A278289 Number of standard Young tableaux of skew shape (2n-1,2n-2,...,2,1)/(n-1,n-2,..,2,1).

1, 1, 16, 101376, 1190156828672, 68978321274090930831360, 40824193474825703180733027309531955200, 440873872874088459550341319780612789503586208384381091840, 140992383930585613207663170866505518985873138480180692888967131590224605582721024

Offset: 0

Author

Alejandro H. Morales, Nov 16 2016

Keywords

• nonn

Examples

			For n = 3 there are a(2) = 16 standard tableaux of shape (3,2,1)/(1).
		

References

• R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Corollary 7.16.3.

Links

• Alejandro H. Morales, Table of n, a(n) for n = 0..22
• A. H. Morales, I. Pak and G. Panova, Asymptotics of the number of standard Young tableaux of skew shape, arXiv:1610.07561 [math.CO], 2016; European Journal of Combinatorics, Vol 70 (2018).
• A. H. Morales, I. Pak and G. Panova, Hook formulas for skew shapes II. Combinatorial proofs and enumerative applications, arXiv:1610.04744 [math.CO], 2016; SIAM Journal of Discrete Mathematics, Vol 31 (2017).
• A. H. Morales, I. Pak and M. Tassy, Asymptotics for the number of standard tableaux of skew shape and for weighted lozenge tilings, arXiv:1805.00992 [math.CO], 2018.
• A. H. Morales and D. G. Zhu, On the Okounkov--Olshanski formula for standard tableaux of skew shapes, arXiv:2007.05006 [math.CO], 2020.
• H. Naruse, Schubert calculus and hook formula, talk slides at 73rd Sém. Lothar. Combin., Strobl, Austria, 2014.
• I. Pak, Skew shape asymptotics, a case-based introduction, 2020.
• Jay Pantone, File with list of n, a(n) for n = 0..438 (warning: file size is 100MB)

Crossrefs

Cf. A005118; for even n the number of terms in Naruse hook length formula is given by A181119 (Corollary 8.1 in arXiv:1610.04744).

Programs

Maple

a:=proc(k) local lam,mu;
lam:=[seq(2*k-i,i=1..2*k-1)];
mu:=[seq(k-i,i=1..k-1),seq(0,i=1..k)];
factorial(binomial(2*k,2)-binomial(k,2))*LinearAlgebra:-Determinant(Matrix(2*k-1, 2*k-1,(i,j)->`if`(lam[i]-mu[j]-i+j<0,0,1/factorial(lam[i]-mu[j]-i+j))));
end proc:
seq(a(n),n=0..5);

Formula

a(n) = ((3*n^2-n)/2)!*det(1/(lambda[i]-mu[j]-i+j)!), where lambda = (2*n-1,2*n-2,...,1) and mu = (n-1,n-2,...,1,0...,0).

There is a constant c such that log(a(k)) = n*log(n)/2 + c*n + o(n) where n = k*(3*k-1)/2 goes to infinity and -0.2368 <= c <= -0.1648. [updated by Alejandro H. Morales, Aug 29 2020]