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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.

Thanks!

--Connor

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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
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1 Answer 1

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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