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Let $n_1$, ..., $n_k$ denote positive integers, and let us write $$ n_i=\prod_{j=1}^m p_j^{\alpha_{ij}}$$ for $1\le i\le k$, where the $p_j$'s are distinct prime numbers, and $\alpha_{ij}\ge 0$ for all $i$ and $j$.

Consider a sum $$N=\sum_{i=1}^k \epsilon_i n_i$$ where $\epsilon_i=\pm 1$ for all $i$.

Can we bound the $p_j$ -adic valuation of $N$ in terms of $k$, $m$ and the $\alpha_{ij}$'s (or more precisely their logarithms)?

I am also interested in special cases, such as:

  • What happens with $m=2$? Or with $k=2$? Or both $k=m=2$?
  • Can we bound the $2$-adic valuation of integers of the form $1\pm 3^{\alpha_1}\pm\dots \pm 3^{\alpha_k}$, $0<\alpha_1<\dots<\alpha_k$?

[EDIT] In particular, a very special case of the latter can be proved (thanks to Gerry Myerson for his comment!):

The maximal $2$-adic valuation of $3^n-1$ is $2+\log(n)$.

Proof. As mentioned at oeis.org/A090740, one has $v_2(3^{2k+1}-1)=1$ for all $k$, and $v_2(3^{2k}-1)=v_2(3^k-1)+1+[\text{$k$ is odd}]$ where $[\text{$k$ is odd}]$ equals $1$ if $k$ is odd and $0$ otherwise. This implies that $v_2(3^n-1)=2+v_2(n)$, whence the bound. $\square$

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    $\begingroup$ The power of $2$ dividing $3^n-1$ is tabulated at oeis.org/A090740 and you will see it follows a simple pattern. $\endgroup$ Mar 7, 2014 at 10:40
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    $\begingroup$ Thanks Gerry! The OEIS gives a recurrence formula that shows that $v_2(3^n-1)=v_2(n)+2$, thus in particular $v_2(3^n-1)\le 2+\log(n)$. $\endgroup$
    – Bruno
    Mar 7, 2014 at 16:03

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