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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.
2
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
0,4
EXAMPLE
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.
PROG
(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
CROSSREFS
Cf. A119805.
Sequence in context: A057021 A152443 A373368 * A300977 A144869 A247564
KEYWORD
easy,nonn
AUTHOR
Leroy Quet, May 24 2006
EXTENSIONS
More terms from R. J. Mathar, May 30 2006
STATUS
approved