A recurrence formula for the Legendre function $P_\mu^\nu(x)$ Im looking for a recurrence formula of type:
$$(\mu-\nu) x P_\mu^\nu(x) + P_{\mu-1}^\nu(x)=?, \quad \mu,\nu\in \mathbb R$$
where $P_\mu^\nu(x)$ is the Legendre function of the first kind (solution to the Legendre differential equation which is regular at the origin). 
My goal is to rewrite the sum in one expression, i.e. $(\mu-\nu) x P_\mu^\nu(x) + P_{\mu-1}^\nu(x)= C P_\alpha^\beta(x)  $
Any useful reference, I will be very grateful. Thank you in advance
 A: Let's take relation 14.10.3 from the NIST Handbook, which, after renaming $\mu \leftrightarrow \nu $ and shifting the new $\mu \rightarrow \mu -2$ reads
$$
(\mu -\nu ) P_{\mu }^{\nu } (x) - (2\mu -1)x P_{\mu -1}^{\nu } (x)
+ (\mu+\nu -1) P_{\mu -2}^{\nu } (x) =0
$$
We can thus isolate the desired l.h.s.,
$$
(\mu -\nu ) xP_{\mu }^{\nu } (x) + P_{\mu -1}^{\nu } (x) =
((2\mu -1)x^2 +1) P_{\mu -1}^{\nu } (x) - (\mu+\nu -1)x P_{\mu -2}^{\nu } (x)
$$
The r.h.s. can be consolidated into an expression containing a single Legendre function by also invoking relation 14.10.4 from the NIST Handbook, which, after renaming $\mu \leftrightarrow \nu $ and shifting the new $\mu \rightarrow \mu -2$ reads
$$
(1-x^2 ) \frac{d}{dx} P_{\mu -2}^{\nu } (x) = 
(\nu-\mu +1) P_{\mu -1}^{\nu } (x) + (\mu -1)x P_{\mu -2}^{\nu } (x)
$$
Solving for $P_{\mu -1}^{\nu } (x)$ and inserting above, we end up with
$$
(\mu -\nu ) xP_{\mu }^{\nu } (x) + P_{\mu -1}^{\nu } (x) =
\left[ \frac{(2\mu -1)x^2 +1}{\nu -\mu +1} \left( (1-x^2 ) \frac{d}{dx}
- (\mu -1)x \right) -(\mu +\nu -1)x \right] P_{\mu -2}^{\nu } (x)
$$
A: See equation 37 here. And now some more characters.
