For $\varepsilon>0$ consider the tridiagonal matrix
$$L_{\varepsilon}=\begin{bmatrix} 0 & 1 & \ & \ & \ & \ & \ & \ \\ 1 & \varepsilon & 1 & \ & \ & \ & \ & \ \\ \ & 1 & 2 \varepsilon & 1 & \ & \ & \ \\ \ & \ & 1 & 3 \varepsilon & 1 & \ & \ & \ \\ \ & \ & \ & 1 & 4 \varepsilon & 1 & \ & \ \\ \ & \ & \ & \ & \ddots & \ddots & \ddots & \ \\ \end{bmatrix}$$
with all blank entries equal to zero. A column vector $$\Phi = \begin{bmatrix} \Phi_0 \\ \Phi_1 \\ \Phi_2 \\ \vdots \end{bmatrix}$$ is an eigenvector $$L_{\varepsilon} \Phi = u \Phi$$ iff the entries $\Phi_h = \Phi_{h}(u; \varepsilon)$ satisfy a three-term recurrence $$\Phi_{h-1} (u; \varepsilon) + h \varepsilon \Phi_h(u; \varepsilon) + \Phi_{h+1}(u; \varepsilon) = u\Phi_{h}(u;\varepsilon)$$ for $h \geq 0$ with convention $\Phi_{-1} = 0$, which is equivalent to $$\Phi_{h-1} (u; \varepsilon) + (h \varepsilon - u) \Phi_h(u; \varepsilon) + \Phi_{h+1}(u; \varepsilon) = 0.$$
Question: Can one identify the $\Phi_h(u; \varepsilon)$ with a known function, possibly a specialization or degeneration of the Gauss hypergeometric function?
As $\varepsilon \rightarrow 0$, the diagonal in $L_{\varepsilon}$ disappears and $\Phi_{h}(u; 0)$ are the Chebyshev polynomials. The recurrence above seems close to that of the Gegenbauer polynomials but is not exactly the same. Any help would be greatly appreciated!