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
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
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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