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

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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.

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

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nice

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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...?

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Fantastic

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

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it would be always positive integer.

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Greaat :) It was very interesting. Thank you !

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great

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