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A109608
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Numbers n such that the number of digits required to write the prime factors of n equals the number of digits of n.
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3
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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
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,1
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COMMENTS
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LINKS
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EXAMPLE
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18775 is a term because it is a 5-digit number with 5 digits in its factorization: 5*5*751 = 18775.
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PROG
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(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
(PARI) nbd(n) = my(f=factor(n)); sum(i=1, #f~, f[i, 2]*#Str(f[i, 1])); \\A076649
(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)
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CROSSREFS
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KEYWORD
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base,easy,nonn
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AUTHOR
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STATUS
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approved
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