## Summary

If you haven’t read everything I wrote last week, but would still like to solve an Erdös-Graham-problem or two with me, here is, basically, everything I know so far:

Let $n$ be a given natural number and $p$ be a prime smaller than $n$. Let $a \in \aleph$ be the unique number smaller than $p$ such that $ap^k \le n < (a+1)p^k$ holds for some $k \in \aleph_0$. Let $L_n$ be the least common multiple of $\{1,2,..,n\}$ and $X_n$ be such that $\dfrac{X_n}{L_n} = \displaystyle \sum_{i=1}^n \dfrac{1}{i}$. And, at last, let $u_{y,z}$ and $v_{y,z}$ be coprime positive integers, such that, for $y,z \in \aleph$$\displaystyle \sum_{i=y}^z \dfrac{1}{i}$ $= \dfrac{u_{y,z}}{v_{y,z}}$. Then:

$\text{ }$

1) $p | X_n$, ﻿﻿﻿﻿ ﻿﻿$p | X_a$ and $p | X_{p-1-a}$ are all equivalent

2) If $p > (1 + o(1))\dfrac{n}{\log{n}}$, then $p$ does not divide $X_n$

3) If $n$ is not a prime power, $X_n \neq X_{n-1} \pmod{p}$ implies $n = ap^k$

4) For every fixed $y$, there is a $z \le 6y$, for which $v_{y,z} < v_{y,z-1}$

$\text{ }$

Btw, one of the problems is: prove or disprove that the greatest common divisor of $X_n$ and $L_n$ is equal to $1$ for infinitely many $n$. Another one is to find the smallest (or at least a better upper bound for) $z = z(y)$, such that $v_{y,z} < v_{y,z-1}$.