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