# Evaluation of a complete homogeneous symmetric polynomial related to Stirling number of 2nd kind

It is well known that the complete homogeneous symmetric polynomial $$h_{n-k}(1,\,2,\,3, ...,\,k-1,\,k)$$ equals $$S(n,\,k)$$ the Stirling number of the second kind. [Wikipedia]

During a research project I stumbled upon the following complete homogeneous symmetric polynomial: $$h_{n-k}(1,\,2,\,3, ...,\,k-1,\,n)$$.

My question is: is the latter symmetric polynomial expressible in nice, simple and/or interpretable terms?

Or is this too much to ask? If so, why?

• How about computing a table for small n and k.and posting it here? – Dima Pasechnik Jan 29 '19 at 20:08

## 1 Answer

Consider the fact that $$\prod_{i=1}^k \frac{1}{1-x_i t} = \sum_j h_j(x_1,\dots,x_k) t^j$$ Writing $$h_j = h_{j+k-k}$$ we get from your first fact: $$\prod_{i=1}^k \frac{1}{1- i t} = \sum_j S(j+k,k) t^j.$$ Now multiply this by $$1/(1-nt)$$. We get $$\sum_j h_j(1,2,\dots,k,n) t^j = \frac{1}{1-nt} \sum_l S(l+k,k) t^l.$$ Comparing the coefficient of $$t^j$$ on both sides we get $$h_j(1,2,\dots,k,n) = \sum_{l+m = j} n^m S(l+k,k).$$ Thus, in your notation, $$h_{n-k}(1,2,\dots,k-1,n) = \sum_{l+m = n-k} n^m S(l+k-1,k-1)$$

• One does not need to use generating functions for this proof; simply note that $h_{n-k}(x_1,\dots,x_k)= h_{n-k}(x_1,\dots,x_{k-1})+x_kh_{n-k-1}(x_1,\dots,x_{k-1})+ x_k^2h_{n-k-2}(x_1,\dots,x_{k-1})+\cdots$. – Richard Stanley Mar 1 '19 at 14:48