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%I #22 Aug 23 2019 22:45:44
%S 1,2,18,1200,735000,4667544000,332086420512000,279394363051195392000,
%T 2892376010829659126572800000,379850021025259936655866602240000000,
%U 648304836222110631242066578424390188032000000
%N 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.
%C Also, determinant of the inverse of the (n+1)-st Hilbert matrix, divided by (2n+1)!. - _Robert G. Wilson v_, Feb 02 2004
%C Also, inverse of determinant of the matrix M_n(i,j) = i*j/(i+j). - _Harry Richman_, Aug 19 2019
%F a(n) = A005249(n)/A000142(n). - _Robert G. Wilson v_, Feb 02 2004
%F a(n) = (n+1)!/(2*n+1)! * Product[Binomial(i,Floor(i/2)), {i,1,2*n+1}]. - _Stefan Steinerberger_, Feb 26 2008
%F a(n) = A163085(2*n+1)/(2*n+1)! = A163085(2*n)/factorial(n)^2. - _Peter Luschny_, Sep 18 2012
%t Table[1/((2n - 1)!Det[Table[1/(i + j - 1), {i, n}, {j, n}]]), {n, 10}] (* _Robert G. Wilson v_, Feb 02 2004 *)
%t 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 *)
%o (PARI) for(n=1,15,print1(1/matdet(matrix(n,n,i,j,i^2/(j+i))),","))
%o (Sage)
%o def A069651(n): return A163085(2*n+1)/factorial(2*n+1)
%o [A069651(n) for n in (0..10)] # _Peter Luschny_, Sep 18 2012
%Y Cf. A000142, A005249, A163085.
%K easy,nonn
%O 0,2
%A _Benoit Cloitre_, Apr 21 2002
%E Edited by _N. J. A. Sloane_, Feb 25 2008
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