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

s_λ(1, q, q², ..., q^n-1) = Π_i<j (q^λ_i+n-i - q^λ_j+n-j) / (q^j-1 - q^i-1).

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

edited Nov 4 2009 at 12:00

asked Oct 15 2009 at 2:40

Armin Straub
802●2●11●19

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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. – Qiaochu Yuan Oct 15 2009 at 18:38

It does seem to be the case that it's always a rational function in $n$. Is this obvious? – Martin Rubey Mar 16 2010 at 20:49

Greg Kuperberg · Answer 1 · 2009-11-04 07:14:41Z UTC

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 s_λ(1, 2,...,n). The answer to that is an easy yes, because you can use the Jacobi-Trudi determinant.

Specializations of Schur functions at consecutive integers

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