A continued fraction for Baxter's four-coloring constant

Denote the continued fraction by $C$, where $a_{2k-1}=4k-2$, $a_{2k}=k+1$. Even contraction gives $$C= \cfrac{2}{ 3-\cfrac{12}{ 10-\cfrac{30}{ 15-\cfrac{56}{ 20-\ddots}}}}$$

Consider the normalized hypergeometric function and $$\mathbf F(a,b;c;z)=\frac{{}_2F_1(a,b;c;z)}{\Gamma(c)}, \quad R=\frac{\mathbf F\left(-\frac16,\frac12;\frac56;\frac14\right)}{\mathbf F\left(-\frac16,\frac32;\frac{11}{6};\frac14\right)}$$

I would like to point out from the outset that this problem is related to the classical Gauss continued fraction, but this does not mean that this particular representation in terms of a two-progression continued fraction can be found in the literature. It may be regarded as a candidate for inclusion in the OEIS (for example, as a comment on A224273) or in databases of polynomial continued fractions $$\frac{\mathbf F(a,b;c;z)} {\mathbf F(a,b+1;c+1;z)}=v_0-\cfrac{w_1}{v_1-\cfrac{w_2}{v_2-\ddots}}$$ where $v_n=c+n+(b-a+n+1)z$ and $w_n=(b+n)(c-a+n)z$

$$\begin{cases} a=-1/6\\ b=1/2\\ c=5/6\\ z=1/4 \end{cases}\Rightarrow \begin{cases} v_n=\dfrac{5(n+1)}{4}\\ w_n=\dfrac{(n+1)(2n+1)}{8} \end{cases}\Rightarrow 4\left( R-\frac{5}{4} \right)=-\cfrac{12}{10-\cfrac{30}{15-\cfrac{56}{20-\ddots}}}$$

Consider the integral $$I_m=\int\limits_0^1 x^{m+1/2}(1-x)^{-2/3}(4-x)^{1/6}dx,\quad m=-1,0$$

It follows from Euler’s integral representation for $\mathbf F$ that $$R=\frac12\frac{I_{-1}}{I_0}\Rightarrow C=\frac{I_0}{I_{-1}-I_0}$$

All that remains is to evaluate the two integrals; we shall use the formula $$J(\alpha,\beta):= \int\limits_0^{\pi/2}\cos^{\alpha-1}t\cos(\beta t)dt=\frac{\pi\Gamma(\alpha)}{2^\alpha\Gamma\left(\frac{\alpha+\beta+1}{2}\right)\Gamma\left(\frac{\alpha-\beta+1}{2}\right)}$$

$$I_{-1}-I_0\overset{x=4\sin^2\theta}{=}4\cdot 2^{1/3} \int\limits_0^{\pi/6}\cos\theta\cos^{1/3}(3\theta)d\theta\overset{t=3\theta}{=}\frac{4\cdot 2^{1/3}}3J\left(\frac43,\frac13\right)=\frac{2\pi}{3}$$

$$I_0=16\cdot 2^{1/3}\int\limits_0^{\pi/6}\frac{\sin^2\theta\cos^2\theta}{\cos^{2/3}(3\theta)}d\theta= \frac{2^{4/3}}3 \left[ J\left(\frac13,0\right)-J\left(\frac13,\frac43\right)\right]=\frac{2\pi\Gamma(1/3)} {3\Gamma(2/3)^2}=\frac{\Gamma(1/3)^3}{2\pi}$$

$$\boxed{С=\frac{\Gamma(1/3)^3/(2\pi)}{2\pi/3}=\frac{3\Gamma(1/3)^3}{4\pi^2}}$$

I am a big fan of continued fractions, and your representation is an equivalent "unfolding" of a special case of Gauss’s continued fraction. DLMF refers here, in particular, to Andrews–Askey–Roy, Special Functions; Lorentzen–Waadeland, Continued Fractions with Applications; and Berndt, Ramanujan’s Notebooks, Part III. If one cannot find this representation in the literature (I searched for it myself), this may mean that it has not attracted much interest or proved particularly useful, or that it is new and possibly has not appeared in print before. However, the identity itself is an immediate consequence of the classical Gauss continued fraction