cp's OEIS Frontend

A291879 Number of monomials of the Schubert polynomial of the permutation 351624 tensor 1^n.

%I A291879 #41 Apr 08 2021 10:21:34
%S A291879 1,8,6720,561120560,4557185891241984,3571558033324129373292768,
%T A291879 269111599998006391761541640176800000,
%U A291879 1945556482213500279178010210766074095827609600000,1347912754604769492992184400055703948513202427323999206349209600
%N A291879 Number of monomials of the Schubert polynomial of the permutation 351624 tensor 1^n.
%C A291879 The permutation 351624 tensor 1^n is the permutation whose permutation matrix is obtained from that of 351624 by replacing each 1 with an n X n identity matrix.
%H A291879 A. H. Morales, I. Pak, G. Panova, <a href="https://arxiv.org/abs/1707.00931">Hook formulas for skew shapes III. Multivariate and product formulas</a>, arXiv:1707.00931 [math.CO], 2017.
%H A291879 R. P. Stanley, <a href="https://arxiv.org/abs/1704.00851">Some Schubert shenanigans</a>, arXiv:1704.00851 [math.CO], 2017.
%F A291879 a(n) = b(n)^5*b(3*n)^2*b(5*n)/(b(2*n)^4*b(4*n)^2) where b(n) = 1!*2!*...*(n-1)! is a superfactorial A000178(n-1). [corrected by _Vaclav Kotesovec_, Apr 08 2021]
%F A291879 a(n) = c(n)*b(3*n)^2*b(6*n)/((7*n^2)!*b(2*n)^2*b(4*n)^2) where b(n) = 1!*2!*...*(n-1)! is a superfactorial A000178(n-1) and c(n) = A291871.
%F A291879 a(n) ~ exp(1/6) * 3^(9*n^2 - 1/6) * 5^(25*n^2/2 - 1/12) / (A^2 * n^(1/6) * 2^(40*n^2 - 2/3)), where A is the Glaisher-Kinkelin constant A074962. - _Vaclav Kotesovec_, Apr 08 2021
%e A291879 For n=1 we have that a(1)=8 since the Schubert polynomial of 351624 equals the following sum of eight monomials: x0^3*x1^3*x2 + x0^3*x1^2*x2^2 + x0^2*x1^3*x2^2 + x0^3*x1^3*x3 + x0^3*x1^2*x2*x3 + x0^2*x1^3*x2*x3 + x0^3*x1^2*x3^2 + x0^2*x1^3*x3^2.
%t A291879 Table[BarnesG[n + 1]^5 * BarnesG[3*n + 1]^2 * BarnesG[5*n + 1] / (BarnesG[2*n + 1]^4 * BarnesG[4*n + 1]^2), {n, 0, 10}] (* _Vaclav Kotesovec_, Apr 08 2021 *)
%o A291879 (Sage) def b(n): return mul([factorial(i) for i in range(1,n)])
%o A291879 def a(n): return b(n)^5*b(3*n)^2*b(5*n)/(b(2*n)^4*b(4*n)^2)
%o A291879 [a(n) for n in range(10)]
%o A291879 (PARI) b(n) = prod(k=1, n-1, k!);
%o A291879 a(n) = b(n)^5*b(3*n)^2*b(5*n)/(b(2*n)^4*b(4*n)^2); \\ _Michel Marcus_, Sep 07 2017
%Y A291879 Cf. A000178, A008793, A291871.
%K A291879 nonn
%O A291879 0,2
%A A291879 _Alejandro H. Morales_, Sep 05 2017