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%I #26 Jul 06 2019 10:42:24
%S 1,-6,-360,252000,2222640000,-258768639360000,-410299414270986240000,
%T 9061429740221589431500800000,2835046804394206618956825845760000000,
%U -12733381268715468286016211650968992153600000000
%N Let M_k be the k X k matrix M_k(i,j)=1/binomial(i+n,j); then a(n)=1/det(M_(n+1)).
%C If k>n+1 det(M_k)=0
%F |a(n)| = det(M^(-1)), where M is an n X n matrix with M[i, j]=i/(i+j-1) (or j/(i+j-1)). |a(n)| = 1/det(HilbertMatrix(n))/n! = A005249(n)/n!. - _Vladeta Jovovic_, Jul 26 2003
%F |a(n)| = Product_{i=1..2n-1} binomial(i,floor(i/2)). - _Peter Luschny_, Sep 18 2012
%F |a(n)| = (Product_{i=1..2n-1} A056040(i))/n! = A163085(2*n-1)/n!. - _Peter Luschny_, Sep 18 2012
%t a[n_] := (-1)^Quotient[n, 2]/(Det[HilbertMatrix[n]] n!); Array[a, 10] (* _Jean-François Alcover_, Jul 06 2019 *)
%o (PARI) for(n=0,10,print1(1/matdet(matrix(n+1,n+1,i,j,1/binomial(i+n,j))),","))
%o (Sage)
%o def A069945(n): return (-1)^(n//2)*mul(binomial(i,i//2) for i in (1..2*n-1))
%o [A069945(i) for i in (1..11)] # _Peter Luschny_, Sep 18 2012
%Y A005249, A056040.
%K easy,sign
%O 1,2
%A _Benoit Cloitre_, Apr 27 2002