Finding a function/differential equation from a 2nd order recurrence relation

Question

Consider the following 2nd order recurrence relation:

$$A_{1}=1,\; A_{2}=z$$

$$A_{n+2}(z)=(b_n+z)A_{n+1}(z)+a_nA_{n}(z),\; n=1,2,3,...$$

where,

$$b_n=n\left(n+m+\frac{1}{2}\right),\; a_n=k^2n\left(n-\frac{1}{2}\right)$$

where $k\neq 0 \in \mathbb{R},m \in \mathbb{Z}$ are constants

What I want to find is $$f(z) \equiv \lim_{n \to \infty}A_n(z)$$

This will be an infinite series. I want to find the function $f(z)$ either in explicit form or a solution to a differential equation.

Naive attempt:

ClearAll["Global`*"]; 
sol[z_]:= RSolveValue[{A[n + 2] == (n (n + m + 1/2) + z) A[n + 1] + 
     k^2 n (n - 1/2) A[n], A[1] == 1, A[2] == z}, A[n], n]

This gives a DifferenceRoot object that I am not able to make much use of.

Edit 1:

As finding explicit $f(z)$ might be too much to ask for, I will be happy to know just how to efficiently determine the zeros of $f(z)$ for a fixed $k$(say $k=1$) and $m$ going from $m=0,\pm 1,\pm2,...,\pm10$ of $f(z)$ with $|z|<500$.

Edit 2:

As numerical estimation seems to be the only way out for now, I have opened a new question that is closer to my eventual goal: Zeros of a rational fraction sequence This post shall be updated if some analytic way is found.

Answer 1

Only a numerical approximation:

EDITED:

f[z_, m_, k_, inf_] := Last@RecurrenceTable[{A[n + 2] == (n (n + m + 1/2) + z) A[n + 1] + 
k^2 n (n - 1/2) A[n], A[1] == 1, A[2] == z}, A[n], {n, 1, inf}];

tab=Table[{m, z /. NSolve[f[z, m, 1, 20] == 0 && RealAbs[z] < 500, z, 
Reals]}, {m, -10, 10, 1}];(* Assume k=1 and inf=20 !!! *)

Visualisation on plots:

Show[ListPlot /@ (Tuples[{{#[[1]]}, #[[2]]}] & /@ tab), PlotRange -> All]

Or:

 Show[ListPlot /@ (Tuples[{#[[2]], {#[[1]]}}] & /@ tab), PlotRange -> All]

for: k=10,inf=25,m in range {-20,20} we see a patterns.