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A076785 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) = denominator of u(n). 1

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

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

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

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

%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) = denominator of u(n).

%C Sequence is 24-periodic

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

%Y Cf. A076784.

%K frac,nonn

%O 1,7

%A _Benoit Cloitre_, Nov 24 2002

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Last modified August 17 07:15 EDT 2024. Contains 375200 sequences. (Running on oeis4.)