A119805 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.

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, 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, 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

Offset: 1

Author

Leroy Quet, May 24 2006

Keywords

• easy
• nonn

Examples

			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.
		

Crossrefs

Cf. A119804.

Programs

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

Extensions

More terms from R. J. Mathar, May 30 2006