A176670 Composite numbers having the same digits as their prime factors (with multiplicity), excluding zero digits.

1111, 1255, 12955, 17482, 25105, 28174, 51295, 81229, 91365, 100255, 101299, 105295, 107329, 110191, 110317, 117067, 124483, 127417, 129595, 132565, 137281, 145273, 146137, 149782, 163797, 171735, 174082, 174298, 174793, 174982, 193117, 208174, 210181, 217894

Offset: 1

Author

Paul Weisenhorn, Apr 23 2010

Keywords

• nonn
• base

Comments

Subsequence of A006753 (Smith numbers).

These numbers still need a better name. - Ely Golden, Dec 25 2016

Terms of this sequence never have more zero digits than their prime factors. - Ely Golden, Jan 10 2017

Examples

			n = 25105 = 5*5021; both n and the factorization of n have digits 1, 2, 5, 5; sorted and excluding zeros.
n = 110191 = 101*1091; both n and the factorization of n have digits 1, 1, 1, 1, 9; sorted and excluding zeros.
n = 171735 = 3*5*107*107; both n and the factorization of n have digits 1, 1, 3, 5, 7, 7; sorted and excluding zeros.
		

Links

• Ely Golden, Table of n, a(n) for n = 1..10000 [Terms 1 through 2113 were computed by Paul Weisenhorn; and terms 2114 to 10000 by Ely Golden, Nov 30 2016]
• Ely Golden, Smith number sequence generator optimized for A176670
• Ely Golden, Proof that A280827(n)>=0 for all n>1
• Eric W. Weisstein, Smith Number

Crossrefs

Cf. A006753.

Programs

Mathematica

fQ[n_] := Block[{id = Sort@ IntegerDigits@ n, s = Sort@ Flatten[ IntegerDigits@ Table[ #[[1]], {#[[2]]}] & /@ FactorInteger@ n]}, While[ id[[1]] == 0, id = Drop[id, 1]]; While[ s[[1]] == 0, s = Drop[s, 1]]; n > 1 && ! PrimeQ@ n && s == id]; Select[ Range@ 200000, fQ]
Select[Range[2*10^5], Function[n, And[CompositeQ@ n, Sort@ DeleteCases[#, 0] &@ IntegerDigits@ n == Sort@ DeleteCases[#, 0] &@ Flatten@ Map[IntegerDigits@ ConstantArray[#1, #2] & @@ # &, FactorInteger@ n]]]] (* Michael De Vlieger, Dec 10 2016 *)

Python

from sympy import factorint, flatten
def sd(n): return sorted(str(n).replace('0', ''))
def ok(n):
  f = factorint(n)
  return sum(f[p] for p in f) > 1 and sd(n) == sorted(flatten(sd(p)*f[p] for p in f))
print(list(filter(ok, range(220000)))) # Michael S. Branicky, Apr 22 2021