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%I #15 Aug 22 2018 12:25:36
%S 1,3,5,15,17,25,27,39,46,49,51,57,58,63,77,85,86,106,141,142,166,175,
%T 177,178,201,202,207,209,226,243,245,255,257,267,278,289,291,298,305,
%U 323,326,329,363,393,394,417,423,519,526,529,533,537,538,553,554,562
%N Number of 1's in binary representation of n equals tau(n), the number of divisors of n.
%H Gheorghe Coserea, <a href="/A071593/b071593.txt">Table of n, a(n) for n = 1..51000</a>
%e 85=1010101 in base 2 and 85 has 4 divisors hence 85 is in the sequence.
%t Select[Range[600],DigitCount[#,2,1]==DivisorSigma[0,#]&] (* _Harvey P. Dale_, Aug 22 2018 *)
%o (PARI) for(n=1,1000,if(sum(i=1,length(binary(n)), component(binary(n),i))==numdiv(n),print1(n,",")))
%o (PARI) Vec(select(x->numdiv(x) == hammingweight(x), vector(562, k, k)) \\ _Gheorghe Coserea_, Oct 26 2016
%K base,easy,nonn
%O 1,2
%A _Benoit Cloitre_, Jun 01 2002
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