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A350126 a(n) = (a(a(n-1)) mod 2) + a(n-2) with a(0) = 0 and a(1) = 1.

%I A350126 #17 Jan 05 2022 15:41:58
%S A350126 0,1,1,2,2,3,2,4,2,5,3,5,4,5,5,6,5,7,5,8,5,9,6,9,7,9,8,9,9,10,10,11,
%T A350126 11,12,11,13,12,13,13,14,14,15,14,16,15,16,16,17,17,18,18,19,18,20,19,
%U A350126 20,20,21,21,22,21,23,22,23,23,24,24,25,25,26,25,27
%N A350126 a(n) = (a(a(n-1)) mod 2) + a(n-2) with a(0) = 0 and a(1) = 1.
%C A350126 It appears that lim_{n->infinity} a(n)/n = 0.33960...
%C A350126 A similar formula is conjectured for A156253: (a(a(n-1)) mod 2) + a(n-2) + 1.
%t A350126 a[0] = 0;
%t A350126 a[1] = 1;
%t A350126 a[n_] := a[n] = Mod[a[a[n - 1]], 2] + a[n - 2]
%t A350126 Array[a, 100, 0]
%o A350126 (Python)
%o A350126 a = [0, 1]
%o A350126 [a.append(a[a[n-1]]%2 + a[n-2]) for n in range(2, 72)]
%o A350126 print(a) # _Michael S. Branicky_, Dec 15 2021
%Y A350126 Cf. A156253.
%K A350126 nonn
%O A350126 0,4
%A A350126 _Jon Maiga_, Dec 15 2021