A380106 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 runs in the subsequence a(m)..a(n-1). Otherwise, a(n+1) = 0.

0, 0, 1, 0, 2, 0, 2, 2, 1, 5, 0, 4, 0, 2, 6, 0, 3, 0, 2, 5, 10, 0, 4, 11, 0, 3, 9, 0, 3, 3, 1, 21, 0, 4, 10, 13, 0, 4, 4, 1, 8, 0, 4, 4, 1, 4, 2, 25, 0, 6, 32, 0, 3, 21, 20, 0, 4, 11, 31, 0, 4, 4, 1, 17, 0, 4, 4, 1, 4, 2, 21, 15, 0, 7, 0, 2, 6, 25, 28, 0, 5, 56

Offset

1,5

Comments

This is a variant of Van Eck's sequence A181391 in which we ask: how many runs (of consecutive equal terms) ago did we last see a(n)?

The longest run in the sequence has length 2.

Links

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A380107, A181391, A380037.

Sequence in context: A112803 A124242 A112274 * A336891 A181391 A333359

Adjacent sequences: A380102 A380103 A380104 * A380107 A380108 A380109

Keyword

nonn

Author

Neal Gersh Tolunsky, Jan 12 2025

Status

approved