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

1,1

Comments

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

Links

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

Example

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

Prog

(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
(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

Crossrefs

Cf. A076649, A280827.

Sequence in context: A122428 A087246 A090421 * A325388 A325405 A118241

Adjacent sequences: A109605 A109606 A109607 * A109609 A109610 A109611

Keyword

base,easy,nonn

Author

Jason Earls, Jul 31 2005

Status

approved