A280827 -id:A280827 - cp's OEIS frontend

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

Paul Weisenhorn, Apr 23 2010

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

			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.

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

Cf. A006753.

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 *)

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

A109608 Numbers n such that the number of digits required to write the prime factors of n equals the number of digits of n.

Original entry on oeis.org

2, 3, 5, 7, 10, 11, 13, 14, 15, 17, 19, 21, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 105, 106, 107, 109, 111, 113, 115, 118, 119, 122, 123, 125, 127, 129, 131, 133, 134, 137, 139, 141, 142, 145, 146, 147, 149, 151, 155

Offset: 1

Jason Earls, Jul 31 2005

Can also be defined as numbers n such that A280827(n) = 0. - Ely Golden, Jan 08 2017

			18775 is a term because it is a 5-digit number with 5 digits in its factorization: 5*5*751 = 18775.

Ely Golden, Table of n, a(n) for n = 1..10000

Cf. A076649, A280827.

nbd(n) = my(f=factor(n)); sum(i=1, #f~, f[i,2]*#Str(f[i,1])); \\ A076649
isok(n) = nbd(n) == #Str(n); \\ Michel Marcus, Oct 11 2021

from sympy import factorint
def ok(n):
    s, f = str(n), factorint(n)
    return n and len(s) == sum(len(str(p))*f[p] for p in f)
print(list(filter(ok, range(156)))) # Michael S. Branicky, Oct 11 2021

def digits(x, n):
    if(x<=0|n<2):
        return []
    li=[]
    while(x>0):
        d=divmod(x, n)
        li.insert(0,d[1])
        x=d[0]
    return li;
def factorDigits(x, n):
    if(x<=0|n<2):
        return []
    li=[]
    f=list(factor(x))
    for c in range(len(f)):
        for d in range(f[c][1]):
            ld=digits(f[c][0], n)
            li+=ld
    return li;
def digitDiff(x,n):
    return len(factorDigits(x,n))-len(digits(x,n))
radix=10
index=1
value=2
while(index<=10000):
    if(digitDiff(value,radix)==0):
        print(str(index)+" "+str(value))
        index+=1
    value+=1
# Ely Golden, Jan 10 2017

cp's OEIS Frontend

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

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