0,3

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

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

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

Sequence in context: A138664 A140357 A089265 this_sequence A113886 A122376 A067240

Adjacent sequences: A113882 A113883 A113884 this_sequence A113886 A113887 A113888

easy,nonn

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