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

  • 6
    $\begingroup$ 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
  • 6
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.