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Given a partition λ = (λ1, λ2, ..., λn) denote with sλ the associated Schur function. There exists a nice product formula for the principal specializations:

sλ(1, q, q2, ..., qn-1) = Πi<j (qλi+n-i - qλj+n-j) / (qj-1 - qi-1).

Is a similar evaluation known for specializations of the type sλ(1, 2, ..., n)?

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    $\begingroup$ Two quick observations: for lambda = (1, 1, ...), e_k(1, 2, ... n) is an unsigned Stirling number of the first kind, and for lambda = (k), h_k(1, 2, ... n) is a Stirling number of the second kind. $\endgroup$ – Qiaochu Yuan Oct 15 '09 at 18:38
  • $\begingroup$ It does seem to be the case that it's always a rational function in $n$. Is this obvious? $\endgroup$ – Martin Rubey Mar 16 '10 at 20:49
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Let's take the strictest meaning of "similar evaluation", i.e., a product formula. The by the remark, the answer is probably no, because for instance the Stirling number s(9,3) = 118124 = 2*2*29531.

A different meaning is whether there is some explicit combination of nested sums and products for a Schur evaluation at consecutive integers. I don't know the answer to that. Unlike for a straight product formula, it is difficult to look for such formulas or rule them out.

The loosest possible meaning is whether there is an efficient way to evaluate sλ(1, 2,...,n). The answer to that is an easy yes, because you can use the Jacobi-Trudi determinant.

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