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