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.
 A: The Central Binomial Coefficients
$$
\binom{2n-2}{n-1}
$$
seem to fit.
Edit: As Nikolay Gromov comments, this is the same sequence as given in his answer, though it is not immediately apparent. This is because
$$
(2n-3)!!=\frac{(2n-2)!}{2^{n-1}(n-1)!}
$$
so that
$$
\frac{2^{n-1}(2n-3)!!}{(n-1)!}=\frac{(2n-2)!}{(n-1)!(n-1)!}=\binom{2n-2}{n-1}
$$
A: 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
A: What about this?
$$\frac{2^{n-1} (2 n-3)\text{!!}}{(n-1)!}$$
which gives
$$
\begin{array}{ccl}
 f(1) & = & 1 \\
 f(2) & = & 2 \\
 f(3) & = & 6 \\
 f(4) & = & 20 \\
 f(5) & = & 70 \\
 f(6) & = & 252 \\
 f(7) & = & 924 \\
 f(8) & = & 3432 \\
 f(9) & = & 12870 \\
 f(10) & = & 48620 \\
 f(11) & = & 184756 \\
 f(12) & = & 705432 \\
 f(13) & = & 2704156 \\
 f(14) & = & 10400600 \\
 f(15) & = & 40116600 \\
 f(16) & = & 155117520 \\
 f(17) & = & 601080390 \\
 f(18) & = & 2333606220 \\
 f(19) & = & 9075135300 \\
 f(20) & = & 35345263800 \\
\end{array}
$$
