The complete question I would like to answer is:
Given positive integers $k,n$, how many descending lists of non-negative integers $(x_1~x_2\ldots x_k)$ are there such that $\sum_{i=1}^k x_i = n$?
As a quick first insight, it's clear that the general problem will come if we can find the solution for $k=n$. It would take a long time to explain the buildup, but I have reduced the problem to calculating the value $f(n,n)$ of a function recursively defined by: $ f(n,m) = f(n,m-1) + f(n-m,\min(m,n-m))~, $ $ f(n,2) = 1+\left\lfloor\frac{n}{2}\right\rfloor~~;~f(n,1) = f(n,0) = 1~. $ For a less clunky reformulation, posit that $f(a,b) = f(a,\min(a,b))$ for each pair $a,b$, and we have $f(n,m) = f(n,m-1) + f(n-m,m)$.
Now my attempt so far is repeatedly applying the recursive rule to the first term to yield $ f(n,m) = 1 + \left\lfloor\frac{n}{2}\right\rfloor + \sum_{i=3}^m f(n-i,i) ~, $ which may itself be repeated to form a summation over no more than $\left\lfloor\dfrac{n-2}{3}\right\rfloor$ variables, but the computations involved (summations over products of floor functions) become extremely tedious.
Does anyone have suggestions as for how to proceed down this particular attempt at solution, a better method of solution, or a different angle on the problem entirely?
For the curious, this number of descending lists is part of a coefficient in the expansion of the expression $D^n[f^k]$, the n$^\text{th}$ derivative of the k$^\text{th}$ power of some function. Attempting this solely out of curiosity. Thanks ahead of time for reading this mess!