In 3-adic valuation of a sum involving binomial coefficients the $p$-adic valuation of $P_n(p)$ has been obtained.
Computations indicate that $M_n(p)=\sum\binom{n}{k}^2(p-1)^k$ for odd $p$ has the same $p$-adic valuation: $$\nu_p (P_n(p)) = \nu_p (M_n(p)).$$

Is this true for all odd $p$?

(Note that $P_n(3)=M _n(3)$).

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    $\begingroup$ I am surprised that this has attracted a vote to close. Would whoever did so like to explain why? $\endgroup$ – Yemon Choi May 7 '17 at 15:21
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    $\begingroup$ This surprises me as well. Moderators should intervene often because I've seen such "unjustified" closes happening quite a bit. $\endgroup$ – T. Amdeberhan May 7 '17 at 15:53

This is a suggestion, in case it helps.

The two sequences satisfy the following recurrences, respectively, \begin{align} nP_n(p)&=p(2n-1)P_{n-1}(p)-(n-1)P_{n-2}(p) \\ nM_n(p)&=p(2n-1)M_{n-1}(p)-(n-1)M_{n-2}(p)\cdot(p-2)^2. \end{align}

  • $\begingroup$ With the method you gave for $p=3$ it is easy to obtain $\nu_p(M_{2n}(p))=\nu_p(P_{2n}(p))=\nu_p(\binom{2n}{n}).$ But for most odd $n$ the valuations of the terms of the right-hand side are equal. $\endgroup$ – Johann Cigler May 8 '17 at 14:19

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