Prove that there exist infinitely many positive integers $n$ such that $a_{n}$ is divisible by $n$ Let $\{a_{n}\}$be a sequence defined by $$a_{0}=0,a_{1}=a_{2}=1,a_{n+3}=a_{n+2}+a_{n+1}+a_{n}$$
Prove that there exist infinitely many positive integers $n$ such that $a_{n}$ is divisible by $n$
I try:let $a,b.c$ be the set of all roots of $X^3-X^2-X-1\in \Bbb{Q}[x]$, we can take $(s,t,u)\in \Bbb{C}^3$, such that $$a_{n}=sa^n+tb^n+uc^n$$ for all $n\in \Bbb{Z}$
 A: $\DeclareMathOperator*{\lcm}{lcm}$We will need a few theorems from "Some properties of a generalized Fibonacci sequence modulo $m$", M.E. Waddill (1978). Let's denote $h(x)$ the Trisano period of the $a_n$ sequence taken modulo $$. To begin with, a few empirical observations:
$$\begin{aligned}
 h(3) &= 13\\
 h(13) &= 168
\end{aligned}$$
Also
$$\begin{align*}
a_{13\cdot 3} &\equiv 0 \pmod{h(3)}\\
a_{13\cdot 3} &\equiv 0 \pmod{h(13)}
\end{align*}$$
This allows us to derive
$$\begin{align*}
a_{13\cdot 3} &\equiv 0 \pmod{\lcm(h(3),h(13))}\\
&\equiv 0 \pmod{h(13\cdot 3)} &\text{(see Waddill Theorem 4)}
\end{align*}$$
And $39|a_{39}$ follows.
The general form for $13\cdot 3^{1+6k}$ with $k$ non-negative integer (as conjectured in the question's comments) poses no major difficulties. We know that
$$a_{13\cdot 3^{1+6k}} \equiv 0 \pmod{h(3^{1+6k})} \tag{1}$$
is true because
$$\begin{aligned}
h(3^{1+6k}) &= 3^{6k}h(3) &\text{(see Waddill Theorem 8)}\\
&= 13\cdot 3^{6k} &\text{which divides $13\cdot 3^{1+6k}$.}
\end{aligned}$$
Taking into account that $3^m$ has period $6$ when taken modulo $168$, we also have
$$a_{13\cdot 3^{1+6k}} \bmod{h(13)} = a_{13\cdot 3} \bmod{h(13)} = 0\tag{2}$$
So we can claim that
$$\begin{align*}
a_{13\cdot 3^{1+6k}} &\equiv 0 \pmod{\lcm(h(3^{1+6k}),h(13))}\\
&\equiv 0 \pmod{h(13\cdot 3^{1+6k})}
\end{align*}$$
and conclude that
$$13\cdot 3^{1+6k} | a_{13\cdot 3^{1+6k}}$$
for all non-negative integer $k$.

On the topic of finding $h(p)$ for $p$ prime: Trisano periods exist for all $p$ (that's Waddill's Theorem 1), however they are no more predictable than Pisano periods are for Fibonacci sequences. One can only compute as many terms as needed to find the first reoccurrence of the first terms. For small primes, using a spreadsheet is fast and simple:

Column A for the indices, column B for the sequence itself, cell D1 for the modulus. Then you copy A4:B4 down a few hundred times, and you will find a new occurrence of $0,1,1$ at index $168$.
