%I #7 Jun 24 2019 04:14:48
%S 1,3,60,10500,18522000,359400888000,81408613942656000,
%T 224737840779305293440000,7812628980363223707442752000000,
%U 3508978524227146242839564498172672000000
%N Inverse determinant of n X n matrix M[i,j] = i*j/(i+j-1).
%F a(n) = 1/Det[ Table[ i*j/(i+j-1), {i, n}, {j, n}]]. a(n+1)/a(n) = A000891[n] = (2n)!(2n+1)! / (n! (n+1)!)^2 = (2n+1)*CatalanNumber[n]^2 = (2n+1)*A000108[n]^2 = C(2n+1,n+1)*CatalanNumber[n] = A001700[n]*A000108[n].
%F a(n) = A163085(2*n)/(2*n)!. - _Peter Luschny_, Sep 18 2012
%t Table[ 1/Det[ Table[ i*j/(i+j-1), {i, n}, {j, n}]], {n,1,12}]
%o (Sage)
%o def A120307(n): return A163085(2*n)/factorial(2*n)
%o [A120307(n) for n in (1..10)] # _Peter Luschny_, Sep 18 2012
%Y Cf. A000108, A000891, A001700.
%K nonn
%O 1,2
%A _Alexander Adamchuk_, Jul 15 2006