A380107 -id:A380107 - cp's OEIS frontend

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

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Neal Gersh Tolunsky, Jan 12 2025

nonn

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.

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

Cf. A380107, A181391, A380037.

cp's OEIS Frontend

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.

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