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
    $\begingroup$ Wolfram|Alpha 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. $\endgroup$ – joriki Sep 29 '12 at 7:46
  • $\begingroup$ Hmm ... thanks, that does seem like a reasonable conclusion. $\endgroup$ – N F Sep 29 '12 at 19:47
  • $\begingroup$ If your still interested: I believe I can answer this using section 1.4 of GouldBK.pdf by Sprugnoli. It used to be available freely but now you have to go through researchgate.net/publication/… . Although I think "fair use" means I could send you a copy. OTOH: I can just write out the solution in "Your Answer". The method falls under "Riordan Array transformations" if you want to search. $\endgroup$ – rrogers Mar 30 at 18:47

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