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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.

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 $x$. As Victor points out in the comments below, this recursion is equivalent to the relation $$D_z D_z f = (x-2) D_x ( x D_x f) + D_x f$$ where $f = \sinh(z)/(\cosh(z)-1+x)$. This latter relation can be verified, without much insight, by typing it into a computer algebra system.

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.

UPDATE Slicker proof for the final steps: $$\alpha_{n+1}=(x-2) D x \beta + \beta = (x-2) x D \beta + (x-1) \beta = \sqrt{x(x-2)} D \left( \sqrt{x(x-2)} \beta \right)$$ Since $\beta$ has $n$ real roots in $[2, \infty)$, and is $O(x^{-2})$ as $x \to \infty$, Rolle's theorem applied to $\sqrt{x(x-2)} \beta$ shows that $\alpha_{n+1}$ has $n+1$ real roots in $[2, \infty)$.

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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 $x$. As Victor points out in the comments below, this recursion is equivalent to the relation $$D_z D_z f = (x-2) D_x ( x D_x f) + D_x f$$ where $f = \sinh(z)/(\cosh(z)-1+x)$. This latter relation can be verified, without much insight, by typing it into a computer algebra system.

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.

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This isn't the sort of

Assuming my pattern is right, here is a proof. The above recursion I want 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 find though -- it expresses $x$.

Inductively, suppose that $\alpha_n$ in terms of has $\alpha_{n-1}$ and its first two derviativesn$real roots,$2 < q_1 < q_2 < \cdots < q_n$. I'd rather express Set$\alpha_n$in terms \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 $\alpha_{n-1}$, \beta$at$\alpha_{n-2}$r_n > q_n$.

Using Rolle's theorem again and their first derivatives; the fact that would be more Sturm like$\lim_{x \to \infty} x \beta =0$ again, we see that $D(x \beta)$ has $n$ real roots, $2 < s_1 < r_1 < s_2 < r_2 < \cdots < s_n < r_n$.

More thoughts later if I 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 any$n+1$ real zeroesin that range.

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