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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.

%I A329981 #13 Nov 29 2019 12:34:49
%S A329981 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,
%T A329981 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,
%U A329981 3,4,3,5,1,5,1,5,1,6,0,4,4,4,4,5,2,5,2
%N 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.
%C A329981 In other words, for n > 0, a(n+1) = floor(o(n) / 3) where o is the ordinal transform of the sequence.
%C A329981 Every nonnegative number appears infinitely many times in the sequence.
%C A329981 The second difference of the positions of the zeros in the sequence appears to be eventually 6-periodic.
%H A329981 Rémy Sigrist, <a href="/A329981/b329981.txt">Table of n, a(n) for n = 1..10000</a>
%H A329981 Rémy Sigrist, <a href="/A329981/a329981.png">Colored scatterplot of the first 10000 terms</a> (where the color denotes the parity of n)
%e A329981 The first terms, alongside their ordinal transform, are:
%e A329981   a   a(n)  o(n)
%e A329981   --  ----  ----
%e A329981    1     0     1
%e A329981    2     0     2
%e A329981    3     0     3
%e A329981    4     1     1
%e A329981    5     0     4
%e A329981    6     1     2
%e A329981    7     0     5
%e A329981    8     1     3
%e A329981    9     1     4
%e A329981   10     1     5
%e A329981   11     1     6
%e A329981   12     2     1
%o A329981 (PARI) o = vector(7); v=0; for (n=1, 87, print1 (v", "); v=o[1+v]++\3)
%Y A329981 See A329982, A329984, A329985 and A330004 for similar sequences.
%K A329981 nonn
%O A329981 1,12
%A A329981 _Rémy Sigrist_, Nov 26 2019