I came up upon the following recurrent sum.
$a\cdot (B_{n-1}\cdot e^{i\cdot k_0 \cdot z}+ B_{n+1}\cdot e^{-i\cdot k_0 \cdot z})=2 \cdot n \cdot B_n$
Where $a$ is complex , $k_0$ and $z$ are real, $n$ is an integer.
In the case where $z=0$ we get the standard Bessel recurrence summation, what happens for general real $z$?
Following @metamorphy suggestion, the trick is the following. We will assume that the structure of the recurring term is the following
$B_n=B_0 \cdot e^{ink_0z}\cdot J_n(a)$
Plugging this solution into the equation we get
$a \cdot (B_0\cdot e^{i(n-1)k_0z}\cdot J_{n-1}(a) \cdot e^{ik_0z}+B_0\cdot e^{i(n+1)k_0z}\cdot J_{n+1}(a) \cdot e^{-ik_0z})=2\cdot n \cdot B_0 \cdot e^{ink_0z}\cdot J_n(a)$
This becomes
$(J_{n-1}(a) +J_{n+1}(a) )=\frac{2n}{a} \cdot J_n(a)$
which is the standard Bessel function recurrence sequence.