Pentagonal Number Theorem

From the previous problems, we have learnt that the Pentagonal Numbers are given by the formula $p_n = \frac{ 3n^2- n} { 2}$ . We have investigated the sequence for positive integers $n$ , which gives us the values:

$1, 5, 12, 22, 35, 51, \ldots$

What happens when we substitute in non-positive integers into this formula? We will get a different sequence, namely:

$0, 2, 7, 15, 26, 40, 57, \ldots$

Together, these numbers are called the Generalized Pentagonal Numbers. They appear in the following theorem:

The Pentagonal Number Theorem states that
$\prod_{n=1}^\infty ( 1 - x^n) = \sum_{k=-\infty}^{\infty} (-1)^k x^{p_k} = 1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} \ldots$

If we take the combinatorial interpretation of the identity, the LHS is the generating function for the number of partitions of $n$ into an even number of distinct parts, minus the number of partitions of $n$ into an odd number of distinct parts. By looking at the coefficients on the RHS, the surprising corollary is that this difference is either -1, 0 or 1 always!

The combinatorial proof of the Pentagonal Number Theorem follows a similar argument. It creates a bijection between the number of even parts, and the number of odd parts, and then accounts for the cases where the bijection fails, namely at the values $p_k = \frac{ 3k^2 - k } { 2 }$ . You can attempt to prove it for yourself, or read the Wikipedia article.

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Pentagonal Number Theorem

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