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%I #54 Apr 22 2021 21:54:33
%S 1111,1255,12955,17482,25105,28174,51295,81229,91365,100255,101299,
%T 105295,107329,110191,110317,117067,124483,127417,129595,132565,
%U 137281,145273,146137,149782,163797,171735,174082,174298,174793,174982,193117,208174,210181,217894
%N Composite numbers having the same digits as their prime factors (with multiplicity), excluding zero digits.
%C Subsequence of A006753 (Smith numbers).
%C These numbers still need a better name. - _Ely Golden_, Dec 25 2016
%C Terms of this sequence never have more zero digits than their prime factors. - _Ely Golden_, Jan 10 2017
%H Ely Golden, <a href="/A176670/b176670.txt">Table of n, a(n) for n = 1..10000</a> [Terms 1 through 2113 were computed by Paul Weisenhorn; and terms 2114 to 10000 by Ely Golden, Nov 30 2016]
%H Ely Golden, <a href="/A176670/a176670_1.sagews.txt">Smith number sequence generator optimized for A176670</a>
%H Ely Golden, <a href="/A280827/a280827.txt">Proof that A280827(n)>=0 for all n>1</a>
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/SmithNumber.html">Smith Number</a>
%e n = 25105 = 5*5021; both n and the factorization of n have digits 1, 2, 5, 5; sorted and excluding zeros.
%e n = 110191 = 101*1091; both n and the factorization of n have digits 1, 1, 1, 1, 9; sorted and excluding zeros.
%e 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.
%t 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]
%t 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 *)
%o (Python)
%o from sympy import factorint, flatten
%o def sd(n): return sorted(str(n).replace('0', ''))
%o def ok(n):
%o f = factorint(n)
%o return sum(f[p] for p in f) > 1 and sd(n) == sorted(flatten(sd(p)*f[p] for p in f))
%o print(list(filter(ok, range(220000)))) # _Michael S. Branicky_, Apr 22 2021
%Y Cf. A006753.
%K nonn,base
%O 1,1
%A _Paul Weisenhorn_, Apr 23 2010
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