Show that the elements of the sequence are divisible by $2^n$ I am trying to prove the following:

Consider the sequence defined by $A_{n+2}=6A_{n+1}+2A_n, A_0=2, A_1=6$. Show that $2^n|A_{2n-1}$ but $2^{n+1}\nmid A_{2n-1}$.

The first terms of this sequence are 2, 6, 40, 252, 1592, 10056, 63520.
In fact, the maximal exponent of $2$ that divides $A_n$ seems to follow a pattern with a period of 4:
1, 1, 3, 2, 3, 3, 5, 4, 5, 5, 7, 6...
But I haven't been able to prove this.

Using induction, it's easy to prove that $A_n=(3+\sqrt{11})^n+(3-\sqrt{11})^n$ (roots of the characteristic polynomial of the corresponding recurrence $t^2-6t-2$ are $3\pm\sqrt{11}$).
Define
$$
B_n:=A_{2n+1}~\text{for}~n\geq 0.
$$
Then,
$$
B_n=(3+\sqrt{11})^{2n+1}+(3-\sqrt{11})^{2n+1}.
$$
It's easy to check that $B_0=6$, $B_1=252$ and
$$
B_{n+2}=40B_{n+1}-4B_n
$$
(because $(3\pm\sqrt{11})^2=20\pm 6\sqrt{11}$ are roots of $t^2-40t+4$).
Now, define $C_n:=\frac{B_n}{2^{n+1}}$. Then, we can rewrite previous equality as
$$
2^{n+3}C_{n+2}=40\cdot 2^{n+2}C_{n+1}-4\cdot 2^{n+1} C_n.
$$
or
$$
C_{n+2}=20C_{n+1}-C_n.
$$
Also, $C_0=3$ and $C_1=63$.
It's clear from the recurrence relation for $\{C_n\}$ that all $C_n$ are integers. Moreover, each $C_n$ is odd (because $C_0,C_1$ are odd and from the recurrence relation: $C_{n+2}\equiv C_n\pmod 2$).
Therefore, $A_{2n+1}=2^{n+1}\cdot C_n$ for some odd $C_n$. Thus, $2^{n+1\mid A_{2n+1}}$ and $2^{n+2}\not\mid A_{2n+1}$, as desired.
