Difference between revisions of "Pure recurrence relation for partition function"
From specialfunctionswiki
(Created page with "==Theorem== The following formula holds: $$p(n)=\displaystyle\sum_{1 \leq \frac{3k^3 \pm k}{2} \leq n} (-1)^{k-1} p \left( n - \dfrac{3k^2 \pm k}{2} \right) = \dfrac{1}{n} \di...") |
(No difference)
|
Revision as of 00:06, 26 June 2016
Theorem
The following formula holds: $$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
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $24.2.1 \mathrm{II}.A.$
