Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

Thanks!

share|improve this question
1  
On a side note: for the TeX: you just have to include the formulae between $'s. –  Pietro Majer Jan 15 '12 at 7:42
1  
Quick comment: your first series should start at $d=0$, and it is the Taylor series for $(1-4x)^{-1/2}$ about $x=0$. –  Alan Haynes Jan 15 '12 at 17:03
    
Thanks for pointing out the typo! –  user20592 Jan 16 '12 at 4:06
2  
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. –  Wadim Zudilin Jan 17 '12 at 2:40
add comment

3 Answers

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.

share|improve this answer
add comment

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

share|improve this answer
2  
Also, the series $\sum_{d\ge0} \frac{(3d)!}{(d!)^3}x^d$ has a bit simpler expression as hypergeometric function $_2F_1(1/3,2/3;1;27x)$. –  Pietro Majer Jan 15 '12 at 8:29
    
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!! –  user20592 Jan 15 '12 at 11:31
1  
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. –  Robert Israel Jan 15 '12 at 19:52
add comment

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.

share|improve this answer
1  
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 $f'(x) = \frac{g(x)-1}{3x}$ –  Robert Israel Jan 16 '12 at 17:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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