Partitions into Parts

Question

Is it possible to express $p_m(n)$, the number of ways to write $n$ as a sum of $m$ parts, as some sum involving only $p(\cdot )$? I never seen such a formula before and was wondering whether this is doable. The reason for my interest in this question is that it will give an infinite series expression for $p_m(n)$ by applying the well-known series for $p(\cdot)$.

Answer 1

Some observations / too long for comments. I'm not an expert so lemme first confirm the terminology:

• $p_m(n) =$ no. of ways to write $n$ as a sum of $m$ positive integers

• $p(n) =$ no. of ways to write $n$ as a sum of any number of positive integers

• So we have $p(n) = \sum_{m=1}^n p_m(n)$

First of all, am I correct?

If so, take any $m$-term summation counted in $p_m(n)$, subtract $1$ from every term, and you have a way to write $n-m$ as a sum of $m$ non-negative integers, some of which can be zero. If $m \ge n-m$ then I think this is also the number of ways to write $n-m$ as a sum of any number of positive integers (since you need at most $n-m$ positive integers and the rest are just padded zeros). So

If $2m \ge n$, then $p_m(n) = p(n-m)$

OTOH if $2m < n$ then this trick only counts the number of ways to write $n-m$ as a sum of up to $m$ positive integers, so it is undercounting $p(n-m)$.

More generally, for any $m,n$ we have $p_m(n) = \sum_{k=1}^m p_k(n-m)$ and the RHS $=p(n-m)$ when $2m \ge n$.

Answer 2

To expand on my comment about this being possible but not useful:

The generating function for $p(n)$ is $$P(x) = \prod_{i\ge 1} (1-x^i)^{-1} = 1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + 11x^6 + 15x^7 + 22x^8 + \cdots$$

The generating function for $p_5(n)$ is $$P_5(x) = x^5 + x^6 + 2x^7 + 3x^8 + 5x^9 + 7x^{10} + 10x^{11} + 13x^{12} + 18x^{13} + \cdots$$

Therefore we find that $$P_5(x) = x^5 P(x) - x^{11}P(x) - x^{12}P(x) - x^{13} P(x) + \cdots$$

I haven't double-checked, but I think that what we're doing is extracting the g.f. of $x^5\prod_{i \ge 6}(1-x^i)^{-1}$.