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Recursive sequence with a(n)=a(a(a(n-2)))+a(n-a(n-1)) and a(0)=a(1)=1.
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%I #1 Feb 24 2006 03:00:00

%S 1,1,2,2,4,3,6,3,9,3,5,8,6,6,15,7,6,10,15,7,9,8,17,9,12,8,21,9,16,8,

%T 23,12,12,14,15,16,12,14,18,15,15,24,18,14,30,14,21,28,18,18,19

%N Recursive sequence with a(n)=a(a(a(n-2)))+a(n-a(n-1)) and a(0)=a(1)=1.

%F a(n) = a(a(a(n-2))) + a(n-a(n-1)) a(0) = a(1) = 1

%e a(5)=3 because a(a(a(5-2)))+a(5-a(5-1)) = ... = 3

%t a[n_] := a[n] = a[a[a[n - 2]]] + a[n - a[n - 1]] a[0] = a[1] = 1; Table[a[n], {n, 100}]

%K easy,nonn

%O 0,3

%A Josh Locker (joshlocker(AT)gmail.com), Jan 28 2006