A147952 a(0) = 0, a(1) = a(2) = 1, and for n >= 3, a(n) = a(a(n-2)) + r(n), where r(n) = a(a(floor(n/3))) when n == 0 or 1 (mod 3) and = a(n - a(floor(n/3))) when n == 2 (mod 3).
0, 1, 1, 2, 2, 3, 2, 3, 4, 3, 3, 5, 3, 4, 5, 4, 5, 7, 4, 4, 6, 4, 4, 8, 4, 6, 6, 4, 4, 8, 4, 6, 10, 5, 6, 7, 4, 5, 9, 5, 5, 8, 6, 7, 7, 5, 5, 10, 6, 6, 7, 5, 6, 8, 4, 6, 8, 4, 6, 8, 4, 6, 10, 4, 5, 8, 5, 6, 8, 6, 8, 6, 6, 4, 10, 4, 5, 8, 5, 6, 13, 4, 6, 8, 4, 6, 8, 6, 8, 6, 6, 4, 10, 4, 5, 8, 6, 7, 10, 6, 6
Offset: 0
Keywords
Programs
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Mathematica
f[0] = 0; f[1] = 1; f[2] = 1; f[n_] := f[n] = f[f[n - 2]] + If[Mod[n, 3] == 0,f[f[n/3]], If[Mod[n, 3] == 1, f[f[(n - 1)/3]], f[n - f[(n - 2)/3]]]]; Table[f[n], {n, 0, 100}]
Formula
a(n) = a(a(n - 2)) + If[Mod[n, 3] == 0, a(a(n/3)), If[Mod[n, 3] == 1, a(a((n - 1)/3)), a(n - a((n - 2)/3))] for n >= 3 with a(0) = 0 and a(1) = a(2) = 1. [edited by Petros Hadjicostas, Apr 13 2020]
Extensions
Name edited by Petros Hadjicostas, Apr 13 2020