# Summing over tuples $(b_1, \ldots, b_n)$ with $\sum ib_i = n$ and $b_j = k$.

Let $$T_n$$ denote the set of $$n$$-tuples $$\left(b_1, \ldots, b_n \right)$$ of non-negative integers such that $$\sum_{i=1}ib_i=n.$$ I am trying to simplify the sum

\begin{align*} \sum_{\underset{b_{j}=k}{\left(b_{i}\right)_{i\le n}\in T_{n}}}\frac{n\left(n-1\right)\cdots\left(n-k-\sum_{i\not=j}b_{i}+2\right)}{\prod_{l\not = j}\left(b_l !\right)}, \end{align*}

where the sum is over all tuples in $$T_n$$ such that the $$j$$th entry $$b_j$$ is equal to $$k$$.

Is there a way to enumerate the elements of $$T_n$$ satisfying $$b_j=k$$? Equivalently, is there a way to enumerate the $$n$$-tuples $$\left(b_1, \ldots, b_j=k, \ldots, b_n\right)$$ such that $$\sum_{i\not =j} ib_i = n-jk$$?

Background: This sum divided by $$k!$$ equals the number of non-crossing partitions on the set $$\lbrace 1, 2, \ldots, n\rbrace$$ with exactly $$k$$ blocks of size $$j$$.

I don't think you can do better than an expression involving the partition function $$p(n)$$, the number of integer partitions of $$n$$, which has some complicated exact formulas or (what is actually used) an infinite recurrence due to Euler.
This $$(b_1, \ldots, b_n)$$ is the "multiplicity representation" of an integer partition of $$n$$: there are $$b_1$$ 1s, $$b_2$$ 2s, etc. Specifying $$b_j = k$$ means the rest of the partition is a partition of $$n-jk$$ with no parts $$k$$. The number of these is $$p(n-jk) - p(n-jk-k)$$ by the following result.
The number of partitions of $$n$$ with at least one part $$m$$ (with $$m \le n$$) is $$p(n-m)$$. Therefore the number of partitions of $$n$$ with no parts $$m$$ is $$p(n) - p(n-m)$$.
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