A119804 - OEIS

%I #7 Apr 09 2014 10:14:38

%S 0,1,1,2,1,3,1,1,1,6,1,1,0,0,1,0,4,9,1,1,1,0,1,0,0,1,0,0,0,0,0,0,13,

%T 14,1,1,1,0,1,0,0,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,36,21,

%U 1,1,1,0,1,0,0,1,0,0,0,1,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0

%N a(0) = 0. For m >= 0 and 0 <= k <= 2^m -1, a(2^m +k) = number of earlier terms of the sequence which equal k.

%e 8 = 2^3 + 0; so for a(8) we want the number of terms among terms a(1), a(2),... a(7) which equal 0. So a(8) = 1.

%o (PARI) A119804(mmax)= { local(a,ncopr); a=[0]; for(m=0,mmax, for(k=0,2^m-1, ncopr=0; for(i=1,2^m+k, if( a[i]==k, ncopr++; ); ); a=concat(a,ncopr); ); ); return(a); } { print(A119804(6)); } - _R. J. Mathar_, May 30 2006

%Y Cf. A119805.

%K easy,nonn

%O 0,4

%A _Leroy Quet_, May 24 2006

%E More terms from _R. J. Mathar_, May 30 2006