I'm noticing a pattern in the numerators. To avoid getting indices wrong, I'll write out an example: $$(22680, 18900, 4410, 255, 1) = (36 \times 630, 16 \times 630+21 \times 420, 9 \times 420 + 10 \times 63, 4 \times 63 + 3 \times 1, 1)$$ Here $(36, 21, 10, 3)$ is the odd indexed triangular numbers and $(16, 9, 4,1)$ are the squares.
Assuming my pattern is right, here is a proof. The above recursion translates into the following relation between the $\alpha$'s: $$\alpha_{n+1} = (x-2) D( x D \alpha_n) + D\alpha_n$$ where $D$ denotes differentiation with respect to $x$.
Inductively, suppose that $\alpha_n$ has $n$ real roots, $2 < q_1 < q_2 < \cdots < q_n$. Set $\beta = D \alpha_n$, then by Rolle's theorem $\beta$ has $n-1$ real roots $2 < r_1 < q_1 < r_2 < q_2 < \cdots < r_{n-1} < q_{n}$. Moreover, $\lim_{x \to \infty} \alpha(x)=0$, forcing another root of $\beta$ at $r_n > q_n$.
Using Rolle's theorem again and the fact that $\lim_{x \to \infty} x \beta =0$ again, we see that $D(x \beta)$ has $n$ real roots, $2 < s_1 < r_1 < s_2 < r_2 < \cdots < s_n < r_n$. We make a little chart of the signs of our functions: $$\begin{array}{rcccccccccc} x: & 2 & r_1 & s_1 & \cdots & s_{n-2} & r_{n-1} & s_{n-1} & r_n & s_n & \gg s_n \\ \beta: & \pm & 0 & \mp & \cdots & - & 0 & + & 0 & - & - \\ (x-2) D(x \beta) : & 0 & \mp & 0 & \cdots & 0 & + & 0 & - & 0 & + \\ \alpha_{n+1} : & \pm & \mp & \mp & \cdots & - & + & + & - & - & + \\ \end{array}$$ To see the bottom right sign, we note that the dominant term of $\alpha_{n+1}$ is $x^{-1}$, which is positive.
So $\alpha_{n+1}$ changes signs $n+1$ times in $[2, \infty)$, and must have $n+1$ real zeroes in that range.