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A076742 Let u(1)=u(2)=1, u(n)=(2^floor(u(n-1)/2)+1)/u(n-2) then a(n) = numerator of u(n). 1

%I #6 Mar 30 2012 18:39:10

%S 1,1,2,3,3,2,4,3,9,1,8,2,27,3,16,4,27,9,8,8,9,27,4,16,3,27,2,8,1,9,3,

%T 4,2,3,3,2,1,1,2,3,3,2,4,3,9,1,8,2,27,3,16,4,27,9,8,8,9,27,4,16,3,27,

%U 2,8,1,9,3,4,2,3,3,2,1,1,2,3,3,2,4,3,9,1,8,2,27,3,16,4,27,9,8,8,9,27,4,16

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

%C Sequence is 36-periodic

%F Period is (1, 1, 2, 3, 3, 2, 4, 3, 9, 1, 8, 2, 27, 3, 16, 4, 27, 9, 8, 8, 9, 27, 4, 16, 3, 27, 2, 8, 1, 9, 3, 4, 2, 3, 3, 2)

%Y Cf. A076780.

%K frac,nonn

%O 1,3

%A _Benoit Cloitre_, Nov 24 2002

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Last modified September 17 16:56 EDT 2024. Contains 375990 sequences. (Running on oeis4.)