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

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==Theorem==
 
==Theorem==
 
The following formula holds:
 
The following formula holds:
$$p(n)=\displaystyle\sum_{1 \leq \frac{3k^3 \pm k}{2} \leq n} (-1)^{k-1} p \left( n - \dfrac{3k^2 \pm k}{2} \right) = \dfrac{1}{n} \displaystyle\sum_{k=1}^n \sigma_1(k) p(n-k), \quad p(0)=1$$
+
$$p(n)=\displaystyle\sum_{1 \leq \frac{3k^3 \pm k}{2} \leq n} (-1)^{k-1} p \left( n - \dfrac{3k^2 \pm k}{2} \right),$$
 
where $p(n)$ denotes the [[partition]] function and $\sigma_1$ denotes the [[sum of divisors]] function.
 
where $p(n)$ denotes the [[partition]] function and $\sigma_1$ denotes the [[sum of divisors]] function.
  
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==References==
 
==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Closed form for partition function with sinh|next=findme}}: $24.2.1 \mathrm{II}.A.$
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* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Closed form for partition function with sinh|next=Recurrence relation for partition function with sum of divisors}}: $24.2.1 \mathrm{II}.A.$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 20:41, 26 June 2016

Theorem

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

Proof

References