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, Table of n, a(n) for n = 1..10000
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
KEYWORD
nonn
AUTHOR
Rémy Sigrist, Nov 26 2019
STATUS
approved