Finding the closed-form formula to a recurrence with summation of terms I am currently trying to work through this recurrence problem but am having a hard time coming up with the solution:
$g\left(n\right)=\left(\sum_{i=1}^{n-1}g\left(i\right)g\left(n-i\right)\right)+1$
Where the base $g(0)=0$.
One thing I noticed was that the values involved in the summation are sort of symmetrical, there would be a $g(2)g(3)$ as well as a $g(3)g(2)$ and this duplicate value would exist for each value of i except for when $i=n-i$, which I wanted to take advantage of. I typically approach closed-form recurrences by creating a polynomial in which I would take a guess for a function and solve for its constants but Im not sure how to extend this idea to something like this.

We can obtain a generating function for your sequence. First we modify the recurrence relation to state $$g(n) = 1 + \sum_{i=0}^ng(i)g(n-i).$$ Note that this gives the same relation because $g(0)=0$. Also note that this equality does not hold for $n=0$ (that gives $g(0) = g(0)^2 + 1$). Now we define the generating function.
Let
$$f(x) = \sum_{n=0}^\infty g(n) \cdot x^n.$$
Then, by taking the Cauchy product of $f$ with itself we see:
\begin{align*}
f(x)^2 &= \sum_{n=0}^\infty \left(\sum_{i=0}^n g(n-i)g(i)\right)x^n\\
&= \sum_{n=1}^\infty \left(\sum_{i=0}^n g(n-i)g(i)\right)x^n\\
&= \sum_{n=1}^\infty \left(g(n) - 1\right)x^n\\
&= \sum_{n=1}^\infty g(n)x^n - \sum_{n=1}^\infty x^n\\
&= f(x) - \frac{x}{1-x}.
\end{align*}
We obtain that the generating function satisfies the relation 
$$f(x)^2 = f(x) - \frac{x}{1-x}.$$ There are two functions satisfying this relation. Namely:
$$f(x) = \frac{1}{2} \left(1-\frac{\sqrt{1-5 x}}{\sqrt{1-x}}\right) \quad\text{ or }\quad f(x) = \frac{1}{2} \left(\frac{\sqrt{1-5 x}}{\sqrt{1-x}}+1\right).$$ 
Only the first choice has its zeroth term equal to zero and thus we conclude that our generating function is $$f(x) = \frac{1}{2} \left(1-\frac{\sqrt{1-5 x}}{\sqrt{1-x}}\right) = x + 2x^2 + 5x^3 + 15x^4 + \cdots.$$ As to a closed form of your sequence. We have now shown that your sequence (up to a different zeroth term because they record the coefficients of $f(x)+1$) is sequence A181768 in the OEIS. You can proceed your research there. The site gives several other ways to calculate the terms of the sequence (see also A007317). A simple formula might be a bit much to ask.
