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A382501 Lexicographically earliest infinite sequence of positive integers such that, for any given k, every subsequence {a(j), a(j+k), a(j+2k)} (j, k >= 1) is unique.

%I A382501 #12 Apr 06 2025 16:49:26
%S A382501 1,1,1,2,1,1,3,1,2,4,3,1,1,4,1,3,2,5,2,4,2,3,4,1,2,5,3,2,4,6,1,3,5,5,
%T A382501 6,1,1,7,2,3,8,4,8,7,1,2,6,5,3,1,4,3,8,7,2,8,2,6,9,1,9,1,4,6,9,4,5,9,
%U A382501 2,7,5,7,3,4,3,10,10,4,9,1,3,6,2,5,8,2,9
%N A382501 Lexicographically earliest infinite sequence of positive integers such that, for any given k, every subsequence {a(j), a(j+k), a(j+2k)} (j, k >= 1) is unique.
%C A382501 Every subsequence {a(n-2k), a(n-k) a(n)} with its corresponding k value (or index spacing) is unique.
%H A382501 Neal Gersh Tolunsky, <a href="/A382501/b382501.txt">Table of n, a(n) for n = 1..10000</a>
%H A382501 Sean A. Irvine, <a href="https://github.com/archmageirvine/joeis/blob/master/src/irvine/oeis/a382/A382501.java">Java program</a> (github)
%e A382501 To find a(10) = 4, we first try 1. We cannot have a(10) = 1 because this would create the subsequence {1,1,1} at i = 6,8,10, which occurred before at i = 1,3,5. In both cases, k = 2, which is not allowed .
%e A382501 a(10) cannot be 2 because then the subsequence {1,1,2} at i = 2,6,10 would be the same as {1,1,2} at  i = 1,5,9. In both cases, k = 4.
%e A382501 a(10) cannot be 3 because {1,1,3} at i = 6,8,10 would be the same as the subsequence at i = 3,5,7. In both cases, k = 2.
%e A382501 When we try a(10) = 4, we see that none of the new subsequences formed have occurred before with the same k value. Since 4 is a first occurrence, every subsequence created is new, and although i = 6,8,10 has the same subsequence {1,1,4} as i = 2,6,10, the k value is different, which is allowed. So a(10) = 4.
%Y A382501 Cf. A364057, A382502.
%K A382501 nonn
%O A382501 1,4
%A A382501 _Neal Gersh Tolunsky_, Mar 29 2025