A peculiar kind of sequences Let $(u_n)$ be a sequence defined by an order $d$ linear homogeneous recurrence relation with constant coefficients. ($d$ is any integer)
Furthermore ($u_n)$ assume only a finite number of values.
What can be told about $u_n$ ?
By pigeonhole principle it is easy to see that there is a constant subsequence of $u_n$ (this is true for for any sequence that assumes only a finite number of values).
I don't know how to deduce other properties of $u_n$ using the fact that it is defined with a linear homogeneous recurrence relation ...
Thanks for you help. 
 A: The solutions have the form
$$u(n)=\sum_{j=1}^J\sum_{i=0}^{I_j-1} c_{ji}n^i\lambda_j^n,$$
where $\lambda_j$ are roots of characteristic polynomial (there're $J$ distinct roots, sorted by their absolute value in decreasing order) of multiplicity $I_j$; $c_{ij}$ are arbitrary constants.
I will outline the major points:
1) Suppose we have $1<|\lambda_1|>|\lambda_2|$ and at least one of $c_{1i}$ is nonzero. Then the leading term will go to infinity, thus the sequence can't take a finite set of values.
2) Suppose we have several eigenvalues of the same absolute value, which is greater than 1. Clearly, we want the powers $n^i$ to be equal, otherwise we have a leading term that goes to infinity. I think it's possible to prove, that such sequence still can't take finite number of values, thanks to the factor $n^i| \lambda_1|^n$ coupled with the changing argument(angle) of the sum. Hence, all coefficients at eigenvalues not in closed unit disk are zero.
3) In the same way, it can be shown that we can't have non-zero coefficients at eigenvalues strictly inside that disk, because the powers would go to zero never actually reaching it.
4) Finally, we can't have terms of the form $n^i \lambda_1^n$ for $| \lambda_1|=1$, $i>0$, for the same reasons: either we have a leading term or we don't have finite number of possible values (or both).
5) At last, we are stuck with sequances  of the form $$u(n)=\sum_{j=1}^J c_{j } \lambda_j^n,\quad | \lambda_j|=1.$$
Even here, we can't have arbitrary eigenvalues, for they must be roots of unity, otherwise we can't have finite number of values.
