A380107 a(1) = 0; for n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and let a(n+1) be the number of distinct runs in the subsequence a(m)..a(n-1). Otherwise, a(n+1) = 0.

0, 0, 1, 0, 2, 0, 2, 2, 1, 4, 0, 4, 2, 4, 2, 2, 1, 5, 0, 6, 0, 2, 5, 4, 7, 0, 5, 4, 4, 1, 8, 0, 5, 5, 1, 4, 5, 3, 0, 6, 11, 0, 3, 4, 6, 5, 6, 2, 12, 0, 7, 13, 0, 3, 9, 0, 3, 3, 1, 13, 6, 10, 0, 6, 3, 6, 2, 11, 14, 0, 6, 5, 14, 4, 15, 0, 6, 6, 1, 13, 13, 1, 2, 11, 11

Offset: 1

Author

Neal Gersh Tolunsky, Jan 12 2025

Keywords

• nonn

Comments

This is a variant of Van Eck's sequence A181391 in which we count distinct runs of consecutive equal values rather than individual terms.

The longest run in the sequence has length 2.

Examples

			a(10)=4: We find that the most recent occurrence of a(n) = a(9) = 1 is a(3) = 1. In between a(3) and a(8), we find 4 distinct runs: [1]; [0]; [2]; [2,2]. So a(10)=4.
		

Links

• Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
• Neal Gersh Tolunsky, Ordinal transform of 100000 terms
• Neal Gersh Tolunsky, First differences of 100000 terms

Crossrefs

Cf. A380106, A268755.