Difference between revisions of "Pure recurrence relation for partition function"

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(Created page with "==Theorem== The following formula holds: $$p(n)=\displaystyle\sum_{1 \leq \frac{3k^3 \pm k}{2} \leq n} (-1)^{k-1} p \left( n - \dfrac{3k^2 \pm k}{2} \right) = \dfrac{1}{n} \di...")
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Revision as of 00:06, 26 June 2016

Theorem

The following formula holds: $$p(n)=\displaystyle\sum_{1 \leq \frac{3k^3 \pm k}{2} \leq n} (-1)^{k-1} p \left( n - \dfrac{3k^2 \pm k}{2} \right) = \dfrac{1}{n} \displaystyle\sum_{k=1}^n \sigma_1(k) p(n-k), \quad p(0)=1$$ where $p(n)$ denotes the partition function and $\sigma_1$ denotes the sum of divisors function.

Proof

References