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A076784 Let u(1)=u(2)=u(3)=1, u(n)=(2^floor(u(n-1)/2)+2^floor(u(n-2)/2)+1)/u(n-3) then a(n) = numerator of u(n). 1

%I #8 May 27 2018 19:51:00

%S 1,1,1,3,4,7,13,13,1,12,12,3,13,91,13,3,12,12,1,13,13,7,4,3,1,1,1,3,4,

%T 7,13,13,1,12,12,3,13,91,13,3,12,12,1,13,13,7,4,3,1,1,1,3,4,7,13,13,1,

%U 12,12,3,13,91,13,3,12,12,1,13,13,7,4,3,1,1,1,3,4,7,13,13,1,12,12,3,13

%N Let u(1)=u(2)=u(3)=1, u(n)=(2^floor(u(n-1)/2)+2^floor(u(n-2)/2)+1)/u(n-3) then a(n) = numerator of u(n).

%C Sequence is 24-periodic

%F Period is (1, 1, 1, 3, 4, 7, 13, 13, 1, 12, 12, 3, 13, 91, 13, 3, 12, 12, 1, 13, 13, 7, 4, 3)

%t PadRight[{},120,{1,1,1,3,4,7,13,13,1,12,12,3,13,91,13,3,12,12,1,13,13,7,4,3}] (* _Harvey P. Dale_, May 27 2018 *)

%Y Cf. A076785.

%K frac,nonn

%O 1,4

%A _Benoit Cloitre_, Nov 24 2002

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