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It is well known that $$Z(x,q) = \prod_{n=1}^\infty\frac{1}{(1-xq^n)} = \sum_{m=0}^\infty\sum_{k=0}^m p_{m,k}x^kq^m$$ is the generating function for the number $p_{m,k}$ of partitions of $m$ in precisely $k$ parts (or equivalently of partitions of $m$ with maximal part exactly $k$). It is also well known that the numbers $p_{m,k}$ satisfy the recurrence relation $$p_{m.k}=p_{m-1,k-1} + p_{m-k,k}$$ where, of course, $p_{m,m+k}=0$ for all $k>0$.

I am interested in the numbers $$Z(x,q)Z(1/x,q) = \sum_{m=0}^\infty\sum_{k=-m}^m c_{m,k}x^kq^m\,.$$ If we say that $p_{-m,k}=0$ for all $m>0$, $p_{m,-k}=0$ for all $k>0$ and $p_{0,k}=\delta_{k,0}$, then we have $$c_{m,k}=\sum_{m'=0}^\infty\sum_{k'=0}^\infty p_{m',k'}p_{m-m',k+k'}\,.$$ Furthermore, we have the recurrence relation $$ c_{m,k}=c_{m-1,k-1}+c_{m-k,k}-c_{m-k-1,k+1} $$ for all $0<k\leq m$. As the definition of the $c_{m,k}$ is symmetric in $k$, this extends in the obvious way to all $-m\leq k<0$. However, I cannot find a recurrence relation for the case $k=0$, i.e. for $c_{m,0}$. Does anyone know such a recurrence relation?

The numbers $c_{m,0}$ are a known sequence in OEIS, but so far the information provided did not help me to come up with a recurrence relation.

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    $\begingroup$ Have you tried rearranging the Jacobi Triple product identity? Two of the terms in it are basically the inverses of $Z(x,q) Z(1/x,q)$. $\endgroup$ Aug 26, 2013 at 5:22
  • $\begingroup$ Thanks, @BenjaminYoung, I tried that. I am not an expert in this, but as far as I can see, it is precisely the fact that I would have to invert the power series, which does prevent (at least) me to get a useful result. The numbers $c_{m,0}$ are listed in the OEIS, but so far, that did not help me either. link $\endgroup$
    – Nithilher
    Aug 29, 2013 at 7:02

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It is not the recurrence that you are looking for but may be helpful.

The generating function for your $c_{m0}$ is $$ f(q):=\sum_{n=0}^\infty\left(\frac{q^n}{(1-q)(1-q^2)\cdots(1-q^n)}\right)^2; $$ now using the Heine transformation one has $$ f(q)=(1-2q+2q^3-2q^6+...+2(-1)^kq^{\frac{k(k+1)}2}+...)\prod_{n=1}^\infty\frac1{(1-q^n)^2}, $$ hence $$ (1-q-q^2+q^5+q^7-q^{12}-q^{15}+q^{22}+...)f(q)=(1-2q+2q^3-2q^6+2q^{10}-2q^{15}+...)\sum_{n=0}^\infty p(n)q^n $$ This gives the equalities $$ c_{m,0}-c_{m-1,0}-c_{m-2,0}+c_{m-5,0}+c_{m-7,0}-c_{m-12,0}-c_{m-15,0}+c_{m-22,0}+\dots=\ p(m)-2p(m-1)+2p(m-3)-2p(m-6)+2p(m-10)-2p(m-15)+... $$

Let me add that the right hand side is OEIS A001522

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