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Let $m_i\geq 2 (1\leq i\leq n)$ be $n$ pairwisely coprime positive integers and let $q_i\geq 2 (1\leq i\leq n)$ be $n$ arbitrary prime powers, let$A=\prod_{i=1}^n(({q_i}^{m_i}-1)/(q_i-1))$. Let $\sigma(A)$ be the number of different prime factors of A, is it true that $\sigma(A)\geq n$? If this is not true, is there a counterexample? Is there a good way to estimate $\sigma(A)$?

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Bad idea to use $\sigma(A)$ for the number of prime factors of $A$ when 1) everyone else uses $\sigma$ for the sum of the divisors and 2) everyone else uses $\omega$ for the number of prime factors. – Gerry Myerson Dec 20 '10 at 21:48
If the q_i are all equal this should follow, at least modulo a few exceptions, from Zsigmondy's theorem:'s_theorem – Qiaochu Yuan Jan 4 '11 at 11:38
@Qiaochu: You don't need them to be equal, you just need them to be powers of the same prime. Also Zsigmondy's theorem helps when $m_i$'s have lots of divisors. – Gjergji Zaimi Jan 4 '11 at 12:04

No it's not true you have the following counterexample: $$ \frac{2^5-1}{2-1} \times \frac{5^3-1}{5-1} = 31^2 $$

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Let take a look to the special case when your $q_i$ are actually $n$ distinct \emph{odd} prime numbers.

I use the standard notations : $\omega(H)$ is the number of distinct prime divisors of $H$ and $\sigma(G)$ is the sum of all positive divisors of $G.$

Put $B$ the product of all the $q_i^{m_i}$

then we have

$$ \sigma(B) = mB $$

if $B$ is an odd $m$-multi-perfect number.

((sure, we do not known concrete examples of this, but...)

So, in this case

$$ \omega(B) = n $$

and you have your lower bound attained.


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However, the question asks about $\omega(\sigma(B)) = \omega(mB)$, not $\omega(B)$. – Greg Martin Sep 4 '11 at 5:43
warn: Huan used sigma'' to denote omega''. Anyway what do you think is the interesting question here ?. No feedback seems known from the OP... – Luis H Gallardo Sep 6 '11 at 18:06

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