Don't use your calculator!

We start by pointing out the case when $n<10$ and $n>19$ . If $n<10$ , then we have that $n+1 < 11 < 20,$ $1 < \frac{11}{n+1} < \frac{20}{n+1}.$ Therefore, $A_{n+1}=\frac{20^{n+1}+11^{n+1}}{(n+1)!} = \frac{20}{n+1} \cdot \frac{20^n}{n!} + \frac{11}{n+1} \cdot \frac{11^n}{n!} > \frac{20^n}{n!}+\frac{11^n}{n!} = A_n.$ It means that $A_1 < A_2 < A_3 < \ldots < A_9 < A_{10}$ . Similarly, if $n > 19$ , we will finally have a converse that $A_{n+1} < A_n$ . Therefore, in the range $n > 19$ , this sequence is decreasing, that is, $A_{20} > A_{21} > A_{22} > \cdots$ . In conclusion, the biggest $A_n$ must be among $A_{10}, A_{11}, \ldots, A_{19},A_{20}$ . The next step is to find relations between these numbers.

Let's say $A_{n+1} > A_n$ , it is true if and only if $\frac{20^{n+1}+11^{n+1}}{(n+1)!} > \frac{20^n+11^n}{n!}$ iff $20^{n+1} + 11^{n+1} > (n+1)(20^n+11^n)$ iff $(19-n) 20^n > (n-10) 11^n.$ It is obvious that $20^n > 11^n$ for all positive integer $n$ . Thus, whenever $19-n > n-10$ , this inequality always holds. It is equivalent to $n < 14.5$ , so we have that $A_{10} < A_{11} < \ldots < A_{15}$ . Now, there are $A_{15}, A_{16}, \ldots, A_{20}$ left.

Moreover, this inequality is equivalent to $(7600-400n) 20^{n-2} > (n-10) 11^n,$ and it is not hard to prove that $20^{n-2} > 11^n$ for all $n \ge 12$ (the proof will be given afterwards). So, whenever $7600-400n > n-10$ , this inequality always holds. We simply get $n<\frac{7610}{401}$ . This works for $n < 19$ , so we have that $A_{15}<A_{16}<\ldots<A_{19}$ . Therefore, we have two candidates left, which are $A_{19}$ and $A_{20}$ .

Lastly, we find that $A_{20} = \frac{20^{20}+11^{20}}{20!} = \frac{20^{19}}{19!}+\frac{11}{20} \cdot \frac{11^{19}}{19!} < \frac{20^{19}}{19!}+\frac{11^{19}}{19!} = A_{19}.$ This concludes that $\boxed{A_{19}}$ is the biggest number in the sequence.

Here's the proof that $20^{n-2} > 11^n$ for all $n \ge 12$ . This inequality is equivalent to $\left( \frac{20}{11} \right)^{n-2} > 121.$ So, if we have that $n \ge 12$ , then $\left( \frac{20}{11} \right)^{n-2} \ge \left( \frac{20}{11} \right)^{10} > (1.8)^{10} = (3.24)^5 > 3^5 = 243 > 121,$ and we are done.