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

Author

Rémy Sigrist, Nov 26 2019

Keywords

• nonn

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.

Examples

			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
		

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)

Crossrefs

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

Programs

PARI

o = vector(7); v=0; for (n=1, 87, print1 (v", "); v=o[1+v]++\3)