Problem 1191

I. Let $p>b$ be prime, let $n=p^{3}$ and $k=p^{2}$. If $p^{3}b L\left(p^{3}+1, p^{2}\right)$.

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 $m !, m !+1, \ldots, m !+m-1$. It follows that $L(m !-1, m+1)=(m !-1) M$, where $M=\operatorname{lcm}(m !, m !+1, \ldots, m !+m-1)$.

Now consider $L(m !, m+1)$. This is $\operatorname{lcm}(M, m !+m)$. But $m !+m=m((m-1) !+1)$, and $m$ divides $M$. Thus $\operatorname{lcm}(M, m !+m) \leq M((m-1) !+1)$, and

$$\frac{L(m !-1, m+1)}{L(m !, m+1)} \geq \frac{m !-1}{(m-1) !+1}$$

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.

To show that for any integer $b$, there exist integers $n$ and $k$ such that $L(n,k) > bL(n+1,k)$, we will proceed as follows:

1. **Choose a Prime $p$:**
Let $p$ be an odd prime such that $p > \max(b, 12)$. This ensures that $p$ is sufficiently large to satisfy the conditions of the problem.

2. **Select an Odd Integer $m$:**
Choose $m$ to be an odd positive integer such that $\frac{p}{2} > m > \frac{p}{3}$. Such an $m$ exists because $p > 12$.

3. **Define $n$ and $k$:**
Set $n = p$ and $k = 4m - p$. This choice of $k$ ensures that $k$ is positive and satisfies the conditions needed for the proof.

4. **Analyze $L(n,k)$:**
Let $A = \text{lcm}(p+1, p+2, \ldots, p+k-1)$. Since $m < \frac{p}{2}$, we have $p + k - 1 < 2p$. Therefore, $p$ does not divide $A$, and thus $L(n,k) = \text{lcm}(p, A) = pA$.

5. **Ensure $k-1 \geq 4$:**
Since $k = 4m - p$, we have $k - 1 = 4m - p - 1 \geq m > \frac{p}{3} \geq 4$. This ensures that in the $k-1$ numbers $p+1, p+2, \ldots, p+k-1$, there is at least one multiple of 4 and one multiple of $m$.

6. **Multiple of 4 and $m$:**
Since $4m$ divides $A$ (because $m$ is odd and there is at least one multiple of 4 in the range), we have $L(n+1,k) = \text{lcm}(A, 4m) = A$.

7. **Compare $L(n,k)$ and $L(n+1,k)$:**
Therefore, $L(n,k) = pL(n+1,k)$. Since $p > b$, we have $pL(n+1,k) > bL(n+1,k)$.

Thus, we have shown that $L(n,k) > bL(n+1,k)$.

$\blacksquare$