A069651 For n >= 1, let M_n be the n X n matrix with M_n(i,j) = i^2/(i+j); then a(n) = 1/det(M_n). Also, a(0) = 1 by convention.

1, 2, 18, 1200, 735000, 4667544000, 332086420512000, 279394363051195392000, 2892376010829659126572800000, 379850021025259936655866602240000000, 648304836222110631242066578424390188032000000

Offset: 0

Author

Benoit Cloitre, Apr 21 2002

Keywords

• easy
• nonn

Comments

Also, determinant of the inverse of the (n+1)-st Hilbert matrix, divided by (2n+1)!. - Robert G. Wilson v, Feb 02 2004

Also, inverse of determinant of the matrix M_n(i,j) = i*j/(i+j). - Harry Richman, Aug 19 2019

Crossrefs

Cf. A000142, A005249, A163085.

Programs

Mathematica

Table[1/((2n - 1)!Det[Table[1/(i + j - 1), {i, n}, {j, n}]]), {n, 10}] (* Robert G. Wilson v, Feb 02 2004 *)
Table[(n + 1)!/(2*n + 1)!*Product[Binomial[i, Floor[i/2]], {i, 1, 2*n + 1}], {n, 0, 10}] (* Stefan Steinerberger, Feb 26 2008 *)

PARI

for(n=1,15,print1(1/matdet(matrix(n,n,i,j,i^2/(j+i))),","))

Sage

def A069651(n): return A163085(2*n+1)/factorial(2*n+1)
[A069651(n) for n in (0..10)] # Peter Luschny, Sep 18 2012

Formula

a(n) = A005249(n)/A000142(n). - Robert G. Wilson v, Feb 02 2004

a(n) = (n+1)!/(2*n+1)! * Product[Binomial(i,Floor(i/2)), {i,1,2*n+1}]. - Stefan Steinerberger, Feb 26 2008

a(n) = A163085(2*n+1)/(2*n+1)! = A163085(2*n)/factorial(n)^2. - Peter Luschny, Sep 18 2012

Extensions

Edited by N. J. A. Sloane, Feb 25 2008