I am currently engaging with the following hypergeometric function as a result of attempting to find a solution for this probability problem for $n$ number of dice:
$$_2F_1\left (\frac{n+k}{2}, \frac{n+k+1}{2};k+1;\frac{4b(d-e+1)}{d^2}\right ); \{n,k,b,d,e\}\in \mathbb{N}^+, z\in \mathbb{R},b < e \leq d$$
I'm trying to determine if there's way to calculate this function that doesn't require infinite sums (ideally, just with elementary functions and finite sums/products).
I first note that this function is of the pattern: $$_2F_1\left (a, a+\frac{1}{2};c;z\right )$$ DLMF 15.9.17 then states that:
$$_2\bar{F}_1\left (a, a+\frac{1}{2}; c; z \right )=2^{c-1}z^\frac{1-c}{2}(1-z)^{-a+\frac{c-1}{2}}P_{2a-c}^{1-c}\left ( \frac{1}{\sqrt{1-z}} \right )$$
where we're using DLMF's definition (14.7.14) of:
$$P_{n}^{m}(z)=\frac{(z^2-1)^{\frac{m}{2}}}{2^nn!}\frac{d^{m+n}}{dx^{m+n}}(z^2-1)^n$$
Okay, that seems reasonable enough to plug in, and if we do, we get:
$$_2\bar{F}_1\left (\frac{n+k}{2}, \frac{n+k+1}{2}; k+1; z \right )=2^{k}z^{-\frac{k}{2}}(1-z)^{-\frac{n}{2}}P_{n-1}^{-k}\left ( \frac{1}{\sqrt{1-z}} \right )$$
So that leaves us with an Associated Legendre Function of negative integer order. DLMF 14.9.3 gives us a recurrence relation that can convert this to an ALF of positive order:
$$P_{\nu}^{-m}\left ( z \right )=\frac{\Gamma(\nu-m+1)}{\Gamma(\nu+m+1)}P_{\nu}^{m}\left ( z \right )$$
But after a lot of work I've realised that this isn't actually suitable for the problem I'm looking at, because for ALFs of integer order and degree, $P_{n}^{m}=0$ where $m>n$, and the problem I'm looking at it's quite common for $|m|>n$, so the easy method is out. I've tried looking at as much of the literature as I can understand, and so far as I can tell, there doesn't seem to be a known identity for $P_{n}^{-m}$ for $|m|>n$ that doesn't involve infinite sums.
So the question I am asking is: Is there a known equation to calculate Associated Legendre Functions of arbitrarily large negative integer order and positive integer degree that doesn't require infinite sums (and, preferably, only requires elementary functions with finite sums/products)?
To evaluate \begin{equation} G={}_2F_1\left (\frac{n+k}{2}, \frac{n+k+1}{2};k+1;x \right) \end{equation} one can recognize the pattern \begin{equation} G={}_2F_1\left (s, s+1/2;2s+1-n;x \right) \end{equation} where $s=(n+k)/2$. Under this form, it can be interpreted as the $n$-th derivative of a simpler hypergeometric function by this relation: \begin{equation} \frac{{\mathrm{d}}^{n}}{{\mathrm{d}z}^{n}}\left(z^{c-1}{}_2F_1\left(a,b;c;z\right)% \right)={\left(c-n\right)_{n}}z^{c-n-1}{}_2F_1\left(a,b;c-n;z\right) \end{equation} ($(.)_n$ correspond to the Pochhamer symbols). Here $a=s,b=s+1/2,c=2s+1$, then \begin{equation} G=\frac{x^{n-2s}}{(2s+1-n)_n}\frac{{\mathrm{d}}^{n}}{{\mathrm{d}x}^{n}}\left(z^{2s}F\left(s,s+1/2;2s+1;x\right)% \right) \end{equation} But this special hypergeometric function is tabulated: \begin{equation} {}_2F_{1}(s,s+{\frac{1}{2}};2\,s+1;z)=2^{2s}\left({\sqrt{1-z}}+1\right)^{-2s} \end{equation} then \begin{equation} G=\frac{2^{n+k}x^{-k}}{(k+1)_n}\frac{{\mathrm{d}}^{n}}{{\mathrm{d}x}^{n}}\left(z^{n+k}\left({1+\sqrt{1-x}}\right)^{-n-k} \right) \end{equation} The inner function \begin{align} z^{n+k}\left({1+\sqrt{1-z}}\right)^{-n-k}&=\left( \frac{z}{1+\sqrt{1-z}} \right)^{n+k}\\ &=\left( 1-\sqrt{1-x} \right)^{n+k}\\ &=\sum_{m=0}^{n+k}\binom{n+k}m (-1)^m\left( 1-x \right)^{m/2} \end{align} As \begin{equation} \frac{{\mathrm{d}}^{n}}{{\mathrm{d}x}^{n}}\left( \left( 1-x \right)^{m/2} \right)=(-1)^n\left( m/2+1-n \right)_n(1-x)^{m/2-n} \end{equation} one obtains \begin{equation} G=\frac{2^{n+k}(-1)^nx^{-k}}{(k+1)_n}\sum_{m=0}^{n+k}(-1)^m\binom{n+k}m \left( m/2+1-n \right)_n(1-x)^{m/2-n} \end{equation} which seems to be numerically correct.