Conjectured continued fraction formula for Catalan's constant

The accepted answer is misleading. The continued fraction may well be found in that reference, but this is not a result from 2002, but rather a trivial consequence of Euler's continued fraction formula from 1748. You should take a look at the wikipedia page:

https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula

Euler's continued fraction formula $$a_0 + a_0 a_1 + \ldots + a_0 \cdots a_n = \frac{a_0}{\displaystyle{1 - \frac{a_1}{\displaystyle{1 + a_1 - \frac{a_2}{\ldots (1 + a_{n-1}) - \frac{a_n}{1 + a_n}}}}}}$$

Now exactly as in the worked example in the wikipedia page for $\tan^{-1}(x)$, you get the completely formal identity: $$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}} {(2n+1)^2} = x + x \left(\frac{-x^2}{3^2}\right) + x \left(\frac{-x^2}{3^2}\right) \left(\frac{-3^2 x^2}{5^2}\right) + x \left(\frac{-x^2}{3^2}\right) \left(\frac{-3^2 x^2}{5^2}\right)\left(\frac{-5^2 x^2}{7^2}\right)+ \ldots$$ $$=\frac{x}{\displaystyle{1 + \frac{x^2}{\displaystyle{9 - x^2 + \frac{(9x)^2}{25 - 9 x^2 + \displaystyle{ \frac{(25 x)^2}{49 - 25 x^2 + \ldots }}}}}}}$$

The case $x=1$ is your example. You can plug in $x=i$ if you want to get a continued fraction for $\pi^2/8$.

There are literally thousands of completely trivial continued fractions on can create in this way; take any infinite sum and just formally write out the corresponding Euler continued fraction, clearing denominators in the obvious way. None of those should be considered anything more than a corollary of Euler's result (given the evaluation of the initial sum). Of course, in this case, the evaluation of the initial sum is that it is $G$ by definition.

(And no, this doesn't give anywhere near good enough convergents to say anything about the rationality or otherwise of $G$.)

This continued fraction can be found in the paper D. Bowman and J. Mc Laughlin, Polynomial continued fractions, Acta Arithmetica 103 (4) 2002, 329–342. See the bottom of page 2 here: https://www.wcupa.edu/sciences-mathematics/mathematics/jMcLaughlin/documents/4paper1.pdf

Too long for a comment.

However, I do not know how to code a continued fraction on Python or Pari/GP (a friend of mine gave it a go, but also to no avail) up to an iteration $n$ without having to write it out manually, which is really tedious. Here is some python code from a friend, coding this fraction up to $12$ iterations to be $\approx 0.9151$, reaching the first three decimal places of $G$.

To address this part, in Python you can do quickly something like this:

from fractions import Fraction

n = 5
a = [1] + [8*(i + 1) for i in range(n)]
b = [1] + [(2*i + 1)**4 for i in range(n)]

x = Fraction(0, 1)
for ai, bi in zip(reversed(a), reversed(b)):
    x = bi / (ai + x)
print(x, float(x))