A062381 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).
1, -2, -360, 16848000, 1897448716800000, -3129723891582775706419200000, -541942196790147039091108680776954796441600000, 66373536294235576434745706427960099542896427384297349714149376000000
Offset: 1
Examples
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.
Links
- Colin Barker, Table of n, a(n) for n = 1..19
- T. M. Richardson, The Filbert Matrix, arXiv:math/9905079 [math.RA], 1999.
Programs
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Mathematica
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 *) -
PARI
vector(8, n, 1/matdet(matrix(n, n, i, j, 1/fibonacci(i+j-1)))) \\ Colin Barker, May 01 2015
Formula
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
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
Extensions
More terms from Vladeta Jovovic, Jul 11 2001
Comments