%I #21 Sep 09 2017 23:28:15
%S 11,125,1255,2510,11009,11099,11255,11379,12326,12955,14379,14397,
%T 15033,15303,16325,17482,21109,25105,31007,31503,33011,35213,37127,
%U 37921,41303,44011,49319,51367,53491,63013,69413,70319,71057,72013,72517,74341,77011,81767
%N Numbers k such that the set of all the decimal digits of k is the same as the set of all the decimal digits of the proper divisors of k.
%C The primes of the form (10^n - 1)/9 are terms (A004022).
%C A majority of numbers of the sequence are semiprimes (so with 3 proper divisors), except 11, 125, 2510, 16325, 21109, 72013, 126530, 132644, 163025, ... with the corresponding number of proper divisors 1, 3, 7, 5, 7, 7, 7, 5, 5, 5, 7, 7, ...
%C The even numbers of the sequence are rarer than the odd numbers: 2510, 12326, 17482, 105002, 123206, ...
%C All terms have a 1 in their decimal representation (A011531). - _Michel Marcus_, Feb 23 2017
%C The union of 11 and A237713. - _R. J. Mathar_, Mar 06 2017
%e 16325 is in the sequence because the set of the digits is E = {1, 2, 3, 5, 6} and the proper divisors (or aliquot parts) of 16325 are 1, 5, 25, 653 and 3265 with the same set of digits.
%p with(numtheory):
%p for n from 1 to 200000 do:
%p z:=convert(n,base,10):n0:=nops(z):lst1:={op(z),z[n0]}:
%p x:=divisors(n):n1:=nops(x):lst:={}:
%p for m from 1 to n1-1 do:
%p y:=convert(x[m],base,10):n2:=nops(y):
%p lst2:={op(y),y[n2]}:lst:=lst union lst2
%p od:
%p if lst1=lst then
%p printf(`%d, `,n):
%p else
%p fi:
%p od:
%t Select[Range[10^5], Function[k, Union@ Flatten@ Map[IntegerDigits, Most@ Divisors@ k] == Union@ IntegerDigits@ k]] (* _Michael De Vlieger_, Feb 25 2017 *)
%Y Cf. A004022, A011531, A032741.
%Y A variant of A237713.
%K nonn,base
%O 1,1
%A _Michel Lagneau_, Feb 21 2017