Limited Proxies

$a_n=f(n)=n-2$

Hence equation translates to:

$n-2=\dfrac{(n-2)!}{(n-3)(n-4)(n-5)(n-6)}$

$\implies \cancel{n-2}=\dfrac{\cancel{(n-2)(n-3)(n-4)(n-5)(n-6)}(n-7)!}{\cancel{(n-3)(n-4)(n-5)(n-6)}}$

$\implies (n-7)!=1$

This has two solutions $n=7,8$ larger of which is $\boxed{\color{#007fff}{8}}$

$\begin{array} {c} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \\ a_n & 1 & 1 & 1 & 1 & 2 & 3 & 4 & 5 & 6 & 14 & 24 & 36 & 50 & 66 & 168 & 312 & 504 & 750 & 1056 & 2856 \end{array}$

Observing the terms early in the sequence, we note that $a_n = (n-2)a_{n-5}=a_{n-5}f(n)$ . Let us prove the claim for all $n \ge 5$ by induction. We first note that $a_4 = \dfrac {2!}{1\cdot 1\cdot1\cdot 1} = 2$ .

The proof

1. For $n = 5$ , as given $a_5 = \dfrac {3!}{2\cdot 1\cdot1\cdot 1} = 3$ . By the claim $a_5 = (5-2)a_0 = 3$ . Therefore, the claim is true for $n = 5$ .

2. Assuming the claim is true for $n$ , for $n+1$ , we have:

$\begin{aligned} \quad \quad a_{n+1} & = \frac {(n-1)!}{a_n a_{n-1} a_{n-2} a_{n-3}} \\ & = \frac {(n-1)(n-2)!a_{n-4}}{a_n a_{n-1} a_{n-2} a_{n-3}a_{n-4}} \\ & = (n-1)a_{n-4} \\ & = (\color{#3D99F6}{n+1}-2)a_{\color{#3D99F6}{n+1}-5}\end{aligned}$

$\quad$ Therefore, the claim is true for $n+1$ and hence true for all $n \ge 5$ .

Now, we have $a_n = a_{n-5}f(n)$ , for $a_n = f(n)$ , $a_{n-5}$ must be equal to 1, the latest term is $a_3=1$ , that is $n=\boxed{8}$ .