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Undid revision 1239027743 by 72.197.182.30 (talk) no, the sum is over positive and negative integers |
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<!-- Note: The following is the same formula as in the source, but in a more compact form. See [[Talk:Partition function (number theory)#Recurrence relations]]. -->
<math display="block">\begin{align}
p(n) &= \sum_{k \in \Z\setminus\{0\}} (-1)^{k+1} (p(n-k(3k-1)/2) + p(n-k(3k+1)/2)) \\
&= p(n-1) + p(n-2)-p(n-5)-p(n-7) +p(n-12) +p(n-15) - p(n-22) -\cdots
\end{align}</math>
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