A253270 -id:A253270 - cp's OEIS frontend

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

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 08 2001

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.

			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.

Colin Barker, Table of n, a(n) for n = 1..19
T. M. Richardson, The Filbert Matrix, arXiv:math/9905079 [math.RA], 1999.

Cf. A000045, A010048.
Cf. A003266, A152686, A253268.
Cf. A062073, A253267, A253270.

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 *)

vector(8, n, 1/matdet(matrix(n, n, i, j, 1/fibonacci(i+j-1)))) \\ Colin Barker, May 01 2015

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

More terms from Vladeta Jovovic, Jul 11 2001

A253268 Product_{k=1..n} Fibonacci(k)^k.

Original entry on oeis.org

1, 1, 8, 648, 2025000, 530841600000, 33309523161907200000, 1259861409934788058133299200000, 76494996113757662751632456649087438028800000, 19375812441937279261781767910157290423127539712000000000000000

Offset: 1

Vaclav Kotesovec, May 01 2015

Cf. A000045, A003266, A062381, A152686, A253270.

Table[Product[Fibonacci[k]^k,{k,1,n}],{n,1,10}]
FoldList[Times,Table[Fibonacci[n]^n,{n,10}]] (* Harvey P. Dale, Oct 17 2019 *)

a(n) ~ c * ((1+sqrt(5))/2)^(n*(n+1)*(2*n+1)/6) / 5^(n*(n+1)/4), where c = A253270 = 1.1188674438617789501735355690557711005492672642422237428157122531318... .

A094221 1/detM(n) where M(n) is the n X n matrix m(i,j)=F(i)/F(i+j-1) and F(i)=i-th Fibonacci number.

Original entry on oeis.org

1, -2, -180, 2808000, 63248290560000, -13040516214928232110080000, -173699422048124050990739961787485511680000, 1013027110717881203216509560866301885575342298295136595148800000

Offset: 1

Benoit Cloitre, May 28 2004

sign

Cf. A062381.

Table[(-1)^Floor[n/2] * Product[Fibonacci[k]^k,{k,1,n-1}] * Product[Fibonacci[k]^(2*n-k),{k,n,2*n-1}] / Product[Fibonacci[k],{k,1,n}] / Product[Product[Fibonacci[k],{k,1,j-1}],{j,1,n}]^2,{n,1,10}] (* Vaclav Kotesovec, May 01 2015 *)

a(n)=1/matdet(matrix(n,n,i,j,fibonacci(i)/(fibonacci(i+j-1))))

a(n) = A062381(n)/A003266(n). - corrected by Vaclav Kotesovec, May 01 2015

a(n) ~ (-1)^floor(n/2) * A253270 * ((1+sqrt(5))/2)^(n*(4*n^2 - 3*n - 1)/6) / (A253267^2 * A062073^(2*n-1)). - Vaclav Kotesovec, May 01 2015

cp's OEIS Frontend

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).

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A253268 Product_{k=1..n} Fibonacci(k)^k.

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A094221 1/detM(n) where M(n) is the n X n matrix m(i,j)=F(i)/F(i+j-1) and F(i)=i-th Fibonacci number.

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