5
$\begingroup$

Does anyone know how to estimate the sums of $p$-powers of $s$-multinomial coefficients, with $p,s\in\mathbb N$ arbitrary? Here's some data that I gathered from OEIS, $$\sum_{a_1+\ldots+a_s=n}\binom{n}{a_1,\ldots,a_s}^p \simeq s^{pn}\sqrt{\frac{K_{sp}}{(\pi n)^{(s-1)(p-1)}}}$$ where $K_{1p}=K_{s1}=1$, $K_{2p}=2^{p-1}/p$, $K_{3p}=1,\frac{27}{16},\frac{81}{16},16,\ldots$ and $K_{s2}=1,1,\frac{27}{16},4,\ldots$

My questions are: (1) is the above formula correct? (2) what's $K_{sp}$, in general?

Many thanks.

Improve this question
$\endgroup$
6
  • $\begingroup$ Have you tried doing the straightforward thing, using Stirling's formula and considering the contribution of the largest terms? $\endgroup$
    – Qiaochu Yuan
    Commented Nov 5, 2013 at 23:33
  • $\begingroup$ There's no such thing as a largest term, unfortunately (try $s=2,p=1$..) they kind of all contribute, it's an analysis problem. Btw (related somehow to your comment) due to "big terms" for small $n$ the convergence is very slow, and part of the above $K_{sp}$ data (that I obtained with $n\leq 50$ or so), is not 100% reliable. $\endgroup$
    – math_guest
    Commented Nov 5, 2013 at 23:45
  • $\begingroup$ Well, it depends on what asymptotic regime you're interested in I guess. I was thinking about $p$ large or even $p \to \infty$, but maybe you want $s, p$ fixed and $n \to \infty$? If so it would be good to indicate this in the problem statement. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 5, 2013 at 23:54
  • $\begingroup$ Okay, in fact I need $p=$ prime, fixed, and $s=q/p$ with $q=p^r$, fixed too, and $n\to\infty$. The above sums count the solutions of $\lambda_1+\ldots+\lambda_{pn}=0$ with $\lambda_i$ roots of unity of order $q$, and I'm interested in estimating the number of these sols, with $n\to\infty$. Quite unobvious. $\endgroup$
    – math_guest
    Commented Nov 5, 2013 at 23:58
  • 1
    $\begingroup$ The sum in question is $n!^p$ times the coefficient of $x^n$ in $$\biggl(\sum_{j=0}^\infty \frac{x^j}{j!^p}\biggr)^s,$$ so the techniques of Section VII of Flajolet and Sedgewick's Analytic Combinatorics (algo.inria.fr/flajolet/Publications/book.pdf) might be helpful. $\endgroup$
    – Ira Gessel
    Commented Nov 8, 2013 at 20:05

1 Answer 1

Reset to default
2
$\begingroup$

$$K_{sp}=\frac{s^{s(p-1)}}{p^{s-1}2^{(p-1)(s-1)}}$$

Improve this answer
$\endgroup$
3
  • $\begingroup$ Misleading was the formula of K_{34}, that's 19.22 instead of 16. $\endgroup$
    – math_guest
    Commented Nov 14, 2013 at 18:40
  • $\begingroup$ I think it is an answer. It would be nice if something about its derivation were included. Gerhard "Unless It Is Utterly Obvious" Paseman, 2013.11.14 $\endgroup$
    – Gerhard Paseman
    Commented Nov 14, 2013 at 19:18
  • $\begingroup$ Holy crap, I forgot about the proof, indeed. That's done in arxiv.org/abs/0807.5028 at p=2, same proof at any p. $\endgroup$
    – math_guest
    Commented Nov 14, 2013 at 19:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .