most recent 30 from http://mathoverflow.net 2013-06-20T08:27:03Z http://mathoverflow.net/feeds/question/49961 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49961/the-number-of-different-prime-factors-of-a-special-class-of-positive-integers Huan Xiong 2010-12-20T13:28:49Z 2011-02-01T13:02:59Z

<p>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 $\sigma(A)$?</p>

http://mathoverflow.net/questions/49961/the-number-of-different-prime-factors-of-a-special-class-of-positive-integers/49968#49968 Luis H Gallardo 2010-12-20T15:08:47Z 2010-12-20T15:08:47Z

<p>Let take a look to the special case when your $q_i$ are actually $n$ distinct \emph{odd} prime numbers.</p> <p>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 $G.$</p> <p>Put $B$ the product of all the $q_i^{m_i}$</p> <p>then we have </p> <p>$$ \sigma(B) = mB $$</p> <p>if $B$ is an odd $m$-multi-perfect number.</p> <p>((sure, we do not known concrete examples of this, but...)</p> <p>So, in this case</p> <p>$$ \omega(B) = n $$</p> <p>and you have your lower bound attained.</p> <p>luis</p> <hr>

http://mathoverflow.net/questions/49961/the-number-of-different-prime-factors-of-a-special-class-of-positive-integers/53985#53985 Esteban Crespi 2011-02-01T13:02:59Z 2011-02-01T13:02:59Z

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