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Let u(1)=u(2)=1, u(n)=(2^floor(u(n-1)/2)+1)/u(n-2) then a(n) = denominator of u(n).
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%I #6 Mar 30 2012 18:39:10

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

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

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

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

%C Sequence is 36-periodic

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

%Y Cf. A076742.

%K frac,nonn

%O 1,3

%A _Benoit Cloitre_, Nov 24 2002