A119804 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.
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, 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, 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
Offset: 0
Examples
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.
Links
- John Tyler Rascoe, Table of n, a(n) for n = 0..8192
Programs
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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
Extensions
More terms from R. J. Mathar, May 30 2006