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%I #18 May 01 2015 13:10:05
%S 1,-2,-360,16848000,1897448716800000,-3129723891582775706419200000,
%T -541942196790147039091108680776954796441600000,
%U 66373536294235576434745706427960099542896427384297349714149376000000
%N Let A_n be the n X n matrix defined by A_n[i,j] = 1/F(i+j-1) for 1<=i,j<=n where F(k) is the k-th Fibonacci number (A000045). Then a_n = 1/det(A_n).
%C In the reference it is proved that not only det(A_n) is a reciprocal of an integer but the inverse matrix (A_n)^(-1) is an integer matrix.
%H Colin Barker, <a href="/A062381/b062381.txt">Table of n, a(n) for n = 1..19</a>
%H T. M. Richardson, <a href="http://arXiv.org/abs/math/9905079">The Filbert Matrix</a>, arXiv:math/9905079 [math.RA], 1999.
%F a(n) = s(n) * f(n) / h(n)^2, where s(n) = (-1)^Floor[n/2], f(n) = Product[Fibonacci[k]^(n-Abs[k-n]),{k,1,2*n-1}], h(n) = Product[Product[Fibonacci[k],{k,1,m-1}],{m,1,n}]. - _Alexander Adamchuk_, May 18 2006
%F a(n) ~ (-1)^floor(n/2) * A253270 * ((1+sqrt(5))/2)^(n*(2*n^2+1)/3) / (A253267^2 * 5^(n/2) * A062073^(2*n-2)). - _Vaclav Kotesovec_, May 01 2015
%e a(3) = -360 because the matrix is / 1,1,1/2 / 1,1/2, 1/3 / 1/2, 1/3, 1/5 / with determinant -1/360.
%t Table[(-1)^Floor[n/2]*Product[Fibonacci[k]^(n-Abs[k-n]),{k,1,2*n-1}],{n,1,10}]/Table[Product[Product[Fibonacci[k],{k,1,m-1}],{m,1,n}],{n,1,10}]^2 (* _Alexander Adamchuk_, May 18 2006 *)
%o (PARI) vector(8, n, 1/matdet(matrix(n, n, i, j, 1/fibonacci(i+j-1)))) \\ _Colin Barker_, May 01 2015
%Y Cf. A000045, A010048.
%Y Cf. A003266, A152686, A253268.
%Y Cf. A062073, A253267, A253270.
%K sign,nice
%O 1,2
%A Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 08 2001
%E More terms from _Vladeta Jovovic_, Jul 11 2001
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