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If we denote the Bernoulli numbers by $B_n$, then $$ {a^{k+1}(a^{2k} - 1) B_{2k} \over 2k}\in \mathbb{Z}$$ for any $a\in \mathbb{Z}$ and $k\in\mathbb{N}$. This fact is often called the Sylvester–Lipschitz theorem, although I think it was known earlier; see for example A Note on Bernoulli Numbers by I. Sh. Slavutskii, J. Number Theory 53 (1995), 309–310. If $a=2$ then (up to a factor of $\pm 2^{k-1}$) there is a combinatorial interpretation of the above formula in terms of alternating permutations.

Is there a combinatorial interpretation for other values of $a$?

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    $\begingroup$ Is there an easy combinatorial explanation why $A_{2k-1}$, the number of alternating permutations of length $2k-1$, is divisible by $2^{k-1}$? $\endgroup$
    – Sam Hopkins
    Commented Oct 20, 2023 at 1:54
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    $\begingroup$ @SamHopkins, if $x$ and $x+1$ are not adjacent, they can be swapped to get another alternating permutation. $x$ can't be adjacent to both $x-1$ and $x+1$ because then you have two consecutive ascents or two consecutive descents. There's a bit of work to be done to show that double-counting doesn't occur, but intuitively this feels like the explanation. $\endgroup$
    – Peter Taylor
    Commented Oct 20, 2023 at 9:30
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    $\begingroup$ For $a=3$ and $1\le k\le10$ and further dividing by 6 we get $1, 3, 39, 1107, 54351, 4085883, 435847959, 62594829027, 11644113200031, 2723549731505163$. Those appear in oeis.org/A080635 at odd positions, i.e. as the "number of permutations on $2k-1$ letters without double falls and without initial falls". $\endgroup$
    – Claude Chaunier
    Commented Oct 23, 2023 at 12:01
  • $\begingroup$ I forgot to mention I also removed the $(-1)^{k-1}$ factor. $\endgroup$
    – Claude Chaunier
    Commented Oct 23, 2023 at 13:22

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