What's the formula for this series for $\pi$? These continued fractions for $\pi$ were given here,
$$\small
\pi = \cfrac{4} {1+\cfrac{1^2} {2+\cfrac{3^2} {2+\cfrac{5^2} {2+\ddots}}}}
= \sum_{n=0}^\infty \frac{4(-1)^n}{2n+1} 
= \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots\tag1
$$
$$\small
\pi = 3 + \cfrac{1^2} {6+\cfrac{3^2} {6+\cfrac{5^2} {6+\ddots}}}
= 3 - \sum_{n=1}^\infty \frac{(-1)^n} {n (n+1) (2n+1)} 
= 3 + \frac{1}{1\cdot 2\cdot 3} - \frac{1}{2\cdot 3\cdot 5} + \frac{1}{3\cdot 4\cdot 7} - \cdots\tag2
$$
$$\small
\pi = \cfrac{4} {1+\cfrac{1^2} {3+\cfrac{2^2} {5+\cfrac{3^2} {7+\ddots}}}}
= 4 - 1 + \frac{1}{6} - \frac{1}{34} + \frac {16}{3145} - \frac{4}{4551} + \frac{1}{6601} - \frac{1}{38341} + \cdots\tag3$$
Unfortunately, the third one didn't include a closed-form for the series. (I tried the OEIS using the denominators, but no hits.)

Q. What's the series formula for $(3)$?  


Given the symmetric continued fraction found in this post
$$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} {3(a+b)+\cfrac{(3a+b)(a+3b)}{5(a+b)+\cfrac{(4a+2b)(2a+4b)}{7(a+b)+\ddots}}}}$$ 
Your continued fraction $(3)$ is a special case when $a=b=1$.
Moreover ,it has a beautiful q-analogue 
$$\begin{aligned}\Big(\sum_{n=0}^\infty q^{n(n+1)}\Big)^2 =\cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2}{1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\ddots}}}}\end{aligned}$$ found here and here.
