Largest prime divisor of a sum - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T00:50:44Z http://mathoverflow.net/feeds/question/60684 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60684/largest-prime-divisor-of-a-sum Largest prime divisor of a sum Woett 2011-04-05T13:28:38Z 2012-01-28T00:22:12Z <p>Let $\{r_i\}_{i \in \mathbb{N}}$ be a periodic sequence of integers. That is, there exists a $t$, such that, for all $i \in \mathbb{N}$, $r_i = r_{i+t}$. What I need is a large prime dividing the numerator of $\displaystyle \sum_{i=1}^n \dfrac{r_i}{i}$. I general, I'd like to know more about the largest prime dividing that sum as $n$ goes to infinity. In particular, I would love to know a proof (or even more spectacular, a counterexample) to the following conjecture:</p> <p>For every $c \in \mathbb{R}$, there exists a positive integer $n \ge 2$ and a prime $p > cn$ that divides the numerator of $\displaystyle \sum_{i=1}^n \dfrac{r_i}{i}$</p> <p>Furthermore, I metaconjecture that my effort over the last 6 months to prove this, will be nullified within an hour or two. </p> <p>If at least one of my conjectures turns out to be false, for which $c > 0$ is my first conjecture true and easily provable?</p> http://mathoverflow.net/questions/60684/largest-prime-divisor-of-a-sum/60808#60808 Answer by Woett for Largest prime divisor of a sum Woett 2011-04-06T11:59:14Z 2011-11-04T21:42:43Z <p>This is partially an answer to Chuck.</p> <p>The denominator is divisible by all primes $p$ such that $\dfrac{n}{\log{n}} \ll p &lt; n$, so the denominator grows exponentially in $n$. Furthermore, I believe (and have been too lazy and incompetent to check) that the sum is (in absolute value) $\gg n^{-t}$. This implies that the numerator also grows exponentially in $n$. So to me this shows that the problem should be easy; how can there ever <em>not</em> be a prime $> cn$ dividing somethig like $e^n$? This is espeically true for $c=1$, since if we know that the numerator is only divisible by primes smaller than $n$, we actually know that all the prime divisors of it must be $\ll \dfrac{n}{\log{n}}$. So, since Gerry pointed out that my metaconjecture has been refuted, I'd like to coin a new one: for $c=1$ the conjecture is provable by a simple counting argument.</p>