Let $P_k(m,n)$ be the number of partitions of the number $m$ having exactly $n$ parts, with each part $\ge k$. Solve $P_k(m,n) = P(x, n)$ for $x.$

Question

I'll try two special cases and see if they generalize.

Let $k = 2, m= 11, n =3.$

This is a problem of putting $11$ identical balls into $3$ identical boxes where each box gets at least two balls. We first put one ball in each box (done in one way), then we simply have to distribute $8$ identical balls to $3$ identical boxes. This implies $P_2(11, 3) = P(8, 3).$

Let $k = 3, m= 12, n = 4.$

This is a problem of putting $12$ identical balls into $4$ identical boxes where each box gets at least three balls. We first put two balls in each box (done in one way), then we simply have to distribute $6$ identical balls to $3$ identical boxes. This implies $P_3(12, 4) = P(6, 3).$

I think we can extrapolate the above to $P_k(m,n)=P(m-(k-1)n, n)$ and so $x = m-(k-1)n.$

Does the above make sense? Thanks.

Answer 1

Yes, it makes perfect sense, this works so nice because there are exactly $n$ summands.

If we worked with partitions $P(m)$ of unrestricted size we would get $P_k(m)=P_{k-1}(m)-P_{k-1}(m-(k-1))$, which is uglier because it is a $2$-dimensional recurrence.