A conjectured continued fraction related to a 2F1 hypergeometric function (and a formula for π)

In a previous question, some months ago, I described a continued fraction for which I was trying to find an identity. I finally found what I was looking for, and I will publish an answer to this question in the days or weeks to come by focusing to what was asked at that time. But now, I am opening a separate question for a new interest: Could the conjectured statements below be proved? Do these identities sound familiar to anyone? etc.

My identity finally is:

$$\begin{array}{lcl} \mathcal{K}\left(a,x\right)&=&\displaystyle\operatorname*{K}_{n=1}^{\infty} \frac{ \left(n+1\right)\left(\left(a-1\right)x-n\right)a}{\left(n+1\right)\left(a+1\right)}\\[12pt] &=& \displaystyle\frac{\left(a-1\right)x-1}{x+1}\; \displaystyle{}_{2}F_{1}\left(1,ax\;; 2+x\;; a^{-1}\right) \end{array} \tag{1} \label{1}$$

Two functional identities

I empirically noticed two functional identities, namely:

$$2+\mathcal{K}\left(a,x\right)+\mathcal{K}\left(1-a,x\right) \;=\; 0 \tag{2} \label{2}$$ and (with $\textrm{B}$ the Beta function): $$\begin{array}{l} 2+\displaystyle\frac{1}{a}\mathcal{K}\left(a,x\right) +\displaystyle\frac{a-1}{a}\mathcal{K}\left(a/\left(a-1\right), \left(a-1\right)x\right)\\[8pt] \qquad\qquad\;=\; ax\left(a-1\right)^{x-1}\left(\displaystyle\frac{a}{a-1}\right)^{ax-2}\textrm{B}\left(x,\left(a-1\right)x\right) \end{array} \tag{3} \label{3}$$ I already told about the latter in another question.

A finite summation

While studying the polynomials from the initial question, I finally found the following summation: $$\mathcal{K}\left(a,x\right)\;=\; \displaystyle\frac{\displaystyle 2\sum_{k=2}^{\left(a-1\right)x-1} \left(k-1\right){{ak/\left(a-1\right)}\choose{k}} {a\left(\left(a-1\right)x-k\right)/\left(a-1\right) \choose{\left(a-1\right)x-k}} } {\displaystyle\left(\left(a-1\right)x-2\right){{ax}\choose{x}}} \tag{4} \label{4}$$ (when $(a-1)x$ is an integer greater than 2).

A formula for $\pi$

I tried several computer algebra systems and found the hyperexpand function in Sympy to be able of simplifying the ${}_2F_1$ hypergeometric function for many rational cases. My favorite case comes from $\mathcal{K}\left(-3, 1/2\right)$, after some cosmetic changes: $$\pi \;=\; \operatorname*{K}_{n=1}^{\infty} \frac{ 81\left(n+1\right)\left(n+2\right)/16}{3\sqrt{3}\left(n+1\right)/2} \tag{5} \label{5}$$ The formula for $\phi^\phi$ in the previous question was also a specific case for the same continued fraction.

• Isn't the continued fraction representation for $\phantom{}_2 F_1$ functions pretty well-known? en.wikipedia.org/wiki/… – Jack D'Aurizio Jun 1 '17 at 19:58
• It looks to me that your $(1)$ follows from Gauss' continued fraction through simple rearrangements. – Jack D'Aurizio Jun 1 '17 at 19:59