%I #7 Apr 09 2014 10:14:38
%S 1,1,2,1,3,1,1,0,5,1,1,0,1,0,0,0,8,1,1,0,1,0,0,1,0,0,0,0,0,0,0,0,12,1,
%T 1,0,1,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,17,1,1,0,
%U 1,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N a(1) = 1. For m >= 0 and 1 <= k <= 2^m, a(2^m +k) = number of earlier terms of the sequence which equal k.
%e 8 = 2^2 + 4; so for a(8) we want the number of terms among terms a(1), a(2),... a(7) which equal 4. So a(8) = 0.
%o (PARI) A119805(mmax)= { local(a,ncopr); a=[1]; for(m=0,mmax, for(k=1,2^m, ncopr=0; for(i=1,2^m+k-1, if( a[i]==k, ncopr++; ); ); a=concat(a,ncopr); ); ); return(a); } { print(A119805(6)); } - _R. J. Mathar_, May 30 2006
%Y Cf. A119804.
%K easy,nonn
%O 1,3
%A _Leroy Quet_, May 24 2006
%E More terms from _R. J. Mathar_, May 30 2006
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