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 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 $p(k,n)$.

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 $p(k,n)$.