%I #10 Jun 21 2014 14:16:47
%S 0,0,0,1,0,0,1,0,1,0,1,2,0,0,1,0,1,0,1,2,0,1,0,1,2,0,1,2,0,1,2,3,0,0,
%T 1,0,1,0,1,2,0,1,0,1,2,0,1,2,0,1,2,3,0,1,0,1,2,0,1,2,0,1,2,3,0,1,2,0,
%U 1,2,3,0,1,2,3,0,1,2,3,4,0,0,1,0,1,0,1,2,0,1,0,1,2,0,1,2,0,1,2,3,0,1,0,1,2,0,1,2,0,1,2,3,0,1,2,0,1,2,3,0
%N Integers from 0 to A000120(n)-1 followed by integers from 0 to A000120(n+1)-1 and so on, starting with n=1.
%F a(n) = n - (1 + A000788(A100922(n-1)-1)).
%e For n=1, also 1 in binary notation, so the count of its 1-bits is 1 (A000120(1)=1), we list numbers from 0 to 0, thus just 0.
%e For n=2, 10 in binary, thus A000120(2)=1, we list numbers from 0 to 0, thus just 0.
%e For n=3, 11 in binary, thus A000120(3)=2, we list numbers from 0 to 1, and so we have the first four terms of the sequence: 0; 0; 0, 1;
%o (Scheme) (define (A243067 n) (- n (+ 1 (A000788 (- (A100922 (- n 1)) 1)))))
%Y Cf. A000788, A100922, A163510, A163511, A227740.
%K nonn
%O 1,12
%A _Antti Karttunen_, Jun 19 2014