# sum of $n^{th}$ powers of prime factors of $x$

Starting with a positive integer $x$, find the sum of the $n^{th}$ powers of the prime factors of $x$, including multiplicities. Then find the sum of the $n^{th}$ prime factors of the result etc. until an $n^{th}$ power of a prime is reached. Will the sequence always terminate, or can it get caught in a loop or diverge to infinity?

python code for creating this sequence:

def factor(n):
m,p,r,k=n,3,7,[]
while m%2==0:
k.append(2)
r=1
m=m/2
while p<=n**0.5 and m!=1:
if m%p==0:
k.append(p)
r=1
m=m/p
else:
p+=2
if r==1:
while p<=n and m!=1:
if m%p==0:
k.append(p)
m=m/p
else:
p+=2
return k
y=1
x,n=int(input("start")),int(input("power"))
while x!=y and x!=0:
y,x=x,0
t=factor(y)
for e in t:
x+=e**n
print(x)

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It is only a puzzle I came up with, and I don't have any ideas for $n>1$. – Angela Richardson Jun 2 '12 at 5:12
Your code doesn't seem to match the question. In the program it excludes primes that appear with multiplicity>1, i.e. unless t.count(e)==1. For example, with start=198 and power=2 it goes to $2^2+11^2=125$ and terminates, whereas from the question statement I would expect the next step to be $2^2+3^2+3^2+11^2=143$. – Zander Jun 2 '12 at 11:02