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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.
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%I #10 Aug 18 2015 00:18:21

%S 1,1,1,2,3,5,18,325,1512,14365,349272,21734245,276623424,6933892901,

%T 577589709312,492757099009565,16532350249637376,1086038875887212525,

%U 1240124656925798848512,1450308695702968720107785

%N 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.

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

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

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

%p 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

%Y Cf. A128531.

%K frac,nonn

%O 1,4

%A _Leroy Quet_, Mar 08 2007

%E More terms from _R. J. Mathar_, Oct 09 2007