Inductivly one can see that $2a_n=5^{n-1}-3^n$ taking modulo $8$ one can see that $v_2(a_n)=1$ if $n$ is even and $v_2(n)=0$ in n is odd.You can also check the sequance just mod $2$ without getting a closed form.

Quote: Inductivly one can see that $2a_n=5^{n-1}-3^n$ taking modulo $8$ one can see that $v_2(a_n)=1$ if $n$ is even and $v_2(n)=0$ in n is odd.You can also check the sequance just mod $2$ without getting a closed form. This is not quite right , go through your solution to fix some minor mistakes that caused you extracting an entirely wrong result.

The correct formula is $a_n=\frac{3\cdot 5^{n-1}-3^{n-1}}{2}$.

You can still use @Taha's method to see that $v_2(a_{2^{2019}})=1$.

math90 wrote: The correct formula is $a_n=\frac{3\cdot 5^{n-1}-3^{n-1}}{2}$. This is not true. $a_1=8$ and not $a_1=6$ as we get by your formula.

$a_n= \frac{5^n-3^n}{2}$

Greece MO 2019 P1 wrote: Define the sequnce ${(a_n)}_{n\ge1}$ by $a_1=1$ and $a_n=5a_{n-1}+3^{n-1}$ for $n\ge2$. Find the greatest power of $2$ that divides $a_{2^{2019}}$. Firstly notice that $a_n=\frac{5^n-3^n}{2}$ with the proof being trivial using induction. We procced with a Claim. Claim: For all $k \in \mathbb{N^*}$ we have that $2^{k+2} \mid \mid 5^{2^k}-3^{2^k}$, where $\mid \mid$ stands for the greatest power of $2$ that divides $5^{2^k}-3^{2^k}$. Proof: Via Induction. For $k=1$, it's obviously true. Let's suppose it's true for a specific $k=m$. I will prove it for $k=m+1$. It suffices to show that $$2^{m+3} \mid \mid (5^{2^m}+3^{2^m})(5^{2^m}-3^{2^m}),$$but since $$2^{m+2} \mid \mid 5^{2^m}-3^{2^m}$$we have to show that $$2^1 \mid \mid 5^{2^m}+3^{2^m},$$which is obvious, since $5^{2^m}+3^{2^m} \equiv 2 \pmod 4 $ $\blacksquare$. Back to our problem, from the Claim it follows that the desired greatest power is $2^{2021}$.

Greece MO 2019 P1 wrote: Define the sequnce ${(a_n)}_{n\ge1}$ by $a_1=1$ and $a_n=5a_{n-1}+3^{n-1}$ for $n\ge2$. Find the greatest power of $2$ that divides $a_{2^{2019}}$. $a_n=5a_{n-1}+3^{n-1}$ is a non-homogeneous recurrence, denote $f_g(n)$ as the general solution to the homogeneous recurrence and $f_p(n)$ as the particular solution $$a_n=5a_{n-1}+3^{n-1} \implies a_{n+1}-5a_n=3^n$$Observe that, if $E$ is the shift operator, $(E-5)$ annihilates $a_{n+1}-5a_n$ and $(E-3)$ annihilates $3^n$ $\implies $ $f_g(n)=A5^n$ for some constant $A$ and $f_p(n)=B3^n$ for some constant $B$ $$a_{n+1}-5a_n=3^n \implies B3^{n+1}-5B3^n=3^n \implies B3^{n+1}=3^n(5B+1) \implies B=-\frac{1}{2}$$$$a_n=f_g(n)+f_p(n)=A5^n-\frac{1}{2}3^n \implies a_1=1=5A-\frac{3}{2} \implies A=\frac{1}{2} \implies \boxed{a_n=\frac{5^n-3^n}{2}}$$Applying Lifting The Exponent Lemma, $$v_2(a_{2^{2019}})=v_2(5^{2^{2019}}-3^{2^{2019}})=v_2(5-3)+v_2(5+3)+v_2(2^{2019})-1=\boxed{2022}$$ In General: $2^{k+3} | 5^{2^k}-3^{2^k}$ which is because: $$v_2(5^{2^k}-3^{2^k})=v_2(5-3)+v_2(5+3)+v_2(2^k)-1=k+3$$