A320063 - OEIS

%I #31 Oct 19 2018 21:48:10

%S 0,1,1,2,2,3,3,3,4,4,4,5,6,6,6,7,8,8,8,8,9,10,10,10,10,11,12,12,13,13,

%T 13,13,14,15,15,15,16,17,17,17,17,18,19,19,19,20,20,21,21,21,21,21,22,

%U 23,23,23,24,24,25,26,27,27,27,27,27,28,28,29,29,29

%N A sequence which changes by one or zero: a(n) = a(n-1-a(a(n-1))) + a(a(a(n-1))) for n > 1, a(n) = n for n < 2.

%C This is similar to a problem that I had in the Monthly 35 years ago. The solution then was by Daniel Kleitman.

%H Alois P. Heinz, <a href="/A320063/b320063.txt">Table of n, a(n) for n = 0..65535</a>

%H Abraham Isgur, Mustazee Rahman, <a href="http://arxiv.org/abs/1407.0425">On variants of Conway and Conolly's Meta-Fibonacci recursions</a>, arXiv:1407.0425 [math.CO], 2014.

%H T. Kubo and R. Vakil, <a href="http://dx.doi.org/10.1016/0012-365X(94)00303-Z">On Conway's recursive sequence</a>, Discr. Math. 152 (1996), 225-252.

%H David Newman and Daniel J. Kleitman, <a href="http://www.jstor.org/stable/2324158">Solution to Problem E3274</a>, Amer. Math. Monthly, 98 (1991), 958-959.

%p a:= proc(n) option remember; `if`(n<2, n,

%p       a(n-1-a(a(n-1)))+a(a(a(n-1))))

%p     end:

%p seq(a(n), n=0..100);  # _Alois P. Heinz_, Oct 08 2018

%t f[0] = 0;

%t f[1] = 1;

%t f[n_] := f[n] = +f[n - 1 - f[f[n - 1]]] + f[f[f[n - 1]]];

%t Table[f[i], {i, 1, 30}]

%Y Cf. A093878.

%K nonn

%O 0,4

%A _David S. Newman_, Oct 04 2018