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

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

Author

Jason Earls, Jul 31 2005

Keywords

• base
• easy
• nonn

Comments

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

Examples

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

Links

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

Crossrefs

Cf. A076649, A280827.

Programs

PARI

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

Python

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

SageMath

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