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).

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, 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, 5, 8, 10, 10, 4, 6, 7, 11, 11, 12, 9, 7, 9, 1, 10, 11, 2

Offset: 1

Author

Peter Kagey, Apr 13 2017

Keywords

• nonn
• look

Comments

Indices of ones are given by A005282.

Conjecture: For all positive i, j, k there exists some n such that a(n) = a(n + j) = a(n + k) = i.

Also, if a(k)=a(j) for any kMarc Morgenegg, May 31 2024

Examples

			For n=1 through n=7, the terms are as follows:
a(1) = 1;
a(2) = 1;
a(3) != 1 because a(1) = a(1+1) = a(1+1) so a(1+1+1) != a(1);
a(3) = 2, the least value such that satisfies the sequence condition;
a(4) = 1;
a(5) != 1 because a(1) = a(1+1) = a(1+3) so a(1+1+3) != a(1);
a(5) = 2, the least value such that satisfies the sequence condition;
a(6) != 1 because a(2) = a(2+2) = a(2+2) so a(2+2+2) != a(2);
a(6) = 2, the least value such that satisfies the sequence condition;
a(7) != 1 because a(1) = a(1+3) = a(1+3) so a(1+3+3) != a(1);
a(7) != 2 because a(3) = a(3+2) = a(3+2) so a(3+2+2) != a(3);
a(7) = 3, the least value such that satisfies the sequence condition.
		

Links

• Peter Kagey, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A005282.