A320297 a(n) = a(n-1-a(a(a(n-1)))) + a(a(a(a(n-1)))) for n > 1, a(n) = n for n < 2.
0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 10, 10, 10, 11, 12, 12, 12, 13, 14, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 20, 20, 20, 21, 22, 22, 23, 23, 23, 24, 25, 25, 26, 26, 26, 26, 27, 28, 28, 29, 29, 29, 29, 30, 31, 31, 32, 32, 32, 32, 33
Offset: 0
Keywords
Links
- Muniru A Asiru, Table of n, a(n) for n = 0..10000
- Abraham Isgur, Mustazee Rahman, On variants of Conway and Conolly's Meta-Fibonacci recursions, arXiv:1407.0425 [math.CO], 2014.
- T. Kubo and R. Vakil, On Conway's recursive sequence, Discr. Math. 152 (1996), 225-252.
- David Newman and Daniel J. Kleitman, Solution to Problem E3274, Amer. Math. Monthly, 98 (1991), 958-959.
Crossrefs
Cf. A320063.
Programs
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GAP
a:=[1,1];; for n in [3..80] do a[n]:=a[n-1-a[a[a[n-1]]]]+a[a[a[a[n-1]]]]; od; Concatenation([0],a); # Muniru A Asiru, Oct 28 2018
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Mathematica
a[0] = 0; a[1] = 1; a[n_] := a[n] = +a[n - 1 - a[a[n - 1]]] + a[a[a[n - 1]]]; Table[a[i], {i, 1, 300}]
Formula
a(n+1) - a(n) = 0 or 1 for all n >= 0.
Comments