A291908 Number of standard Young tableaux of skew shape lambda/mu where lambda is the staircase (4*n-1,4*n-2,...,2,1) and mu is the square n^n.

1, 16, 4362327818240, 19265181532031090042534736325278852710400, 830325323503973129435791248069702287019820905338483131168940909920954227594481411031040

Offset

0,2

Comments

The number of standard Young tableaux of a fixed skew shape has a determinantal formula, the Jacobi-Trudi formula. It is rare when a family of skew shapes has a product formula for the number of standard Young tableaux. This product formula has independently been proved using P-Schur functions (by DeWitt) and using the Naruse hook-length formula for skew shapes (by Morales, Pak and Panova).

Links

Table of n, a(n) for n=0..4.

E. A. DeWitt, Identities Relating Schur s-Functions and Q-Functions, Ph.D. thesis, University of Michigan, 2012, 73 pp.

A. H. Morales, I. Pak, G. Panova, Hook formulas for skew shapes III. Multivariate and product formulas, arXiv:1707.00931 [math.CO], 2017.

Formula

a(n) = (binomial(4*n,2)-n^2)!*b(n)^3*b(3*n)*c(n)*c(3*n)/(b(2*n)^3*c(2*n)^2*c(4*n)) where b(n) = 1!*2!*...*(n-1)! is the superfactorial A000178(n-1), and c(n) = 1!!*3!!*...*(2*n-3)!! is super doublefactorial A057863(n-1).

a(n) ~ sqrt(Pi) * 3^(9*n^2 - 3*n/2 - 1/24) * 7^(7*n^2 - 2*n + 1/2) * exp(7*n^2/2 - 2*n + 23/56) * n^(7*n^2 - 2*n + 7/8) / (A^(3/2) * 2^(33*n^2 - 6*n - 7/8)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Apr 08 2021

Example

a(1)=16 since there are 16 standard Young tableaux of skew shape 321/1 since this is the same as the number of standard Young tableaux of straight shape 321 given by the hook-length formula: 16 = 6!/(3^2*5).

Maple

b:=n->mul(factorial(i), i=1..n-1):
c:=n->mul(doublefactorial(2*i-1), i=1..n-1):
a:=n->factorial(binomial(4*n, 2)-n^2)*b(n)^3*b(3*n)*c(n)*c(3*n)/(b(2*n)^3*c(2*n)^2*c(4*n)):
seq(a(n), n=0..9);

Prog

(Sage)
def b(n): return mul([factorial(i) for i in range(1, n)])
def d(n): return factorial(n+1)/(2^((n+1)/2)*factorial((n+1)/2))
def c(n): return mul([d(2*i-1) for i in range(1, n)])
def a(n):
    return factorial(binomial(4*n, 2)-n^2)*b(n)^3*b(3*n)*c(n)*c(3*n)/(b(2*n)^3*c(2*n)^2*c(4*n))
[a(n) for n in range(10)]

Crossrefs

Cf. A000178, A057863, A008793, A291871, A000085, A061343, A005118.

Sequence in context: A013878 A058418 A372150 * A059933 A002488 A330716

Adjacent sequences: A291905 A291906 A291907 * A291909 A291910 A291911

Keyword

nonn

Author

Alejandro H. Morales, Sep 05 2017

Status

approved