, 2012

I. Let $p > b$ be prime, let $n = p^3$ and $k = p^2$. If $p^3 < i < p^3 + p^2$, then no power of $p$ greater than 1 divides $i$, while $p$ divides $p^3 + p$. It follows that $L(p^3, p^2) = p^2 L(p^3 + 1, p^2 - 1)$. A similar calculation shows that $L(p^3 + 1, p^2) = p L(p^3 + 1, p^2 - 1)$. Thus $L(p^3, p^2) = p L(p^3 + 1, p^2) > b L(p^3 + 1, p^2)$.

II. Let $m > 1$. Then $L(m! - 1, m + 1)$ is the least common multiple of the integers from $m! - 1$ to $m! + m - 1$. But $m! - 1$ is relatively prime to all of $m!$, $m! + 1, \dots, m! + m - 1$. It follows that $L(m! - 1, m + 1) = (m! - 1) M$, where $M = \text{lcm}(m!, m! + 1, \dots, m! + m - 1)$.
Now consider $L(m!, m + 1)$. This is $\text{lcm}(M, m! + m)$. But $m! + m = m((m - 1)! + 1)$, and $m$ divides $M$. Thus $\text{lcm}(M, m! + m) \le M((m - 1)! + 1)$, and
$$\frac{L(m! - 1, m + 1)}{L(m!, m + 1)} \ge \frac{m! - 1}{(m - 1)! + 1}.$$
Since $m$ can be arbitrarily large, so can $L(m! - 1, m + 1)/L(m!, m + 1)$. Therefore taking $n = m! - 1$ for sufficiently large $m$, and $k = m + 1$, works.