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Topic: Sigma(n) and r(n) (Read 709 times) |
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ThudnBlunder
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Sigma(n) and r(n)
« on: Mar 15th, 2009, 6:36pm » |
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Let sigma(n) denote the sum of the positive integer n
and
Let r(n) be the number of such divisors.
Prove sigma(n) is prime implies r(n) is also prime.
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Immanuel_Bonfils
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Re: Sigma(n) and r(n)
« Reply #1 on: Mar 15th, 2009, 6:49pm » |
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Please, what is sum of the positive integer and the number of such divisors?
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ThudnBlunder
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Re: Sigma(n) and r(n)
« Reply #2 on: Mar 15th, 2009, 7:14pm » |
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on Mar 15th, 2009, 6:49pm, Immanuel_Bonfils wrote:| Please, what is sum of the positive integer and the number of such divisors? |
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Sorry,
let sigma(n) denote the sum of the positive divisors of the positive integer n
and
let r(n) be the number of such divisors.
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| « Last Edit: Mar 17th, 2009, 2:41am by ThudnBlunder » |
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Eigenray
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Re: Sigma(n) and r(n)
« Reply #3 on: Mar 18th, 2009, 7:50am » |
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| hidden: | Since sigma is multiplicative, and sigma(n)>1 for n>1, it's clear that if sigma(n) is prime, then n=pk is a prime power, in which case r(n) = k+1. Now if k+1 = ab, with a,b > 1, then
sigma(n) = 1+p+p2+...+pab
= (1+p+..+pa-1)(1+pa+...+pa(b-1))
would be composite, a contradiction. |
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