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This comes purely out of curiosity and experiments. I'm not sure if the literature has any coverage.

Let $p(n)$ be the number of integer partitions of $n$. Then, we have the well-known generating function $$\sum_{n\geq0}p(n)x^n=\prod_{k=1}^{\infty}\frac1{1-x^k}.$$

Question. For each fixed $k\in\mathbb{N}$, is the following set finite? $$\mathcal{A}_k:=\{(n,m)\in\mathbb{Z}_{\geq0}^2: p(n)+k=m^2\}.$$

Update. Is it even known that $p(n)$ is ever a perfect square beside $n=0, 1$?

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    $\begingroup$ What if $k=0$? Is the answer clear in this case? $\endgroup$
    – Joël
    Commented Feb 25, 2017 at 18:00
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    $\begingroup$ Is $p(n)$ ever a square, for $n\ge2$? $\endgroup$
    – Gerry Myerson
    Commented Feb 25, 2017 at 22:04
  • $\begingroup$ I don't think so, but it's unclear to me if such is known. $\endgroup$
    – T. Amdeberhan
    Commented Feb 25, 2017 at 22:07
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    $\begingroup$ $p(n)$ is the sequence of coefficients of a weakly holomorphic modular form of weight $-1/2$. If you consider instead the question for the sequence of coefficients of a holomorphic modular form of integral weight say $\geq 4$, you get another question of which I don't know what to expect. $\endgroup$
    – Joël
    Commented Feb 25, 2017 at 22:49
  • $\begingroup$ @Gerry Myerson Probably $p(n)$ can't be any perfect power. $\endgroup$
    – Alexey Ustinov
    Commented Feb 26, 2017 at 8:52

1 Answer 1

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Numerical results give random picture. If it is really the case then usual heuristic arguments confirm your conjecture.

ListPlot[Table[FractionalPart[Sqrt[PartitionsP[i]]], {i, 1, 5000}]]

enter image description here

Distance $0$ number is 1: $p(1)=1^2$. Distance 1 numbers: \begin{gather}p(2)=1^2+1,\quad p(3)=2^2-1,\quad p(4)=2^2+1,\\p(7)=4^2-1,\quad p(13)=10^2+1,\quad p(35)=122^2-1.\end{gather} Distance 2 numbers: $$p(5)=3^2-2,\quad p(6)=3^2+2,\quad p(20)=25^2+2.$$

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    $\begingroup$ Is it clear that these are the only ones? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 26, 2017 at 9:06
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    $\begingroup$ @მამუკა ჯიბლაძე I've checked $p(n)$ for $n\le 10\,000$. So it is only a reasonable conjecture. $\endgroup$
    – Alexey Ustinov
    Commented Feb 26, 2017 at 9:42

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