Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)
share|improve this question
    
In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. If this is homework, please add the [homework] tag. Also, many find the use of imperative ("Find") to be rude when asking for help; please consider rewriting your post. –  Zev Chonoles Jun 2 '12 at 5:08
1  
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.