<s>[INST] figuring out an integer function $f(1) = 1\\
f(2) = 2\\
f(3) = 6\\
f(4) = 20\\
f(5) = 70\\
f(6) = 252\\
f(7) = 924\\
f(8) = 3432\\
f(9) = 12870$
Then what is $f(n)$ (where $n > 0$)?
I though about many many possibilities but still cannot figure out the expression.
 [/INST] You could approach this kind of question in this way. You'd like to find what is the relation between $f_n$ and $f_{n+1}$ (you believe that there is some)  so that's why most of the time it's a good idea to analyze expressions like $\frac{f_{n+1}}{f_n}$ or $f_{n+1}-f_n$. Let's look at $\frac{f_{n+1}}{f_n}$. You may notice that $\frac{f_{n+1}}{f_{n}}\sim 4$ as the sequence grows, more precisely:
$$
\begin{array}{ccl}
 \frac{f_2}{f_1}& = & 2 \\
 \frac{f_3}{f_2} & = & 3 \\
 \frac{f_4}{f_3} & = &3.(3) \\
 \frac{f_5}{f_4} & = & 3.5 \\
 \frac{f_6}{f_5} & = & 3.6 \\
 \frac{f_7}{f_6} & = & 3.(6)\\
 \frac{f_8}{f_7} & = & 3.(714285) \\
 \frac{f_9}{f_8} & = & 3.75 \\
\end{array}
$$
So you might expect that $\frac{f_{n+1}}{f_{n}} = 4 - g_n$ where $g_n \to 0$ as $n\to +\infty$. Upon more precise calculation we get: 
$$
\begin{array}{ccl}
 \frac{f_2}{f_1}& = & 4-\frac{2}{1} \\
 \frac{f_3}{f_2} & = & 4-\frac{2}{2} \\
 \frac{f_4}{f_3} & = &4-\frac{2}{3} \\
 \frac{f_5}{f_4} & = &4-\frac{2}{4} \\
 \frac{f_6}{f_5} & = &4-\frac{2}{5} \\
 \frac{f_7}{f_6} & = & 4-\frac{2}{6}\\
 \frac{f_8}{f_7} & = & 4-\frac{2}{7} \\
 \frac{f_9}{f_8} & = & 4-\frac{2}{8} \\
\end{array}
$$
So we could see that $\frac{f_{n+1}}{f_n} = 4-\frac{2}{n} = \frac{2(2n-1)}{n}$. From here 
$$\frac{f_{n+1}}{1} =\frac{f_{n+1}}{ f_n } \times \frac{f_{n}}{f_{n-1}} \times \dots \times \frac{f_2}{f_1} = \frac{2(2n-1)}{n}\times \frac{2(2n-3)}{n-1} \dots  = \frac{2^n(2n-1)!!}{n!}$$
which is $\binom{2n}{n}$ as already mentioned
</s>