The pattern for the continued fractions below is quite straightforward. $F_1$ has numerators with all the integers but,
$F_2\; \text{is missing}\; 2m+1 = 3,5,7,\dots\\ F_3\; \text{is missing}\; 3m+1 = 4,7,10\dots\\ F_4\; \text{is missing}\; 4m+1 = 5,9,13\dots$
These cfracs have their uses. $F_1$ appears in an identity of Ramanujan's, $F_2$ is in this, $F_3$ is in a cubic version of Ramanujan's identity, and closed-forms are known for them,
\begin{align} F_1 &= \cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\cfrac{4}{1+\cfrac{5}{1+\ddots}}}}} = 0.5251\dots = \frac1{\sqrt{\frac{\pi\,e}2}\operatorname{erfc}\big(\sqrt{\frac12}\big)} -1\\ F_2 &= \cfrac{1}{1+\cfrac{2}{1+\cfrac{\color{red}4}{1+\cfrac{6}{1+\cfrac{8}{1+\ddots}}}}} = 0.5456\dots = \sqrt[4]{\frac{\pi^2\,e}{16}}\operatorname{erfc}\big(\tfrac12\big)\\ F_3 &= \cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\cfrac{\color{red}5}{1+\cfrac{6}{1+\ddots}}}}} = 0.5078\dots = \sqrt[3]{\frac{e}9}\,\Gamma\big(\tfrac13,\tfrac13\big)\\ F_4 &= \cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\cfrac{4}{1+\cfrac{\color{red}6}{1+\ddots}}}}} = \, ?? = \,?? \end{align}
An alternative expression for the first is $F_1 = -1+\large{\frac1{\sqrt{\frac{e}2}\,\Gamma\big(\frac12,\frac12\big)}}$ and note its affinity to $F_3 = \sqrt[3]{\frac{e}9}\,\Gamma\big(\tfrac13,\tfrac13\big)$.
Question: Does $F_4,\, F_5$, or general $F_n$ have closed-forms? Or, at the very least, what are the numerical values of $F_4,\, F_5$, etc for at least 50 decimal digits? (The Inverse Symbolic Calculator might then prove useful.)
P.S. I find it hard to apply Wolfram's ContinuedFractionK command because these cfracs $F_n$ for $n>1$ skips certain numerators.
