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0, 0, 0, 0, 1, 0, 2, 0, 2, 1, 3, 0, 4, 2, 2, 2, 5, 1, 5, 2, 4, 4, 6, 0, 6, 5, 5, 3, 7, 2, 8, 4, 6, 6, 6, 2, 9, 7, 7, 3, 9, 4, 10, 6, 6, 8, 10, 2, 10, 7, 9, 7, 11, 5, 9, 6, 10, 10, 12, 2, 12, 10, 8, 8, 11, 7, 13, 9, 11, 7, 13, 4, 14, 12, 10, 10, 12, 8, 14, 6, 12, 13, 15, 5, 13, 13, 13, 9, 15, 6, 14
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OFFSET
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1,7
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COMMENTS
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It appears that the indices of the zeros in the sequence are in A018253. - Omar E. Pol, Oct 22 2013
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LINKS
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EXAMPLE
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n=15: sequence of D1 = {floor(15/j)} = {15,7,5,3,3,2,2,1,1,1,1,1,1,1,1}, Union(D1) = {15,7,5,3,2,1} = divisors(15) and {7,2}, a(15)=2 the number of terms beyond divisors.
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MATHEMATICA
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Table[Length[Union[Table[Floor[w/j], {j, 1, w}]]] -DivisorSigma[0, w], {w, 1, 128}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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