cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A336674 Number of positive terms of the Okounkov-Olshanski formula for the number of standard tableaux of skew shape (n+3,n+2,...,1)/(n-1,n-2,...,1).

Original entry on oeis.org

1, 1, 5, 65, 1757, 87129, 7286709, 965911665, 193387756045, 56251615627273, 23021497112124901, 12903943243053179681, 9680994096074346690365, 9530338509606467082850745, 12099590059386455266220499477
Offset: 0

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Author

Alejandro H. Morales, Jul 29 2020

Keywords

Comments

a(n) is also the number of semistandard Young tableaux of skew shape (n+3,n+2,...,1)/(n-1,n-2,...,1) such that the entries in row i are at most i for i=1,...,n+3.
a(n) is also the number of semistandard Young tableaux T of shape (n-1,n-2,...,1) such that j-i < T(i,j) <= n+3 for all cells (i,j).

Examples

			For n=2 the a(2)=5 semistandard Young tableaux of skew shape (5,4,3,2,1)/(1) are determined by their first column which are [1,2,3,4], [1,2,3,5], [1,2,4,5], [1,3,4,5], and [2,3,4,5]. Also, the a(2)=5 semistandard Young tableaux of shape (1) with entries between 0 and 5 are [1], [2], [3], [4], and [5]. Also, the a(3)=70-5=65 are the semistandard Young tableaux of shape (2,1) with entries at most 6 excluding the five tableaux whose entry in the first row and first column is 1: [[1,1],[2]], [[1,1],[3]], [[1,1],[4]], and [[1,1],[5]].
		

Crossrefs

A110501, A005700 gives the number of terms of the Naruse hook length formula for the same skew shape.

Programs

  • Maple
    b := proc(n)
        return 2*(-1)^n*(1-4^n)*bernoulli(2*n)/factorial(2*n);
    end proc:
    a := proc(n)
        return factorial(2*n+4)*factorial(2*n+6)*(b(n+1)*b(n+3)-b(n+2)^2)/6;
    end proc:
    seq(a(n),n=0..10);
  • Sage
    def b(n):
        return 2*(-1)^n*(1-4^n)*bernoulli(2*n)/factorial(2*n) ;
    def a(n):
        return factorial(2*n+4)*factorial(2*n+6)*(b(n+1)*b(n+3)-b(n+2)^2)/6;
    [a(i) for i in range(10)]

Formula

a(n) = ((2*n+4)!*(2*n+6)!/3!)*(b(n+1)*b(n+3)-b(n+2)^2) where b(n)=A110501(n)/(2*n)!.