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According to the pentagonal number theorem:

$\prod_{n=1}^{\infty} (1-q^{n}) = \sum_{k=-\infty}^{\infty} (-1)^{k}q^{k(3k-1)/2}$

Now the reciprocal of this has the partition numbers $p(k)$ in its power series representation:

$\prod_{n=1}^{\infty} \frac{1}{1-q^{n}} = \sum_{k=0}^{\infty} p(k)q^{k}$

What strikes me about this pair is that there is information about the partitions encoded in the generalized pentagonal numbers.

  1. Are there other pairs of $q$ products and reciprocals for which there is a lot of curiosity about the reciprocal because some number-theoretic information is encoded in the coefficients of the power series representation, and the non-reciprocated power series representation has fairly straightforward exponents?

  2. If the answer to 1 is yes, then also: how do those exponents relate to the information being encoded by the coefficients of the power series representation of the reciprocal?

  • 2
    Despite the fact that you allude to that generating functions and power series have a combinatorial interpretation -and- there is some relevant category theory there, if you say 'power series' or 'generating function', category theory is nowhere near the main thing that comes to mind in any sense (and vice versa) so it is very misleading. Stick with just number theory.2011-04-29

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The number of partitions of $n$ into distinct parts is equal to the number of partitions of $n$ into odd parts:

$\sum_{n=0}^\infty p_d(n)q^n = (-q;q)_\infty = \frac{(q^2;q^2)_\infty}{(q;q)_\infty} = \frac{1}{(q;q^2)_\infty} = \sum_{n=0}p_o(n)q^n.$

You can produce many like this, and it is good practice to prove them.


Most incredible and beautiful is Ramanujan's Identity: $\sum_{n=0}^\infty p(5n+4) q^n = 5\frac{(q^5;q)^5}{(q;q)^6}.$ This proves that $p(5n+4) \equiv 0 \pmod 5$!