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%I #12 Jan 27 2015 01:04:19
%S 5,6,39,43,45,46,51,53,54,57,58,60,287,303,311,315,317,318,335,343,
%T 347,349,350,359,363,365,366,371,373,374,377,378,380,399,407,411,413,
%U 414,423,427,429,430,435,437,438,441,442,444,455,459,461,462,467,469,470
%N Numbers n for which r(n) = 3 * S(n)/2, where r(n) = number of digits of n and S(n) = sum of digits of n; n in binary notation.
%C Integers n such that in Base 2, number of `1`'s = twice number of `0`'s. IntegerDigits[43,2]={1,0,1,0,1,1},IntegerDigits[60,2]={1,1,1,1,0,0},... - _Vladimir Joseph Stephan Orlovsky_, Jul 21 2009
%t f0[n_]:=DigitCount[n,2,0]; f1[n_]:=DigitCount[n,2,1]; f[n_]:=f1[n]/f0[n]; lst={};Do[If[f[n]==2,AppendTo[lst,n]],{n,6!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Jul 21 2009 *)
%t Select[Range[500],DigitCount[#,2,1]==2*DigitCount[#,2,0]&] (* _Harvey P. Dale_, May 22 2013 *)
%o (PARI) is(n)=hammingweight(n)==2/3*#binary(n) \\ _Charles R Greathouse IV_, May 28 2013
%Y Cf. A000120, A070939, A225222.
%K easy,nonn,base
%O 1,1
%A _Ctibor O. Zizka_, Mar 11 2008
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