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A119805
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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.
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2
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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
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OFFSET
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1,3
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LINKS
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EXAMPLE
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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.
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PROG
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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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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