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

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, 11, 12, 11, 13, 12, 13, 13, 14, 14, 15, 14, 16, 15, 16, 16, 17, 17, 18, 18, 19, 18, 20, 19, 20, 20, 21, 21, 22, 21, 23, 22, 23, 23, 24, 24, 25, 25, 26, 25, 27

Offset

0,4

Comments

It appears that lim_{n->infinity} a(n)/n = 0.33960...

A similar formula is conjectured for A156253: (a(a(n-1)) mod 2) + a(n-2) + 1.

Links

Table of n, a(n) for n=0..71.

Mathematica

a[0] = 0;
a[1] = 1;
a[n_] := a[n] = Mod[a[a[n - 1]], 2] + a[n - 2]
Array[a, 100, 0]

Prog

(Python)
a = [0, 1]
[a.append(a[a[n-1]]%2 + a[n-2]) for n in range(2, 72)]
print(a) # Michael S. Branicky, Dec 15 2021

Crossrefs

Cf. A156253.

Sequence in context: A129294 A363719 A363723 * A323914 A350068 A300230

Adjacent sequences: A350123 A350124 A350125 * A350127 A350128 A350129

Keyword

nonn

Author

Jon Maiga, Dec 15 2021

Status

approved