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A071593
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Number of 1's in binary representation of n equals tau(n), the number of divisors of n.
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5
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1, 3, 5, 15, 17, 25, 27, 39, 46, 49, 51, 57, 58, 63, 77, 85, 86, 106, 141, 142, 166, 175, 177, 178, 201, 202, 207, 209, 226, 243, 245, 255, 257, 267, 278, 289, 291, 298, 305, 323, 326, 329, 363, 393, 394, 417, 423, 519, 526, 529, 533, 537, 538, 553, 554, 562
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OFFSET
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1,2
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LINKS
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EXAMPLE
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85=1010101 in base 2 and 85 has 4 divisors hence 85 is in the sequence.
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MATHEMATICA
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Select[Range[600], DigitCount[#, 2, 1]==DivisorSigma[0, #]&] (* Harvey P. Dale, Aug 22 2018 *)
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PROG
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(PARI) for(n=1, 1000, if(sum(i=1, length(binary(n)), component(binary(n), i))==numdiv(n), print1(n, ", ")))
(PARI) Vec(select(x->numdiv(x) == hammingweight(x), vector(562, k, k)) \\ Gheorghe Coserea, Oct 26 2016
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CROSSREFS
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KEYWORD
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base,easy,nonn
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AUTHOR
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STATUS
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approved
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