Is there some well known efficient way to change this recurrence formula to a somewhat different one? \begin{align}
& J_0 = 2\sin\alpha. \qquad J_1 = 4\sin\alpha-4\alpha\cos\alpha. \\[6pt]
\text{For }n\ge2,\quad & J_n = 2n\Big( (n-1)J_{n-1} - (n-2)\alpha^2 J_{n-2} \Big). \qquad\qquad\qquad\qquad
\end{align}
(This sequence occurs in Mary Cartwright's proof of the irrationality of $\pi,$ which, if I'm not mistaken, is (along with Ivan Niven's proof) a simplification of Charles Hermite's proof.)
Obviously if I iterate this I can write $J_n$ as a function of $J_{n-2}$ and $J_{n-3},$ but I would like to express it in terms of $J_{n-2}$ and $J_{n-4}.$
Probably I can do this if I fiddle with it for a while, but at this point I suspect that there is some widely known way of quickly and efficiently seeing whether this can be done, and of doing it if possible. Maybe something involving generating functions? All of my knowledge of generating functions is rusty except for a few special instances.
So my question is: Is there a widely known efficient way to transform this to a relation expressing the $n$th term as a function of the $(n-2)$th and $(n-4)$th term? Or if none is widely known, is there at least a quick efficient way to do it?
(It just now occurs to me that there may be a somewhat efficient way to do this that works only for this particular sequence. But I'm not sure it will work.)
 A: For any second order linear recurrence relation, we can use the following procedure to obtain a relation between the $n$-th, $(n-2)$-th, and $(n-4)$-th term.
Suppose the original relation is $$a_n = f_n a_{n-1} + g_n a_{n-2} + h_n\label{1}\tag{1}\ ,$$ then substituting $n-1$ and $n-2$ for $n$, we get the following relations,
\begin{align}
a_{n-1} & = f_{n-1} a_{n-2} + g_{n-1} a_{n-3} + h_{n-1}\ ,\label{2}\tag{2}\\
a_{n-2} & = f_{n-2} a_{n-3} + g_{n-2} a_{n-4} + h_{n-2}\ .\label{3}\tag{3}
\end{align}
Combined with the original relation, we have three equations, from which we can eliminate two variables.
Multiplying \eqref{2} by $f_{n}$ and adding it to \eqref{1} gets us
$$
a_n = (g_n + f_n f_{n-1}) a_{n-2} + f_n g_{n-1} a_{n-3} + (h_n + f_n h_{n-1})\label{4}\tag{4}\ .
$$
(Or in other words just substitute \eqref{2} into \eqref{1} to eliminate $a_{n-1}$. You seem to have already tried this for the sequence $J_n$.)
Multiplying \eqref{3} by $f_n g_{n-1}/f_{n-2}$ and subtracting it from \eqref{4} gives us
$$
a_n = \left(g_n + f_n f_{n-1} + \frac{f_n g_{n-1}}{f_{n-2}}\right) a_{n-2} - \frac{f_n g_{n-1} g_{n-2}}{f_{n-2}} a_{n-4} + \left(h_n + f_n h_{n-1} + \frac{f_n g_{n-1}}{f_{n-2}} h_{n-2} \right)\ .
$$
The result of this procedure in your case is a rather complicated expression, and @JeanMarie's transformation doesn't improve the result by much. In any case, we end up with
$$
(n-2) J_n = \left[4n(n-1)^2(n-2)^2 - 2\alpha^2 n (2n^2 -6n+5)\right] J_{n-2} - 4\alpha^4 n(n-4)(n-1)^2 J_{n-4}\ .
$$
