L I need to find the explicit recurrence relation (which should be linear and homogeneous, with rational coefficients in $a,b,c,z$) between:
$$F_1={_2 F_1} (a,b,c,z) \\ F_2={_2 F_1} (a + 1,b + 1,c + 2,z) \\ F_3={_2 F_1} (a - 1,b - 1,c - 2,z)$$
I know it's supposed to be easy, just using Gauss' contiguous relations and the hypergeometric equation.
However, I'm struggling to derive the explicit relation.
I found the following:
$$F_2= \frac{c^2}{a b (c-a) (c-b) z} \left((cz+z-c) \frac{d}{dz}+ab \right) F_1$$
Also, by using the so-called Q-form, I have been able to derive the simple equations for the auxiliary functions:
$$z^2(1-z)^2 \frac{d^2}{dz^2} u_1+f(z) u_1=0 \\ z^2(1-z)^2 \frac{d^2}{dz^2} u_2+f(z) u_2=c(1-z)u_2 \\ z^2(1-z)^2 \frac{d^2}{dz^2} u_3+f(z) u_3=-(c-2)(1-z)u_3$$
Where:
$$v u_1=F_1 \\ \frac{v}{z} u_2=F_2 \\ z v u_3 = F_3$$
$$v=z^{-c/2} (1-z)^{(c-a-b-1)/2}$$
$$f(z)=\frac{1}{4} \left(z^2(1-(a-b)^2)+z(2c(a+b-1)-4ab)-c(c-2) \right)$$
Trying for a generalting function, we have:
$$G(a,b;c;z;x)=\sum _{n=0}^{\infty } \frac{ (a)_n (b)_n }{n! (c)_{2 n}} \, _2F_1(a+n,b+n;c+2 n;z)x^n= \\ = \frac{1}{B(b,c-b)} \int_0^1 t^{b-1} (1-t)^{c-b-1} (1-z t)^{-a} {_1 F_1} \left(a;c-b; x\frac{t(1-t)}{1-z t} \right) dt$$
We know that $w={_1F_1} (\alpha,\beta;y)$ obeys a simple ODE:
$$y \frac{d^2 w}{d y^2}+(\beta-y) \frac{d w}{d y}- \alpha w=0$$
If we find the corresponding ODE for $G$ as a function of $x$, we will recover our recurrence.
Hint:
$\dfrac{d}{dz}{_2}F_1(a,b;c;z)=\dfrac{ab}{c}{_2}F_1(a+1,b+1;c+1;z)$