A128531 a(n) = numerator of r(n): r(n) is such that the continued fraction (of rational terms) [r(1);r(2),...r(n)] equals the n-th Fibonacci number, for every positive integer n.

1, 1, -2, 3, -10, 6, -65, 378, -5525, 16632, -1278485, 25147584, -1012815817, 8022079296, -2114837334805, 570081043090944, -60533314393713485, 1256458618972440576, -4540728540084435567025, 1677888660820605842036736, -466914087740138106185288665

Offset: 1

Author

Leroy Quet, Mar 08 2007

Keywords

• frac
• sign

Examples

			The 5th Fibonacci number = 5 = 1 +1/(1 +1/(-2 +1/(3/2 -3/10))).
The 6th Fibonacci number = 8 = 1 +1/(1 +1/(-2 +1/(3/2 +1/(-10/3 +5/6)))).
		

Crossrefs

Cf. A128532.

Programs

Maple

L2cfrac := proc(L,targ) local a,i; a := targ ; for i from 1 to nops(L) do a := 1/(a-op(i,L)) ; od: end: A128531 := proc(nmax) local b,n,bnxt; b := [1] ; for n from nops(b)+1 to nmax do bnxt := L2cfrac(b,combinat[fibonacci](n+1)) ; b := [op(b),bnxt] ; od: [seq( numer(b[i]),i=1..nops(b))] ; end: A128531(22) ; # R. J. Mathar, Oct 09 2007

Mathematica

r[n_] := r[n] = Switch[n, 1, 1, 2, 1, 3, -2, _, -Fibonacci[n]/(Fibonacci[n-3]*r[n-1])];
a[n_] := Numerator[r[n]];
Table[a[n], {n, 1, 21}] (* Jean-François Alcover, Sep 24 2024 *)

Formula

For n>=4, r(n) = -F(n)/(F(n-3) r(n-1)), where F(n) is the n-th Fibonacci number.

Extensions

More terms from R. J. Mathar, Oct 09 2007