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
Keywords
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
Comments