Given $a_0=1$ and $a_n=a_{n-1}+a_{\lfloor n/3\rfloor}$, for primes $p\le 13$ prove there are infinitely many terms divisible by $p$ I have been able to solve the cases for $p=2$ and $3$. For $p\ge 5$, I suspect that a general formula giving an infinite sequence of $k$ such that $p|a_k$ is not possible; however I may be wrong.
The sequence does not appear on OEIS.
Edit: As suggested below, here I'll repeat the question in the description:
There is a sequence of integers $\{a_n\}$ where $n\ge 0$, which satisfies $a_0=1$ and the recurrence $a_n=a_{n-1}+a_{\left\lfloor\frac{n}{3}\right\rfloor}$. For every prime $p$ where $p\le 13$, prove that there exist infinitely many $k$ such that $p|a_k$.
Edit: As suggested by a comment, I’ll add some details/context: 
This is a problem in one of my exercises, in a high-school-level maths olympic training. I don’t know anything about the author/country/source etc.
Here are my solutions for $p=2$ and $3$:
For $p=2$, suppose that (for contradiction) for every $k\ge N$, $2\nmid a_k$. Then consider $a_{3k}=a_{3k-1}+a_k$. As both terms on the right side are odd, the left side must be even. Contradiction.
For $p=3$ the problem is very obvious, since $a_{3k+2}=a_{3k-1}+3a_k$, and $a_2$ is $3$, which gives $\forall k\ 3|k\implies 3|a_k$

Sequence $\{a_n\}$ is in OEIS, see A005704.
Proof: $a_1=2$, $a_2=3$, $a_3=5$, $a_4=7$, $a_{11}=33=3*11$, $a_{20}=117=9*13$.
If $a_m$ is divisible by prime $p$, then $a_{3m+2}\equiv{a_{3m+1}}\equiv{a_{3m}}\equiv{a_{3m-1}}\mod p$. Suppose $a_{3m-1}\equiv{r}\mod p$. Thus $a_{9m-3}\equiv{a_{9m-4}+a_{3m-1}}\equiv{a_{9m-4}+r}\mod p$, $a_{9m-2}\equiv{a_{9m-3}+a_{3m-1}}\equiv{a_{9m-4}+2r}\mod p$, ..., $a_{9m+8}\equiv{a_{9m+7}+a_{3m+2}}\equiv{a_{9m-4}+12r}\mod p$.
If $r=0$, then $a_{3m-1}$ is divisible by $p$, otherwise there exists $k\in\{9m-4,9m-3,9m-2,...,9m+8\}$ such that $a_k$ is divisible by $p$, where prime $p\le 13$.
