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.

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, 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

See A329982, A329984, A329985 and A330004 for similar sequences.

Sequence in context: A247977 A359239 A143232 * A096030 A025815 A219199

Adjacent sequences: A329978 A329979 A329980 * A329982 A329983 A329984

Keyword

nonn

Author

Rémy Sigrist, Nov 26 2019

Status

approved