A303921 Main diagonal of triangle A303920: a(n) = A303920(n,n) for n>=0.

1, 1, 6, 56, 722, 12012, 246092, 6002824, 170048394, 5489377628, 198966923232, 8002061191632, 353657146741108, 17038311744899928, 888756685396257456, 49903123853737160256, 3001090647251938886634, 192456294604677056842812, 13110208254597852188752232, 945417747582856587884200944, 71952514694665595216762956518, 5763451519600988678663191769380

Offset: 0

Author

Paul D. Hanna, May 02 2018

Keywords

• nonn

Comments

G.f. of A303920: (1-y) * Sum_{n>=0} y^n * (1 + x*(1-y)^2)^(n^2) = Sum_{n>=0} Sum_{k=0..2*n} A303920(n,k)*x^n*y^k; this sequence is A303920(n,n) for n>=0.

Examples

			Triangle A303920 begins:
[1];
[0, 1, 1];
[0, 0, 6, 6, 0];
[0, 0, 4, 56, 56, 4, 0];
[0, 0, 1, 117, 722, 722, 117, 1, 0];
[0, 0, 0, 126, 2982, 12012, 12012, 2982, 126, 0, 0];
[0, 0, 0, 84, 6916, 79548, 246092, 246092, 79548, 6916, 84, 0, 0]; ...
the main diagonal of which forms this sequence; note that the row sums of A303920 equals A001813(n) = (2*n)!/n!.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..50

Crossrefs

Cf. A303920, A303922.

Formula

a(n) ~ sqrt(3) * 2^(2*n + 1/2) * n^(n - 1/2) / (sqrt(Pi) * exp(n)). - Vaclav Kotesovec, Oct 08 2019