6
$\begingroup$

We learn in calculus how to obtain a sum of binomial coefficients $\frac{(2d)!}{(d!)^2}$ in terms of a generating function

$\sum_{d \geq 0} \frac{(2d)!}{(d!)^2} x^d$

by the Taylor series of $(1-4x)^{-1/2}$ at $x=0$.

My question: is there a way to do this for trinomial coefficients? In particular what is

$\sum_{d \geq 1} \frac{(3d-1)!}{(d!)^3} x^d =?$

I can't imagine this not being studied before, but can not find a specific answer after a few futile hours of searching.

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ Your function naturally occurs in Mahler measure evaluations, see $n(\alpha)$ and $g(\alpha)$ in arxiv.org/abs/1012.3036v3, as well as many nontrivial integral expressions for $g$ in Section 4 there. $\endgroup$
    – Wadim Zudilin
    Commented Jan 17, 2012 at 2:40

3 Answers 3

Reset to default
6
$\begingroup$

What is "a way"? Of course, your question (in even more general form) was asked centuries ago and gave rise to hypergeometric series, series of the form $\sum_n c_n$ with ratio $c_{n+1}/c_n$ being a rational function of index $n$. The most convenient form is therefore the hypergeometric $_4F_3$ series expression given in Robert's answer. Note that the hypergeometric function is not algebraic (i.e., transcendental) in this case, so that no formula like $(1-4z)^{-1/2}$ can be given; all algebraic hypergeometric instances are now tabulated thanks to a fantastic result of Beukers and Heckman.

Dyson's famous 1962 paper Statistical theory of the energy levels of complex systems originated the study of constant term identities. In particular, Dyson's ex-conjecture states that for $a_1,\dots,a_n$ nonnegative integers $$ \text{constant term} \prod_{1\leq i\neq j\leq n} \biggl(1-\frac{x_i}{x_j}\biggr)^{a_i} =\frac{(a_1+a_2+\cdots+a_n)!}{a_1!a_2!\cdots a_n!}\,. $$ This could serve a different basis for other type generating functions.

Improve this answer
$\endgroup$
2
$\begingroup$

Maple writes this as $2\ x\ {}_4F_3(1,1,4/3,5/3;\ 2,2,2;\ 27 x)$.

Improve this answer
$\endgroup$
3
  • 3
    $\begingroup$ Also, the series $\sum_{d\ge0} \frac{(3d)!}{(d!)^3}x^d$ has a bit simpler expression as hypergeometric function $_2F_1(1/3,2/3;1;27x)$. $\endgroup$
    – Pietro Majer
    Commented Jan 15, 2012 at 8:29
  • $\begingroup$ Thanks to both Mr. Israel and Majer for the prompt help. I really appreciate it. Actually I got this summation from a solution of a specific hypergeometric equation, so I knew it is a hypergeometric function. What I don't know is whether it can be written in terms of other "known" analytic functions. In fact, what I am most concerned is its analytic property. For example, it seems that this hypergeometric function, is log of an analytic function. However, browsing through Ramanujan's notebooks did not find anything relevant. I should have made it clearer in the post. Thanks anyway!! $\endgroup$
    – user20592
    Commented Jan 15, 2012 at 11:31
  • 1
    $\begingroup$ According to Maple, $\sum_{d \ge 0} \frac{(3d)!}{(d!)^3} x^d$ can also be written, for example, as $LegendreP(-1/3,1-54 x)$, which is a Legendre function of the first kind. $\endgroup$
    – Robert Israel
    Commented Jan 15, 2012 at 19:52
2
$\begingroup$

For what it is worth (probably very little!), $$\sum_{d\ge 0}\binom{3d}{d,d,d} x^d$$ is the coefficient of $y^0z^0$ in $$\frac{1}{1-(yz+1/y+1/z)^3x},$$ and the original series is $\frac13$ of the derivative.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ You mean, if this is $g(x)$ and the original series $f(x) = \sum_{d \ge 1} \frac{(3d-1)!}{(d!)^3} x^d$, that $f'(x) = \frac{g(x)-1}{3x}$ $\endgroup$
    – Robert Israel
    Commented Jan 16, 2012 at 17:52

You must log in to answer this question.

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