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.

More general approach is to use generating functions. In this case, this means finding $$f(x)=g(0)+g(1)x+g(2)x^2+\dots.$$ Then perhaps from the form of $f(x)$ we could extract the coefficients in some way. To find the $f(x)$, first notice that your recurrence definition looks very similar to convolution of two series, i.e. if you multiply $\sum_{n \geq 0} a_nx^n \cdot \sum_{n \geq 0} b_nx^n=\sum_{n \geq 0} c_nx^n$, then 
$$c_n=a_0b_n+a_1b_{n-1}+\dots+a_nb_0=\sum_{i=0}^{n}a_ib_{n-i},$$ 
which suggests we should take something like $f(x)f(x)$ to get $f(x)$. To see it, notice that since $g(0)=0$, we can write the recurrence as $g(n)-1=\sum_{i=0}^{n}g\left(i\right)g\left(n-i\right)$ for $n\geq 1$. Also notice for $n=0$ the equality does not hold (left side is $-1$, right side $0$), we can correct for this by adding $1$. This means that 
$$
1+(g(0)-1)+(g(1)-1)x+(g(2)-1)x^2+\dots=f(x)f(x)=f(x)^2.
$$
Left side is now equal to $1+f(x)-(1+x+x^2+\dots)=1+f(x)-\frac{1}{1-x}=f(x)-\frac{x}{1-x}$, and so we have found out
$$
f(x)-\frac{x}{1-x}=f(x)^2.
$$
By solving this as a quadratic equation in $f$, we have
$$
f(x)=\frac{1\pm \sqrt{1-4\frac{x}{1-x}}}{2}=\frac{1}{2}\pm\frac{1}{2}\frac{\sqrt{1-5x}}{\sqrt{1-x}}.
$$
Now we expect $f(0)=g(0)=0$, so from this we see 
$$f(x)=\frac{1}{2}-\frac{1}{2}\frac{\sqrt{1-5x}}{\sqrt{1-x}}.$$
Having a generating function, we can try and deduce the coefficients in a non-recursive way. To do this, we will use series expansions that we already know, in this case binomial series is very useful since we can write $f(x)=\frac{1}{2}-\frac{1}{2}(1-5x)^{1/2}(1-x)^{-1/2}$. So by the binomial series we get
$$
(1-5x)^{1/2}=\sum_{k=0}^{\infty}\binom{1/2}{k}(-1)^k5^kx^k=1-\frac{1}{2}5x+\frac{1/2\cdot -1/2}{2}5^2x^2+\dots\\
(1-x)^{-1/2}=\sum_{k=0}^{\infty}\binom{-1/2}{k}(-1)^kx^k=1+\frac{1}{2}x+\frac{-1/2\cdot -3/2}{2}x^2+\dots\\
$$ 
and so 
\begin{align}
f(x)&=\frac{1}{2}-\frac{1}{2}\sum_{n=0}^{\infty}\sum_{i=0}^{n}\binom{1/2}{i}(-1)^i5^i\cdot \binom{-1/2}{n-i}(-1)^{n-i} x^n\\
&=-\frac{1}{2}\sum_{n=1}^{\infty}\sum_{i=0}^{n}5^i\binom{1/2}{i}(-1)^i\cdot \binom{-1/2}{n-i}(-1)^{n-i} x^n\\
&=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2}\sum_{i=0}^{n}5^i\binom{1/2}{i}\binom{-1/2}{n-i}x^n.\\
\end{align}
Now we can read off the coefficient at $x^n$ to get $g(n)$, so 
$$
\bbox[#ffd,10px]{g(n)=\frac{(-1)^{n+1}}{2}\sum_{i=0}^{n}5^i\binom{1/2}{i}\binom{-1/2}{n-i}.}
$$
You can possibly further simplify this writting out the generalized binomial coefficients inside the sum, but I wouldn't expect a closed form without the sum involved. For example, you should be able to arrive at 
$$
g(n)=\sum_{k=0}^{n-1}(-1)^k3^{n-k-1}\binom{n-1}{k}\binom{k}{\lfloor k/2 \rfloor}
$$
as mentioned in https://oeis.org/A007317, which has the additional benefit that all its terms are integers (and so might be for example more suitable for investigating divisibility properties of the sequence).
