Number of unordered factorizations into $k$ distinct parts

Asked Jul 12 '15 at 21:11

Active Jul 18 '15 at 16:48

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Let $H_d(n)$ denote the number of distinct ordered factorizations of $n$ and $H_d(n,k)$ the number of ordered factorizations of $n$ into $k$ distinct parts. We have the following recurrence: $$H_d(n,k+1) = k! \sum_{j=0}^k \frac{(-1)^j}{(k-j)!} \sum_{d,d|n} H_d(n/d, k-j),$$ where the inside sum is taken over all $d$ such that $d|n$ and for $d \geq 2$, $d$ is a ($j+1$)-st power. We then obtain $H_d(n)$ by summing over the $k$: $H_d(n) = \sum_{k=1}^n H_d(n,k)$.

Let $P_d(n,k)$ be the number of unordered factorizations into $k$ distinct parts. We know that $H_d(n) = \sum_{k=1}^n k! P_d(n,k)$ and it turns out that $H_d(n,k) = k! P_d(n,k)$ which lead to $$P_d(n,k) = \frac{H_d(n,k)}{k!}.$$

Since $H_d(n,k)$ contains a product by $k!$, I thought $P_d(n,k)$ could be simplified but I do not know how to proceed due to the use of $k$ in the recursive call.

Any help to get a nice formula for $P_d(n,k)$ would be appreciated!

edited Jul 18 '15 at 16:48

JMP

22,637

asked Jul 12 '15 at 21:11

lbeziaud

As a start, for a fixed prime $p$, can you get a formula for $P_d(p^r,k)$ as a function of $r$? – Greg Martin Jul 15 '15 at 17:19
@GregMartin $P_d(p^r,k)=Q(r,k)$ where $Q(n,k)$ denote the number of ways of partitioning $n$ into exactly $k$ distinct parts. The formula is given by the equation 19 here. – lbeziaud Jul 15 '15 at 19:02
This MSE link may prove useful reading. – Marko Riedel Jul 16 '15 at 20:18
@MarkoRiedel I am aware of this answer but thought that a "nicer" formula could be deduced from $H_d$. – lbeziaud Jul 17 '15 at 11:18

Number of unordered factorizations into $k$ distinct parts

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