A237713 -id:A237713 - cp's OEIS frontend

A282755 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.

11, 125, 1255, 2510, 11009, 11099, 11255, 11379, 12326, 12955, 14379, 14397, 15033, 15303, 16325, 17482, 21109, 25105, 31007, 31503, 33011, 35213, 37127, 37921, 41303, 44011, 49319, 51367, 53491, 63013, 69413, 70319, 71057, 72013, 72517, 74341, 77011, 81767

Offset: 1

Michel Lagneau, Feb 21 2017

The primes of the form (10^n - 1)/9 are terms (A004022).
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, ...
The even numbers of the sequence are rarer than the odd numbers: 2510, 12326, 17482, 105002, 123206, ...
All terms have a 1 in their decimal representation (A011531). - Michel Marcus, Feb 23 2017
The union of 11 and A237713. - R. J. Mathar, Mar 06 2017

			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.

Cf. A004022, A011531, A032741.

A variant of A237713.

with(numtheory):
for n from 1 to 200000 do:
z:=convert(n,base,10):n0:=nops(z):lst1:={op(z),z[n0]}:
x:=divisors(n):n1:=nops(x):lst:={}:
  for m from 1 to n1-1 do:
   y:=convert(x[m],base,10):n2:=nops(y):
   lst2:={op(y),y[n2]}:lst:=lst union lst2
   od:
   if lst1=lst then
   printf(`%d, `,n):
   else
  fi:
od:

Select[Range[10^5], Function[k, Union@ Flatten@ Map[IntegerDigits, Most@ Divisors@ k] == Union@ IntegerDigits@ k]] (* Michael De Vlieger, Feb 25 2017 *)

cp's OEIS Frontend

A282755 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.

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