A128532 a(n) = denominator 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, 1, 2, 3, 5, 18, 325, 1512, 14365, 349272, 21734245, 276623424, 6933892901, 577589709312, 492757099009565, 16532350249637376, 1086038875887212525, 1240124656925798848512, 1450308695702968720107785

Offset: 1

Author

Leroy Quet, Mar 08 2007

Keywords

• frac
• nonn

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. A128531.

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: A128532 := 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( denom(b[i]),i=1..nops(b))] ; end: A128532(22) ; # R. J. Mathar, Oct 09 2007

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