login
A329981
a(1) = 0, and for n > 0, a(n+1) = floor(k / 3) where k is the number of terms equal to a(n) among the first n terms.
5
0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 2, 0, 2, 0, 2, 1, 2, 1, 2, 1, 3, 0, 2, 2, 2, 2, 3, 0, 3, 1, 3, 1, 3, 1, 4, 0, 3, 2, 3, 2, 3, 2, 4, 0, 3, 3, 3, 3, 4, 1, 4, 1, 4, 1, 5, 0, 4, 2, 4, 2, 4, 2, 5, 0, 4, 3, 4, 3, 4, 3, 5, 1, 5, 1, 5, 1, 6, 0, 4, 4, 4, 4, 5, 2, 5, 2
OFFSET
1,12
COMMENTS
In other words, for n > 0, a(n+1) = floor(o(n) / 3) where o is the ordinal transform of the sequence.
Every nonnegative number appears infinitely many times in the sequence.
The second difference of the positions of the zeros in the sequence appears to be eventually 6-periodic.
LINKS
Rémy Sigrist, Colored scatterplot of the first 10000 terms (where the color denotes the parity of n)
EXAMPLE
The first terms, alongside their ordinal transform, are:
a a(n) o(n)
-- ---- ----
1 0 1
2 0 2
3 0 3
4 1 1
5 0 4
6 1 2
7 0 5
8 1 3
9 1 4
10 1 5
11 1 6
12 2 1
PROG
(PARI) o = vector(7); v=0; for (n=1, 87, print1 (v", "); v=o[1+v]++\3)
CROSSREFS
See A329982, A329984, A329985 and A330004 for similar sequences.
Sequence in context: A247977 A359239 A143232 * A096030 A025815 A219199
KEYWORD
nonn
AUTHOR
Rémy Sigrist, Nov 26 2019
STATUS
approved