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A083567
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Let B(k) be the number of binary digits in k equal to 1. This is the sequence of positive integers k such that 2B(k)=B(k^2).
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8
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21, 37, 42, 45, 53, 69, 73, 74, 81, 83, 84, 90, 106, 133, 137, 138, 141, 146, 148, 155, 161, 162, 165, 166, 168, 177, 180, 211, 212, 261, 265, 266, 269, 273, 274, 276, 281, 282, 289, 291, 292, 295, 296, 299, 310, 321, 322, 324, 330, 332, 336, 354, 359, 360
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OFFSET
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1,1
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COMMENTS
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This includes all k > 1 such that the average of ones in the binary expansion of k is the same of the average of ones in binary expansion of k^2; these are the values in the sequence with sqrt(2)*2^j < a(k) < 2^(j+1) for some j. - Corrected by Franklin T. Adams-Watters, Aug 23 2012
Conjecture: The counting function p(n) satisfies p(n)=c n/log n + o(n/log n).
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LINKS
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EXAMPLE
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21 is in the sequence because 21=10101_2 (3 1's) and 441=110111001_2 (6 1's).
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MAPLE
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select(t -> 2*convert(convert(t, base, 2), `+`) = convert(convert(t^2, base, 2), `+`), [$1..1000]); # Robert Israel, Aug 27 2015
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MATHEMATICA
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f[n_] := Total@ IntegerDigits[n, 2]; Select[Range@ 360, 2 f@ # == f[#^2] &] (* Michael De Vlieger, Aug 27 2015 *)
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PROG
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(PARI) isok(n) = norml2(binary(n^2)) == 2*norml2(binary(n)) \\ Michel Marcus, Jun 20 2013
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CROSSREFS
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KEYWORD
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easy,nonn,base
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AUTHOR
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STATUS
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approved
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