most recent 30 from http://mathoverflow.net 2013-05-26T07:53:36Z http://mathoverflow.net/feeds/question/549 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/549/specializations-of-schur-functions-at-consecutive-integers Armin Straub 2009-10-15T02:40:16Z 2009-11-04T12:00:19Z

<p>Given a partition &lambda; = (&lambda;₁, &lambda;₂, ..., &lambda;_n) denote with s_&lambda; the associated Schur function. There exists a nice product formula for the principal specializations: <p>s_&lambda;(1, q, q², ..., q^n-1) = &Pi;_i&lt;j (q^{&lambda;_i+n-i} - q^{&lambda;_j+n-j}) / (q^j-1 - q^i-1). <p>Is a similar evaluation known for specializations of the type s_&lambda;(1, 2, ..., n)?</p>

http://mathoverflow.net/questions/549/specializations-of-schur-functions-at-consecutive-integers/4061#4061 Greg Kuperberg 2009-11-04T07:14:41Z 2009-11-04T07:14:41Z

<p>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.</p> <p>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.</p> <p>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.</p>