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

%I A109608 #21 Oct 11 2021 08:46:20
%S A109608 2,3,5,7,10,11,13,14,15,17,19,21,23,25,29,31,35,37,41,43,47,49,53,59,
%T A109608 61,67,71,73,79,83,89,97,101,103,105,106,107,109,111,113,115,118,119,
%U A109608 122,123,125,127,129,131,133,134,137,139,141,142,145,146,147,149,151,155
%N A109608 Numbers n such that the number of digits required to write the prime factors of n equals the number of digits of n.
%C A109608 Can also be defined as numbers n such that A280827(n) = 0. - _Ely Golden_, Jan 08 2017
%H A109608 Ely Golden, <a href="/A109608/b109608.txt">Table of n, a(n) for n = 1..10000</a>
%e A109608 18775 is a term because it is a 5-digit number with 5 digits in its factorization: 5*5*751 = 18775.
%o A109608 (SageMath)
%o A109608 def digits(x, n):
%o A109608     if(x<=0|n<2):
%o A109608         return []
%o A109608     li=[]
%o A109608     while(x>0):
%o A109608         d=divmod(x, n)
%o A109608         li.insert(0,d[1])
%o A109608         x=d[0]
%o A109608     return li;
%o A109608 def factorDigits(x, n):
%o A109608     if(x<=0|n<2):
%o A109608         return []
%o A109608     li=[]
%o A109608     f=list(factor(x))
%o A109608     for c in range(len(f)):
%o A109608         for d in range(f[c][1]):
%o A109608             ld=digits(f[c][0], n)
%o A109608             li+=ld
%o A109608     return li;
%o A109608 def digitDiff(x,n):
%o A109608     return len(factorDigits(x,n))-len(digits(x,n))
%o A109608 radix=10
%o A109608 index=1
%o A109608 value=2
%o A109608 while(index<=10000):
%o A109608     if(digitDiff(value,radix)==0):
%o A109608         print(str(index)+" "+str(value))
%o A109608         index+=1
%o A109608     value+=1
%o A109608 # _Ely Golden_, Jan 10 2017
%o A109608 (PARI) nbd(n) = my(f=factor(n)); sum(i=1, #f~, f[i,2]*#Str(f[i,1])); \\ A076649
%o A109608 isok(n) = nbd(n) == #Str(n); \\ _Michel Marcus_, Oct 11 2021
%o A109608 (Python)
%o A109608 from sympy import factorint
%o A109608 def ok(n):
%o A109608     s, f = str(n), factorint(n)
%o A109608     return n and len(s) == sum(len(str(p))*f[p] for p in f)
%o A109608 print(list(filter(ok, range(156)))) # _Michael S. Branicky_, Oct 11 2021
%Y A109608 Cf. A076649, A280827.
%K A109608 base,easy,nonn
%O A109608 1,1
%A A109608 _Jason Earls_, Jul 31 2005