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I am familiar with the partition function p(k,n) where p is the number of partitions of n using only natural numbers at least as large as k. Is there a way of determining if p(k1, n1) > p(k2, n2) that does not actually use the partition function? To clarify, I want to know if there is a quick way to tell if the number of partitions of k1, n1 is greater than or less than the number of partitions of k2, n2 without using the partition function.



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"at least as large as k" should read "no bigger than k", right? – Jonah Ostroff Dec 7 '09 at 14:01

No, he probably means exactly what he said. That is the way the partition function is usually defined. But either way, the answer is no.

If $q(k,n)$ counts partitions of n into integers no bigger than k, as Jonah suggests, then note that $q(2,2m) = m+1$ for every $m$. (A partition is determined by the number of 2's.) So being able to compare values of $q(k,n)$ would in particular entail being able to compare $q(k,n)$ to any given integer.

As for the question as actually asked, note that $p(2k,4k-1)=k+1$ for every $k$. Once again, knowing the relative sizes of all $p(k,n)$ is tantamount to knowing whether $p(k,n)$ is more or less than each integer, i.e. knowing the values of $p(k,n)$.

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You have a broken sentence (and a parenthesis that doesn't close): "(A partition is determined by So being able to" ... – Scott Morrison Dec 14 '09 at 5:41
Thanks Scott, fixed now. – Cap Khoury Dec 27 '09 at 22:24

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