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A078117 Let u(1)=n, u(2)=n+1, v(1)=n+2, v(2)=n+3, u(k)=abs(u(k-1)-v(k-2)), v(k)=abs(v(k-1)-u(k-2)), then a(n) is the smallest integer such that for any k>=a(n), v(k)=u(k). 0

%I #6 Mar 30 2012 18:39:11

%S 8,11,15,17,15,11,12,14,12,14,9,14,12,17,21,23,21,16,18,20,18,20,15,

%T 20,18,23,27,29,27,22,24,26,24,26,21,26,24,29,33,35,33,28,30,32,30,32,

%U 27,32,30,35,39,41,39,34,36,38,36,38,33,38,36,41,45,47,45,40,42,44,39,44

%N Let u(1)=n, u(2)=n+1, v(1)=n+2, v(2)=n+3, u(k)=abs(u(k-1)-v(k-2)), v(k)=abs(v(k-1)-u(k-2)), then a(n) is the smallest integer such that for any k>=a(n), v(k)=u(k).

%F a(n)/n -> 1/2; for n>= 7, a(n) = (1/2)*(n+b(n)) where b(n) is the 12-periodic sequence (17, 20, 15, 18, 7, 16, 11, 20, 27, 30, 25, 14)

%K nonn

%O 1,1

%A _Benoit Cloitre_, Dec 05 2002

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