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Added a missing term in the recurrence relation sum. The p(n) sum, as previously described, does not yield the series (p(n-1) + p(n-2) - p(n-5) - p(n-7)) that is below it. Also made changes that improve readability. Tag: Reverted |
Undid revision 1226346618 by 60.248.110.152 (talk) perhaps you didn't notice that the sum is over NEGATIVE as well as positive indices |
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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-1) + p(n-2)-p(n-5)-p(n-7) +p(n-12) +p(n-15) - p(n-22) -\cdots
\end{align}</math>
As base cases, <math>p(0)</math> is taken to equal <math>1</math>, and <math>p(k)</math> is taken to be zero for negative <math>k</math>. Although the sum on the right side appears infinite, it has only finitely many nonzero terms,
coming from the nonzero values of <math>k</math> in the range
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