, 2006

Let $g(n)=n f(n)$ for $n \geq 1$ and $g(0)=0$. We note that, for $k=1, \ldots, n$,
$$\left\lfloor\frac{n}{k}\right\rfloor-\left\lfloor\frac{n-1}{k}\right\rfloor=0$$
if $k$ is not a divisor of $n$ and
$$\left\lfloor\frac{n}{k}\right\rfloor-\left\lfloor\frac{n-1}{k}\right\rfloor=1$$
if $k$ divides $n$. It therefore follows that if $d(n)$ is the number of positive divisors of $n \geq 1$ then
$$\begin{aligned} g(n) & =\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+\cdots+\left\lfloor\frac{n}{n-1}\right\rfloor+\left\lfloor\frac{n}{n}\right\rfloor \\ & =\left\lfloor\frac{n-1}{1}\right\rfloor+\left\lfloor\frac{n-1}{2}\right\rfloor+\cdots+\left\lfloor\frac{n-1}{n-1}\right\rfloor+\left\lfloor\frac{n-1}{n}\right\rfloor+d(n) \\ & =g(n-1)+d(n) \end{aligned}$$
Hence
$$g(n)=g(n-1)+d(n)=g(n-2)+d(n-1)+d(n)=\cdots=d(1)+d(2)+\cdots+d(n)$$
meaning that
$$f(n)=\frac{d(1)+d(2)+\cdots+d(n)}{n} .$$
In other words, $f(n)$ is equal to the arithmetic mean of $d(1), d(2), \ldots, d(n)$. In order to prove the claims, it is therefore sufficient to show that $d(n+1)>f(n)$ and $d(n+1)<f(n)$ both hold infinitely often.

We note that $d(1)=1$. For $n>1, d(n) \geq 2$ holds, with equality if and only if $n$ is prime. Since $f(6)=7 / 3>2$, it follows that $f(n)>2$ holds for all $n \geq 6$.

Since there are infinitely many primes, $d(n+1)=2$ holds for infinitely many values of $n$, and for each such $n \geq 6$ we have $d(n+1)=2<f(n)$. This proves claim (b).

To prove (a), notice that the sequence $d(1), d(2), d(3), \ldots$ is unbounded (e. g. $d\left(2^{k}\right)=k+1$ for all $k$ ). Hence $d(n+1)>\max \{d(1), d(2), \ldots, d(n)\}$ for infinitely many $n$. For all such $n$, we have $d(n+1)>f(n)$. This completes the solution.