A083567 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).

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

Offset: 1

Author

Giuseppe Melfi, Jun 13 2003

Keywords

• easy
• nonn
• base

Comments

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).

Examples

			21 is in the sequence because 21=10101_2 (3 1's) and 441=110111001_2 (6 1's).
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Giuseppe Melfi, Su alcune successioni di interi, 2° Incontro Italiano di Teoria dei Numeri, Parma, 13-15 novembre 2003.
• Giuseppe Melfi, On certain positive integer sequences, arXiv:math/0404555 [math.NT], 30 Apr 2004.

Crossrefs

Cf. A000120, A077436, A159918.

Programs

Maple

select(t -> 2*convert(convert(t,base,2),`+`) = convert(convert(t^2,base,2),`+`), [$1..1000]); # Robert Israel, Aug 27 2015

Mathematica

f[n_] := Total@ IntegerDigits[n, 2]; Select[Range@ 360, 2 f@ # == f[#^2] &] (* Michael De Vlieger, Aug 27 2015 *)

PARI

isok(n) =  norml2(binary(n^2)) == 2*norml2(binary(n)) \\ Michel Marcus, Jun 20 2013