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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.

%I A320297 #21 Dec 24 2018 06:21:15
%S A320297 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,
%T A320297 14,14,14,15,15,16,16,16,17,17,18,18,18,19,19,20,20,20,21,22,22,23,23,
%U A320297 23,24,25,25,26,26,26,26,27,28,28,29,29,29,29,30,31,31,32,32,32,32,33
%N 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.
%C A320297 This sequence hits every positive integer.
%H A320297 Muniru A Asiru, <a href="/A320297/b320297.txt">Table of n, a(n) for n = 0..10000</a>
%H A320297 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 A320297 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 A320297 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.
%F A320297 a(n+1) - a(n) = 0 or 1 for all n >= 0.
%t A320297 a[0] = 0; a[1] = 1; a[n_] := a[n] = +a[n - 1 - a[a[n - 1]]] + a[a[a[n - 1]]];
%t A320297 Table[a[i], {i, 1, 300}]
%o A320297 (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
%Y A320297 Cf. A320063.
%K A320297 nonn
%O A320297 0,4
%A A320297 _David S. Newman_, Oct 09 2018