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In general, if $a(n)$ is an integer sequence with generating function $A(t)$ and $b(n)$ is an integer sequence with generating function $B(t)$, it is not easy to find the generating function $C(t)$ for $c(n)=a(n)b(n)$ in terms of $A$ and $B$, i.e., to find the Hadamard product of $A$ and $B$. However, it is not impossible to do so in some special cases.

I am interested in the case where $a(n) = \binom{N}{n}$ and $b(n) = \binom{M}{n}$, i.e., where the sequences are binomial coefficients. In this case $A(t) = (1+t)^N$ and $B(t) = (1+t)^M$. But what is $C$?

Thanks in advance for any advice or references!

  • 2
    [Wolfram|Alpha](https://www.wolframalpha.com/input/?i=sum+of+binom+%28N%2Cn%29+binom+%28M%2Cn%29+x^n+for+n%3D0..infinity) gives $_2F_1(-M,-N,;1;x)$; I suspect that if there were a way to express that more elementarily W|A would know about it.2012-09-29
  • 0
    Hmm ... thanks, that does seem like a reasonable conclusion.2012-09-29

0 Answers 0