# 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.

Note by Calvin Lin
4 years, 8 months ago

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nice

- 4 years, 8 months ago

Does it work for n-gon as well? I haven't tried to generalize it yet........

- 4 years, 8 months ago

Does what work for n-gon?

If you're referring to finding the formula for the number of dots in Figure K of an N-gon diagram, then yes you can use similar approaches. The easiest will be to find the Arithmetic Progression, and then find the sum of that. I wanted to show a different approach, which uses the idea of Mathematical Induction, or Method of Differences.

Staff - 4 years, 8 months ago

I meant the number of dots only. Thanks, sir.

- 4 years, 8 months ago

Why don't you investigate and then write up a note? I'd be interested in your results.

Staff - 4 years, 8 months ago

great

- 4 years, 8 months ago

Greaat :) It was very interesting. Thank you !

- 4 years, 8 months ago

it would be always positive integer.

- 4 years, 8 months ago

i will continue my thinking unless i will got it. Thanks

- 4 years, 7 months ago

Fantastic

- 4 years, 6 months ago

Nice one sir! Can I suggest something? Knowing 1st pentagonal number is one, we can see that number of dots to be added to form the next pentagonal number are in A.P. - where a=4, and common difference d=3.... Thus can't we express pentagonal numbers as sums of the increasing A.P. 1,4,7,10; i.e. S1, S2, S3 and so on, where S(n) is the sum to 'n' terms of the A.P. 1,4,7,10...?

- 4 years, 3 months ago