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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?

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    $\begingroup$ How about computing a table for small n and k.and posting it here? $\endgroup$ – Dima Pasechnik Jan 29 '19 at 20:08
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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, $$ h_{n-k}(1,2,\dots,k-1,n) = \sum_{l+m = n-k} n^m S(l+k-1,k-1) $$

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

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