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A281511 The lexicographically earliest sequence of positive integers such that for all k >= j >= 1, if a(n) = a(n + j) = a(n + k) then a(n + j + k) != a(n).

%I A281511 #23 Jul 23 2024 21:36:48
%S A281511 1,1,2,1,2,2,3,1,3,2,4,3,1,4,4,2,5,3,3,4,1,4,5,2,5,5,6,6,7,5,1,3,2,4,
%T A281511 6,3,6,7,7,5,6,7,8,4,1,8,8,2,9,7,3,5,6,8,9,6,7,9,4,8,10,9,9,2,10,1,3,
%U A281511 5,8,10,10,4,6,7,11,11,12,9,7,9,1,10,11,2
%N A281511 The lexicographically earliest sequence of positive integers such that for all k >= j >= 1, if a(n) = a(n + j) = a(n + k) then a(n + j + k) != a(n).
%C A281511 Indices of ones are given by A005282.
%C A281511 Conjecture: For all positive i, j, k there exists some n such that a(n) = a(n + j) = a(n + k) = i.
%C A281511 Also, if a(k)=a(j) for any k<j, j-k is a unique result for the value of a(k). - _Marc Morgenegg_, May 31 2024
%H A281511 Peter Kagey, <a href="/A281511/b281511.txt">Table of n, a(n) for n = 1..10000</a>
%e A281511 For n=1 through n=7, the terms are as follows:
%e A281511 a(1) = 1;
%e A281511 a(2) = 1;
%e A281511 a(3) != 1 because a(1) = a(1+1) = a(1+1) so a(1+1+1) != a(1);
%e A281511 a(3) = 2, the least value such that satisfies the sequence condition;
%e A281511 a(4) = 1;
%e A281511 a(5) != 1 because a(1) = a(1+1) = a(1+3) so a(1+1+3) != a(1);
%e A281511 a(5) = 2, the least value such that satisfies the sequence condition;
%e A281511 a(6) != 1 because a(2) = a(2+2) = a(2+2) so a(2+2+2) != a(2);
%e A281511 a(6) = 2, the least value such that satisfies the sequence condition;
%e A281511 a(7) != 1 because a(1) = a(1+3) = a(1+3) so a(1+3+3) != a(1);
%e A281511 a(7) != 2 because a(3) = a(3+2) = a(3+2) so a(3+2+2) != a(3);
%e A281511 a(7) = 3, the least value such that satisfies the sequence condition.
%Y A281511 Cf. A005282.
%K A281511 nonn,look
%O A281511 1,3
%A A281511 _Peter Kagey_, Apr 13 2017