**The number sequence $1,9,8,2...$ is defined so that$t_n$ is the last digit of $t_{n-1}+t_{n-2}+t_{n-3}+t_{n-4}$ for $n>4$
Will the numbers $3,0,4,4$ ever occur in the sequence order?
To examine this I first noted that if we continue the rule forward we get $3,0,4,4,1,9,8,2$
So we repeat the sequence from the start.
It is clear from the pigeon-hole principle that the sequence must repeat at some point. However, I am not sure how to prove that we do not hit some intermediate loop and therefore never return to the beginning?
$\endgroup$ 201 Answer
$\begingroup$The sequence is given by
$$t_n = t_{n-1}+t_{n-2}+t_{n-3}+t_{n-4} \pmod {10}$$
This allows you to calculate the sequence in the forward direction. You can however rewrite the above as
$$t_{n-4} = t_n - t_{n-1}-t_{n-2}-t_{n-3} \pmod {10}$$
With this you can go backwards in the sequence, too. For example the term before $1,9,8,2$ is $2-8-9-1\equiv 4 \bmod{10}$.
You know that the sequence repeats eventually, so for some positive integers $N$ and $p$ we have $t_{i}=t_{i+p}$ for all $i>N$. But now you can go backwards from both those points to get $t_{i-1}=t_{i-1+p}$. And then backwards again, and again until you arrive at $t_{0}=t_{p}$. So the sequence repeats from the beginning.
If you have a sequence which does not repeat from the start but from some later point, then that point will have two distinct predecessors - one from the starting point and one from the repeating loop. That is not possible in this case because we can travel backwards in the sequence, calculating the unique predecessor at each step.
$\endgroup$ 4
