First four numbers of a sequence are $2,0,1,8,$. Each next is the last digit of the sum of the preceding four numbers. Will $2,0,1,8$ show up again? Let $a_n$ is a sequence of numbers. The firs four numbers are $2,0,1,8$ and each following number is the last digit of the sum of the preceding four numbers. The first ten numbers are $2,0,1,8,1,0,0,9,0,9$. 
Will the succession $2,0,1,8$ show up again? Will the succession $2,0,1,9$ show in this sequence of numbers? 
I couldn't do more. 
$2+0+1+8=11$, $\enspace$ $11 \enspace mod \enspace 10=1$,$\enspace$ $a_5=1$ 
$a_n=(a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4})\enspace mod\enspace10$
Could someone give me an idea?

Notice that given any four sequential numbers in the sequence, say $a_n, a_{n+1}, a_{n+2}, a_{n+3}$ with $n>1$, $a_{n-1}$ is determined uniquely; in fact, $a_{n-1} = a_{n+3} - a_{n+2} - a_{n+1} - a_{n} \bmod 10$.
Since there are only finitely many four-tuples of the digits $0-9$, the sequence of four-tuples must eventually repeat.  That is, there are $m$ and $n$ with $m<n$ and $a_{m+i} = a_{n+i}$ for $i = 0,1,2,3$.  Let $k$ be the least such $m$.  We claim $k=1$.  If not, we must have $a_{k+i} = a_{n+i}$ for $i = -1, 0, 1, 2$, contradicting the minimality of $k$.  This shows that the sequence eventually produces $2,0,1,8$ again.
On the other hand, the sequence will never produce $2,0,1,9$ because modulo $2$ the sequence $0,0,1,0,1$ repeats indefinitely, so two odd digits can never appear in succession.
