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2 votes
0 answers
147 views

Could this yield a formula for the Partition numbers?

Background: Lately, I have fallen down the rabbit hole of partition numbers. Specifically the partition function, $p(n)$. It's well known that no closed-form expression (with only finitely many ...
Graviton's user avatar
  • 4,472
3 votes
0 answers
44 views

By which scheme should I add the elements in series $(\sum n^{-2})^2$ and $\sum n^{-4}$ to show their rational equivalence?

We know that $\sum n^{-2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots=\frac{\pi^2}{6}$ and $\sum n^{-4}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\dots=\frac{\pi^4}{90}$ from "high mathematics" ...
sanaris's user avatar
  • 253
4 votes
1 answer
430 views

How To Apply and Understand the Generating Function for Number Partitioning

The function p(n) counts the number of ways a number can be made up of smaller numbers. For example, the p(5) = 7 because you ...
Code Whisperer's user avatar
7 votes
1 answer
233 views

What is the significance of this identity relating to partitions?

I was watching a talk given by Prof. Richard Kenyon of Brown University, and I was confused by an equation briefly displayed at the bottom of one slide at 15:05 in the video. $$1 + x + x^3 + x^6 + \...
augurar's user avatar
  • 1,866
0 votes
0 answers
94 views

Truncation of partitions generating function question

$A (x)$ is the generating function for partitions. $B(x)=\sum_{n=0}^{\infty}b_nx^n $ $$b_n =\binom{\text{number of partitions of }n}{\text{into an even number of parts}}-\binom{\text{number of ...
Mathy Person's user avatar
  • 1,725
2 votes
1 answer
943 views

Proof that the series for the generating function of the partition function converges?

For $|q| < 1$, the generating function of the partition function $p(n)$ is given by $$ \sum_{n=0}^\infty p(n) q^n = \prod_{k=1}^\infty {1 \over 1-q^k}. \tag{1} $$ I have an intuitive ...
A l'Maeaux's user avatar