%I
%S 1,9,10,12,27,28,30,36,37,39,81,82,84,85,90,91,93,94,108,109,111,112,
%T 117,118,120,243,244,246,247,252,253,256,270,271,273,274,279,280,282,
%U 283,324,325,327,328,333,334,336,337,351,352,354,355,360,361,363
%N Positive numbers having the same set of digits in base 2 and base 3.
%C From _Alonso del Arte_, Sep 10 2017: (Start)
%C Neither binary repunits (A000225 without the initial 0) nor ternary repunits (A003462 without the initial 0) can be in this sequence, except for 1.
%C The ternary repunits are numbers of the form (3^k  1)/2. If k is odd, then (3^k  1)/2 is even and therefore its binary representation ends in 0. If k is even, then (3^k  1)/2 = 1 mod 4, which means its binary representation ends in 01.
%C For much more obvious reasons, numbers with even just one 2 in their ternary representations (A074940) can't be in this sequence. (End)
%H John Cerkan, <a href="/A037408/b037408.txt">Table of n, a(n) for n = 1..10000</a>
%e 9 is 1001 in binary and 100 in ternary. In both representations, the set of digits used is {0, 1}, hence 9 is in the sequence.
%e 10 is 1010 in binary and 101 in ternary. In both representations, the set of digits used is {0, 1}, hence 10 is in the sequence.
%e 11 is 1011 in binary and 102 in ternary. Clearly the binary representation can't include the digit 2, hence 11 is not in the sequence.
%p filter:= proc(n) local F;
%p F:= convert(convert(n,base,3),set);
%p if has(F,2) then return false fi;
%p evalb(F = convert(convert(n,base,2),set))
%p end proc:
%p select(filter, [$1..1000]); # _Robert Israel_, Sep 18 2017
%t Select[Range[400], Union[IntegerDigits[#, 2]] == Union[IntegerDigits[#, 3]] &] (* _Vincenzo Librandi_ Sep 09 2017 *)
%o (PARI) isok(n) = vecsort(digits(n, 2),,8) == vecsort(digits(n, 3),,8); \\ _Michel Marcus_, Jan 05 2017
%K nonn,base
%O 1,2
%A _Clark Kimberling_
%E Initial 0 added by _Alonso del Arte_, Sep 10 2017
%E Initial 0 removed by _Georg Fischer_, Oct 30 2020
