What is the closed form of this quickly growing sequence? Is there a pattern in the following sequence:
$$3,19,451,22051,\dots$$
I tried setting some $f(n)$ to this sequence, where $n$ is the term number. Then I basically treated it like a matrix equation: multiplying, adding, subtracting and dividing to hopefully obtain a simple, known result. Like the Fibonacci sequence, or something else of the sort. I've just put some randomly growing numbers together in hopes of finding a representation of some kind. This is not homework.
 A: If I subtract $1$ from each term, I see $2,18,450,22050,...$. The ratio of successive terms is $9,25,49,...$
And now I think I see it : the ratio of successive terms, minus $1$, gives the list of odd squares, but starting from $3^2 = 9$.
In other words, if $f(n)$ denotes the $n$th number of the sequence, then $(f(n)-1) = (f(n-1) - 1)(2n-1)^2$, along with $f(1) = 3$. For example, $$
f(2) - 1 = (f(1) - 1) \times 9 = 18 \\
f(3) -1 = (f(2) - 1) \times 25 = 450 \\
\vdots
$$
It follows from here that $$(f(n) - 1) = (f(1) - 1) \times 9 \times 25 \times ... \times (2n-1)^2 \\ = 2 \times \prod_{k=1}^n (2k-1)^2$$.
(The $k=1$ just gives $1$ so we include it).
Turns out this can be simplified :
\begin{align}
\prod_{k=1}^n (2k-1) & = 1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1) \\  & = \frac{1 \cdot 2 \cdot 3 \cdot ... \cdot 2k}{2 \cdot 4 \cdot 6 \cdot ... \cdot 2k} \\ &= \frac{(2k)!}{2^kk!}
\end{align}
So, after squaring and completing the formalities :
$$\bbox[yellow,5px,border:2px solid red]{
\color{brown}{f(n) = 1 + \frac{((2n)!)^2}{2^{2n-1}(n!)^2}}}
$$
Trying this out: at $n=1$, we have $1+\frac{2!^2}{2^11!^2} = 1+2 = 3$.
At $n=2$ we have $1 + \frac{4!^2}{2^32!^2} = 19$.
At $n=3$ we have $1 + \frac{6!^2}{2^53!^2} = 451$.
And so on.
