A071593 Number of 1's in binary representation of n equals tau(n), the number of divisors of n.
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
Offset: 1
Examples
85=1010101 in base 2 and 85 has 4 divisors hence 85 is in the sequence.
Links
- Gheorghe Coserea, Table of n, a(n) for n = 1..51000
Programs
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Mathematica
Select[Range[600],DigitCount[#,2,1]==DivisorSigma[0,#]&] (* Harvey P. Dale, Aug 22 2018 *)
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PARI
for(n=1,1000,if(sum(i=1,length(binary(n)), component(binary(n),i))==numdiv(n),print1(n,",")))
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PARI
Vec(select(x->numdiv(x) == hammingweight(x), vector(562, k, k))) \\ Gheorghe Coserea, Oct 26 2016