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A319579 Lexicographically earliest infinite sequence of positive terms such that for any n >= 0, a(n+1) = a(n + a(n)) - a(n). 1

%I #44 Oct 01 2019 10:12:09

%S 2,2,4,6,2,8,10,2,6,8,2,5,7,18,14,8,12,10,2,25,27,2,18,20,2,4,6,12,22,

%T 10,24,32,18,14,15,2,5,7,45,34,38,12,3,22,52,25,2,29,31,17,32,3,53,15,

%U 56,2,2,4,6,4,46,10,10,50,10,74,49,8,71,52,27,20,60

%N Lexicographically earliest infinite sequence of positive terms such that for any n >= 0, a(n+1) = a(n + a(n)) - a(n).

%C The smallest legal term is 2, otherwise for a(n) = 1: a(n+1) = a(n + 1) - 1.

%H Rémy Sigrist, <a href="/A319579/b319579.txt">Table of n, a(n) for n = 0..10000</a>

%H Rémy Sigrist, <a href="/A319579/a319579.txt">C++ program for A319579</a>

%H Rémy Sigrist, <a href="/A319579/a319579.png">Scatterplot of the first 100000 terms</a>

%F a(n+1) = a(n + a(n)) - a(n).

%e a(0) = 2, because it's the smallest positive integer that satisfies the rule a(n+1) = a(n + a(n)) - a(n).

%e a(1) = 2, because we have again free choice inside the rules.

%e a(2) = 4, because a(1) = a(0 + a(0)) - a(0) = a(0 + 2) - a(0) = a(2) - 2 = 2.

%e a(3) = 6, because a(2) = a(1 + a(1)) - a(1) = a(1 + 2) - a(1) = a(3) - 2 = 4.

%e a(6) = 10, because a(3) = a(2 + a(2)) - a(2) = a(2 + 4) - a(2) = a(6) - 4 = 6.

%e a(4) = 2, because we have again free choice inside the rules.

%e And so on.

%o (C++) See Links section.

%Y Cf. A309681.

%K nonn

%O 0,1

%A _Marc Morgenegg_, Aug 27 2019

%E Name amended by _Rémy Sigrist_, Oct 01 2019

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