$a_0$ to $a_5$ are set independently, all positive.
Then let us define the sequence as $a_n=\frac{\sum_{i=1}^6a_{n-i}}{6},\;\forall n\geq6$.
How do I find $\lim_{n\rightarrow\infty}a_n$?
Or, will it converge at all? Or will it just be exactly the limit value after some $n$?
Hint: Note that if $$x_n = \begin{pmatrix} a_{n-5}\\ a_{n-4}\\ a_{n-3}\\ a_{n-2}\\ a_{n-1}\\ a_{n} \end{pmatrix} ; \, \, \, \, \, A = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1& 0 & 0 \\ 0 & 0 & 0 & 0 & 1& 0 \\ 0 & 0 & 0 & 0 & 0 & 1\\ 1/6 & 1/6 &1/6 &1/6 &1/6 &1/6 \end{pmatrix}, $$ We have $x_n = Ax_{n-1}$. You can check that $1$ is the single eigenvalue of $A$ with greatest absolute, and its associated eigenvector is $$v = \begin{pmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ 1 \end{pmatrix}.$$ From here, if $a_0+\dots+a_5 \ne 0$, then $x_n$ converges to a multiple of $v$, say $x_n \rightarrow c v$ for some $c$. This proves that if $a_0+\dots+a_5 \ne 0$, the sequence converges.
To find $c$, keeping in mind that $w = \frac{1}{21} (1, 2, 3, 4, 5, 6)$ is the left eigenvector associated with eigenvalue $1$, try to show that $$c = w\, x_5,$$ which means $a_n \rightarrow \frac{1}{21} (a_0 + 2a_1+\dots+6a_5)$.