%I #17 Apr 09 2016 14:25:25

%S 18,24,36,42,48,56,70,72,84,90,96,98,112,120,135,140,144,150,154,168,

%T 170,175,180,182,186,192,196,198,204,220,224,225,234,240,242,245,248,

%U 266,270,276,279,280,286,288,294,300,304,306,308,310,312,315,322,330,336,338

%N Numbers n such that at least one other integer m exists with the same smallest prime factor, same largest prime factor, and same set of binary digits as n.

%C Binary digits of m are a permutation of binary digits of n.

%C Conjecture: there is X such that among integers bigger than X more than 50% are in the sequence.

%C Since a set is an unordered collection of distinct elements, one should say "same multiset (or bag) of binary digits as n." - _Daniel Forgues_, Mar 31 2016

%H Robert Israel, <a href="/A214620/b214620.txt">Table of n, a(n) for n = 1..10000</a>

%e 18 and 24 have the same set of binary digits: 10010 and 11000, same smallest prime factor 2, and same largest prime factor 3, so both 18 and 24 are in the sequence.

%p N:= 10: # to get all terms < 2^N

%p H:= proc(n) local F, B;

%p F:= numtheory:-factorset(n);

%p B:= sort(convert(n,base,2));

%p [min(F),max(F),op(B)];

%p end proc:

%p T:= {}:

%p for n from 1 to 2^N-1 do

%p h:= H(n);

%p if assigned(R[h]) then T:= T union {n,R[h]}

%p else R[h]:= n

%p fi

%p od:

%p sort(convert(T,list)); # _Robert Israel_, Apr 01 2016

%t nn = 360; TakeWhile[Union@ Flatten@ Table[Select[Complement[Range[3 n], {n}], And[Length@ Union[DigitCount[#, 2] & /@ {#, n}] == 1, Length@ Union[{First@ #, Last@ #} &@ Map[First, FactorInteger@ #] & /@ {#, n}] == 1] &] /. i_ /; MissingQ@ i -> Nothing, {n, nn}], # <= nn &] (* _Michael De Vlieger_, Apr 01 2016, Version 10.2 *)

%Y Cf. A214619, A214621.

%K nonn,base

%O 1,1

%A _Alex Ratushnyak_, Jul 23 2012