Conditions for 'extraordinary' natural numbers

Question

We say that a composite number is 'extraordinary' if: $$\tau((n - 1)!) \ | \ \tau(n!)$$ and $\tau : \mathbb{N}^* \to \mathbb{N}^*$ represents the arithmetic function called 'number of divisors'. Show that are finitely many 'extraordinary numbers'

My ideea was that if: $$n = \prod p_i^{\beta_i}$$

We may define: $$\alpha_i = v_{p_i} ((n - 1)!)$$

So an 'extraordinary' number has the following property: $$\frac{\prod (\alpha_i + \beta_i)}{\prod \alpha_i} \in \mathbb{N}^*$$

So, it turns into a product divisibility problem, however, I couldn't link the idea of $(n - 1)!$ being a factorial number so that I could proceed with my solution by finding some conditions for $(\alpha_i)$ and $(\beta_i)$.

But we have: Theorem. For a natural number $n$, we have $\tau(n!)=2\tau((n-1)!)$ iff $n$ is prime or $n=4$., see here. So all prime numbers $n=p$ are extraordinary, hence infinitely many already. Where is the mistake? — Dietrich Burde, Mar 01 '22 at 19:13
@DietrichBurde I'm sorry! I had a mistake. $n$ is 'extraordinary' if $n$ is also composite. — MathStackExchange, Mar 01 '22 at 20:12
@janmarqz $v_p(n)$ := the exponent of prime number $p$ in number $n$. It is commonly called 'p-adic (e)valuation' — MathStackExchange, Mar 01 '22 at 20:14

score 1 · Accepted Answer · answered Mar 02 '22 at 04:30

Let $n=p_1^{\alpha_1}..p_k^{\alpha_k}$ where either $k \ge 2$ or $k=1, \alpha_1 \ge 2$; we will show that if $(n-1)!=p_1^{\beta_1}..p_k^{\beta_k}q, (q,n)=1$, then $\beta_r \ge \alpha_r$ unless $n=4$.

This means that if $n \ne 4$ any divisor of $n!$ that is not a divisor of $(n-1)!$ which perforce must be $p_1^{\beta_1+\delta_1}..p_k^{\beta_k+\delta_k}s, 0 \le \delta_k \le \alpha_k, s|q$ with at least one $\delta_r >0$, can be associated with $p_1^{\beta_1-\delta_1}..p_k^{\beta_k-\delta_k}s$ which is a well defined divisor of $(n-1)!$ and clearly the association is injective and misses $(n-1)!$ as at least one of the $\delta_r \ge 1$. But this means that $\tau(n!) \le 2\tau(n-1)!-1$ and since $\tau(n!) >\tau(n-1)!$ the result follows unless $n=4$ (and of course $n=p$ prime which is excluded by hypothesis)

Notice that if $k \ge 2, p_r^{\alpha_r} \le n-1$ so $\beta_r \ge \alpha_r$, while if $k=1, \alpha_1=\alpha \ge 2$ and $p_1=p \ge 3, n=p^{\alpha}$, then $p^{\alpha-1} \ne 2p \le n-1$ so $\beta \ge \alpha$ while if $p=2, \alpha \ge 3, n \ge 8$, one has $2^{\alpha-1} \ne 6 \le n-1$ so $\beta \ge \alpha$ in this case too and we are done!

Conditions for 'extraordinary' natural numbers

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