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The large-parts formula for p(n)

Published 7 Feb 2010 in math.CO | (1002.1458v1)

Abstract: A new formula for the partition function p(n)p(n) is developed. We show that the number of partitions of nn can be expressed as the sum of a simple function of the two largest parts of all partitions. Specifically, if $a_1 + >... + a_k = n$ is a partition of nn with a1≤...≤aka_1 \leq ... \leq a_k and a0=0a_0 = 0, then the sum of ⌊(ak+ak−1)/(ak−1+1)⌋\lfloor(a_k + a_{k-1}) / (a_{k-1} + 1)\rfloor over all partitions of nn is equal to $2p(n) - 1$.

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