%I #20 Jan 14 2024 12:38:08
%S 4,8,9,16,32,49,64,128,133,256,259,512,961,1024,2048,2059,2449,3713,
%T 4096,4681,4867,6169,6241,8192,8401,8773,9353,10261,10561,12307,12449,
%U 16129,16384,16459,16531,16771,18467,20491,24649,24721,24961,25217
%N Composite numbers such that all divisors >1 have the same number of 1's in binary representation.
%C A000120(d)=constant for all d with 1<d<=a(n) and d|a(n).
%C Are there terms with more than 2 distinct prime factors?
%C No terms with omega(n)>2 up to 10000000. - _Michel Marcus_, Jun 05 2013
%C From _Robert Israel_, Dec 01 2015: (Start)
%C The only term divisible by 3 is 9.
%C The terms divisible by 2 are 2^k for k > 1.
%C There are no terms divisible by 5. (End)
%H Harvey P. Dale and Robert Israel, <a href="/A089042/b089042.txt">Table of n, a(n) for n = 1..1000</a> (a(1) to a(100) from Harvey P. Dale)
%e Divisors >1 of 259: 7, 37 and 259, which have all three 1's in binary: 7->'111', 37->'100101' and 259->'100000011', therefore 259 is a term.
%p A000120:= proc(n) convert(convert(n,base,2),`+`) end proc:
%p filter:= proc(n) local t,f;
%p if isprime(n) then return false fi;
%p if n::even then return evalb(n = 2^ilog2(n)) fi;
%p if n mod 3 = 0 then return evalb(n = 9) fi;
%p t:= A000120(n);
%p for f in numtheory:-divisors(n) minus {1,n} do
%p if A000120(f) <> t then return false fi;
%p od;
%p true
%p end proc:
%p select(filter, [$4..10^5]); # _Robert Israel_, Dec 01 2015
%t dn1Q[n_]:=!PrimeQ[n]&&Length[Union[(DigitCount[#,2,1]&/@Rest[Divisors[ n]])]] == 1; Select[Range[26000],dn1Q] (* _Harvey P. Dale_, Oct 03 2013 *)
%o (PARI) isok(n) = {if (isprime(n) || n==1, return (0), my(nb = norml2(binary(n))); fordiv(n, d, if (d!=1 && norml2(binary(d)) != nb, return (0))); return (1););} \\ _Michel Marcus_, Jun 05 2013
%Y Cf. A007088, A002808.
%K nonn,base
%O 1,1
%A _Reinhard Zumkeller_, Dec 02 2003