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A069214 Let u(n,k) be the recursion defined by u(n,1)=1, u(n,2)=2, u(n,3)=n, and u(n,k+3) = (u(n,k+2) + u(n,k+1))/u(n,k) if u(n,k) divides u(n,k+2) + u(n,k+1), u(n,k+3) = u(n,k) otherwise. Then u(n,k) is periodic and a(n) = Max(u(n,k), k=1,2,3,4,...). 0

%I

%S 5,4,5,6,8,8,11,10,14,12,17,14,20,16,23,18,26,20,29,22,32,24,35,26,38,

%T 28,41,30,44,32,47,34,50,36,53,38,56,40,59,42,62,44,65,46,68,48,71,50,

%U 74,52,77,54,80,56,83,58,86,60,89,62,92,64,95,66,97,68,100,70,103,72

%N Let u(n,k) be the recursion defined by u(n,1)=1, u(n,2)=2, u(n,3)=n, and u(n,k+3) = (u(n,k+2) + u(n,k+1))/u(n,k) if u(n,k) divides u(n,k+2) + u(n,k+1), u(n,k+3) = u(n,k) otherwise. Then u(n,k) is periodic and a(n) = Max(u(n,k), k=1,2,3,4,...).

%C Let p(n) denote the period of u(n,k) (i.e., p(n) is the smallest integer such that u(n,k) = u(n, k+p(n))). p(n) = 22,12,22,21,15,9,15,9,15,9,... for n = 1,2,3,4,5,6,.... Hence for n > 4, p(n) = 15 if n is odd; p(n) = 9 if n is even.

%F a(1)=5, a(2n)=2n+2, a(2n+1)=3n+2.

%e E.g., for k=1..15, u(7, k) = 1, 2, 7, 9, 8, 7, 9, 2, 7, 1, 4, 7, 11, 4, 7; hence a(7)=11.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Apr 11 2002

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