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Let $P(n,d)$ denote the number of partitions of $n$ with Durfee square of size $d$ (the largest square that fits inside the diagram of the partition).

The sequence $P(n,d)$ is known to satisfy the recursion

$P(n,d) = 2P(n-d,d) + P(n-2d+1,d-1) - P(n-2d,d)$.

How does one translate the above into a polynomial recursion for the generating function

$\displaystyle P_n(t)= \sum_{d=0}^{\lfloor \sqrt{n} \rfloor} P(n,d) t^d$ ?

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Define $\, f(t,x) := \sum_{n,d=0}^\infty P(n,d) t^d x^n = \sum_{n=0}^\infty P_n(t) x^n. \,$ The recursion directly translates into $\, f(t,x) = 2 f(t x,x) - (1 - t x) f(t x^2,x). \,$ The OEIS sequence A115994 is the sequence of $\,P(n,d)\,$ with $\,n,d>0.\,$ While OEIS sequence A115720 is the sequence of $\,P(n,d)\,$ with $\,n,d\ge0.\,$

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  • $\begingroup$ Thanks, but is it possible to express this in terms of $P_n(t)$ only, i.e. extracting the coefficient in front of x^n? $\endgroup$ – user353673 Aug 13 '18 at 13:45

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