NMO Selection Tests for JBMO

Let $\tau(n) = k$ and let $d$ be a divisor of $n$, such that $d \le \sqrt{n}$. Since $\frac{n}{d}$ also divides $n$ and $\frac{n}{d} \ge \sqrt{n}$, we infer that $\tau(n) \le 2\sqrt{n}$, for any $n \in \mathbb{N} \setminus \{0\}$, (1).

Also, if $1 = d_1 < d_2 < \dots < d_k = n$ are the divisors of $n$, then
$$2\sigma(n) = (d_1 + d_k) + (d_2 + d_{k-1}) + \dots + (d_k + d_1) = \left(d_1 + \frac{n}{d_1}\right) + \dots + \left(d_k + \frac{n}{d_k}\right).$$
But, if $d$ divides $n$, then $1 + n - d - \frac{n}{d} = \frac{n(d-1)}{d} - (d-1) = \frac{(n-d)(d-1)}{d} \ge 0$. Consequently, $\sigma(n) \le \frac{1}{2}\tau(n) \cdot (n+1)$, (2).

We infer that $n = \sigma(\tau(n)) \le \frac{1}{2}\tau(\tau(n)) \cdot (\tau(n)+1) \le \frac{1}{2} \cdot 2\sqrt{\tau(n)} \cdot (2\sqrt{n}+1) \le \sqrt{2\sqrt{n}} \cdot (2\sqrt{n}+1)$.

Successively, this leads to
$$n^2 \le 2\sqrt{n}(4n + 4\sqrt{n} + 1) \Leftrightarrow n\sqrt{n} \le 2(4n + 4\sqrt{n} + 1) \Leftrightarrow \sqrt{n} \le 8 + \frac{8}{\sqrt{n}} + \frac{2}{n}.$$
Suppose, by contradiction, that $n \ge 81$. Then $\sqrt{n} \ge 9$ and $8 + \frac{8}{9} + \frac{2}{81} < 9 \le \sqrt{n}$, which is a contradiction. Hence, $n \le 80$, which means that $n$ can only have one of the following forms: $1, p, p^2, p^3, p^4, p^5, pq, p^2q, p^3q, p^4q, p^2q^2, p^3q^2, pqr, p^2qr$, where $p, q, r$ are prime numbers.

Every such possible $n$ has at most 12 divisors. Studying all cases for $\tau(n) \in \{1, 2, \dots, 12\}$, we easily find that the only solutions are $n = 1, n = 3, n = 4$ and $n = 12$.