}2^{v_k} ) $$

where m is a positive integer, and

$v_1 = 0 $

$v_{k+1} = v_k +1$ for $1 \leq k < m-1 $

$v_{m} > v_{m-1} $

For a fixed m, can we find a relationship between h and $v_m$, such that x is an integer ?

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For $k\leq m-1$, we have $v_k = k-1$. The expression for $x$ is then reduced to $$x=\frac{1}{3^m}\left( 2^h - \sum_{k=1}^{m-1} 3^{m-k}2^{k-1} - 2^{v_m}\right)=\frac{1}{3^m}\left( 2^h - 3^m + 2^m + 2^{m-1} - 2^{v_m} \right)$$ which is integer as soon as $$2^{v_m} \equiv 2^h + 2^m + 2^{m-1} \pmod{3^m}.$$ Taking discrete logarithm (if it exists) modulo $3^m$ will give a solution.

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