%I #35 Apr 25 2016 16:11:15
%S 1,2,59,1292,53862,3615208,392961340,68986099580,19595297946515,
%T 9048133666290540,6832278662513786160,8489106538840284343800,
%U 17456177529017536829265000,59700294731704834466701403040,340945552945616104095546549396336,3261527521637774696821080128931389072
%N 'Core' alternating sign n X n matrices, i.e., those that are not 'blown up' from a smaller matrix by inserting row i, column j with a_ij = 1 and all other entries in that row and column equal to 0.
%D R. K. Guy, Unsolved Problems in Number Theory, D1.
%H Alois P. Heinz, <a href="/A003827/b003827.txt">Table of n, a(n) for n = 3..100</a>
%F Let b(n) = Product_{i=0..n-1} (3*i+1)!/(n+i)! be the number of alternating sign n X n matrices (i.e., sequence A005130), and a(n) the number of core alternating sign n X n matrices considered here, with the sequence [1,2,59,...] starting at offset n=3. Then it is not hard to show that for n>3: a(n) = b(n) - n! - Sum_{k=1..n-3} binomial(n,k)^2 * k! *a(n-k). - _Christine Bessenrodt_, Oct 02 2015
%F a(n) ~ exp(1/36) * Pi^(1/3) * 2^(5/12 - 2*n^2) * 3^(-7/36 + 3*n^2/2) / (A^(1/3) * Gamma(1/3)^(2/3) * n^(5/36)), where A = A074962 is the Glaisher-Kinkelin constant. - _Vaclav Kotesovec_, Apr 25 2016
%o (PARI) \\ rather inefficient, should use memoization
%o b(n) = prod(i=0, n-1, (3*i+1)!/(n+i)! );
%o a(n) = b(n) - n! - sum(k=1, n-3, binomial(n,k)^2 * k! *a(n-k) );
%o vector(20,n,a(n)) \\ _Joerg Arndt_, Oct 03 2015
%Y Cf. A051055, A005130.
%K nonn
%O 3,2
%A _Christine Bessenrodt_
%E Corrected and extended by _Christine Bessenrodt_, Oct 02 2015