I need to calculate in how many ways can I add the integers in the interval $[1,9]$ in groups of $P$ elements, so that they sum to $N$. The set of groups must not contain sums that have the same elements (in the sense that 1+1+3, 1+3+1, and 3+1+1 would all count as 1 variation).
We may call this a specific partition Q, so I'm trying to figure out:
$$Q_p(n) = ?$$
$Q$ can be defined as $0$ if $\frac{n}{p}<1$ or $\frac{n}{p}>9$, since we cannot generate a feasible solution.
For instance $Q_2(8) = 4$, because we can write 8 as the following sums: $(1+7, 2+6, 3+5, 4+4)$
The solution of $Q_3(8) = 5$ and the sums are: $(1+1+6,1+2+5,1+3+4,2+2+4,2+3+3)$
I've calculated programmatically the values up to 9, but I can't seem to figure out a pattern.
1
1 1
1 1 1
1 2 1 1
1 2 2 1 1
1 3 3 2 1 1
1 3 4 3 2 1 1
1 4 5 5 3 2 1 1
1 4 7 6 5 3 2 1 1
It is clear for me that:
$Q_1(N)=1$
$Q_2(N) = \lfloor N/2 \rfloor$ (only for N < 19, thanks Mike)
$Q_{N-1}(N)=1$
$Q_N(N)=1$
But I can't figure out either a generic way to represent all of them, nor even the central diagonals. Is this a known problem? Any help or hints highly appreciated
EDIT: this seems like a variation of the restricted part size or number of parts that can be solved recursively as (but will fail for N>10 because of the restriction interval): $$Q_p(n) = Q_p(n-p) + Q_{p-1}(n-1)$$
Letting $Q^m_p(n)$ be the number of partitions of $n$ with $p$ parts whose sizes are between $1$ and $m$, you want $Q^9_p(n)$. This can be computed via $$ Q^m_p(n) = Q^{m-1}_{p}(n-p)+Q^m_{p-1}(n-1). $$ This is a known problem. It is related to the Gaussian binomial coefficients. Specifically, $$ \sum_{n=p}^{mp}Q^{m}_{p}(n)q^n=q^p\binom{m+p-1}{p}_q. $$