Regarding $p$-adic valuation of Lucas sequences, a quite precise result is given in [1].

**Theorem.** Let $(u_n)_{n \geq 0}$ be a nondegenerate Lucas sequence with $u_0 = 0$, $u_1 = 1$, and $u_{n+2} = a u_{n+1} + b u_n$ for all $n \geq 0$, where $a$ and $b$ are two integers. Furthermore, let $p$ be a prime number not dividing $b$.

Then for any positive integer $n$ we have 
$$v_p(u_n) = \begin{cases}
v_p(n) + v_p(u_p) - 1 & \text{ if } p \mid \Delta ,\; p \mid n, \\
0 & \text{ if } p \mid \Delta ,\; p \nmid n, \\
v_p(n) + v_p(u_{p\tau(p)}) - 1 & \text{ if } p \nmid \Delta ,\; \tau(p) \mid n ,\; p \mid n, \\
v_p(u_{\tau(p)}) & \text{ if } p \nmid \Delta ,\; \tau(p) \mid n ,\; p \nmid n, \\
0 & \text{ if } p \nmid \Delta ,\; \tau(p) \nmid n ,
\end{cases}$$
where $\Delta := a^2 + 4b$ and $\tau(p)$ is the rank of apparition of $p$ in $(u_n)_{n \geq 0}$, i.e., the least positive integer $m$ such that $p \mid u_m$. Moreover, if $p \geq 3$ then
$$v_p(u_n) = \begin{cases}
v_p(n) + v_p(u_p) - 1 & \text{ if } p \mid \Delta ,\; p \mid n, \\ 
0 & \text{ if } p \mid \Delta ,\; p \nmid n, \\ 
v_p(n) + v_p(u_{\tau(p)}) & \text{ if } p \nmid \Delta ,\; \tau(p) \mid n , \\ 
0 & \text{ if } p \nmid \Delta ,\; \tau(p) \nmid n ,
\end{cases}$$
while if $p \geq 5$ then
$$v_p(u_n) = \begin{cases}
v_p(n) & \text{ if } p \mid \Delta , \\ 
v_p(n) + v_p(u_{\tau(p)}) & \text{ if } p \nmid \Delta ,\; \tau(p) \mid n , \\ 
0 & \text{ if } p \nmid \Delta ,\; \tau(p) \nmid n .
\end{cases}$$
Actually, in [1] the theorem is stated for $a$ and $b$ relatively prime. 
However, as explained in [2], the result holds even if $a$ and $b$ are not coprime.

[1] C. Sanna, *The $p$-Adic Valuation of Lucas Sequences*, Fibonacci Quart. **54** (2016), no. 2, 118–124.

[2] N. Murru, C. Sanna, *On the k-regularity of the k-adic valuation of Lucas sequences* http://arxiv.org/abs/1603.09310