login
A291871
Number of standard Young tableaux of skew shape (3*n^(2*n), 2*n^n)/(n^n).
2
1, 42, 14617044842400, 5458228515594914179387748450655273588000, 864891828322912925373153355728411014930091519471102108791040960580578545212124160000
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 a combinatorial model for the Selberg integral (by Kim and Oh) and using the Naruse hook-length formula for skew shapes (by Morales, Pak and Panova).
LINKS
J. S. Kim and S. Oh, The Selberg integral and Young books, arXiv:1409.1317 [math.CO], 2014.
J. S. Kim and S. Oh, The Selberg integral and Young books, J. Combin. Theory Ser. A 145 (2017), 1-24.
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) = (7*n^2)!*b(n)^5*b(5*n)/(b(2*n)^2*b(6*n)) where b(n) = 1!*2!*...*(n-1)! is a superfactorial A000178(n-1).
a(n) = (7*n^2)!*c(n)*b(n)^2*b(2*n)*b(5*n)/(b(6*n)*b(3*n)) where b(n) = 1!*2!*...*(n-1)! is a superfactorial A000178(n-1) and c(n) = A008793.
log a(n) = 7*n^2*log(n) + (75/2 - 15*log(2) - 18*log(3) + 25/2*log(5) + 7*log(7))*n^2 + O(n*log(n)). (See Example 6.2 in Morales et al.)
a(n) ~ sqrt(Pi) * 5^(25*n^2/2 - 1/12) * 7^(7*n^2 + 1/2) * exp(7*n^2/2 + 1/4) * n^(7*n^2 + 3/4) / (A^3 * 2^(22*n^2 - 3/4) * 3^(18*n^2 - 1/12)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Apr 08 2021
EXAMPLE
a(1)=42 since there are 42 standard Young tableaux of skew shape 332/1 since this is the same as the number of standard Young tableaux of straight shape 332 given by the hook-length formula: 42 = 8!/(2^2*3*4^2*5).
MAPLE
b:=n->mul(factorial(i), i=1..n-1):
a:=n->factorial(7*n^2)*b(n)^5*b(5*n)/(b(2*n)^2*b(6*n)):
seq(a(i), i=0..9);
MATHEMATICA
b[n_] := Product[i!, {i, n - 1}]; Table[(7 n^2)!*b[n]^5*b[5 n]/(b[2 n]^2*b[6 n]), {n, 0, 4}] (* Michael De Vlieger, Sep 10 2017 *)
PROG
(Sage) def b(n): return mul([factorial(i) for i in range(1, n)])
def a(n): return factorial(7*n^2)*b(n)^5*b(5*n)/(b(2*n)^2*b(6*n))
[a(n) for n in range(10)]
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved