Unique inner products of integer vectors ** Update **
I calculated $V(m,n)$ by enumeration. It turned out that for $n\geq N_0$, $V(m,n)$ is a algebraic series with common difference $m^2-1$.
For $m=1,2$, $N_0=1$; for $3\leq m \leq 31$, $N_0=2$; for $32\leq m\leq 50$, $N_0=3$. I have not calculated the cases where $m>50$.

$A(m,n)$ is the set of all vectors that have $n$ components, each of which is a integer between $1$ and $m$. So $|A(m,n)|=m^n$. Take two vectors from $A(m,n)$ and calculate their inner products, there are $m^{2n}$ such combinations and $V(m,n)$ unique results. 
Is there a formula to calculate $V(m,n)$ with minimal enumeration (by recursion or by direct calculation)? Or is there an OEIS page for this? I noticed that $V(2,n)=3n$ for $n\leq 12$, does this hold for all $n$? Thanks!
 A: Here's a relatively simple way to recursively calculate $V(m,n)$ in $n$ (though you need the actual dot products not just the number of distinct dot products at each step).
Fix $m, n$ and calculate the set $S(m,n)$ of all dot products of pairs of element in $A(m,n)$.  Now note that if $x = (x_1, \ldots, x_{n+1})$ and $y=(y_1, \ldots, y_{n+1})$ in $A(m,n+1)$, the dot product
$$ x \cdot y = \sum x_i y_i = x' \cdot y' + x_{n+1}y_{n+1}$$
where $x' = (x_1, \ldots, x_n)$ and $y' = (y_1, \ldots, y_n)$.
Consequently
$$ S(m,n+1) = \{ s + ab : s \in S(m,n), 1 \le a \le b \le m \} = \{ s + t :
s \in S(m,n), t \in S(m,1) \}.$$
Then just count $V(m,n) = | S(m,n)|$.
I think this approach should prove $V(2,n) = 3n$ for you, and maybe give you formulas for larger $m$ as well.  Namely, note that $S(m,n)$ is just the set obtained from adding $S(m,1)$ to itself $n$ times.  First compute $S(2,1) = \{ 1, 2, 4\}$ and $S(2,2) = \{ 2, 3, 4, 5, 6, 8 \}$.  I didn't work this out, but it looks like you should be able to show $S(2, n) = \{ n, n+1, \ldots, 4n-2, 4n \}$ by induction. 
Edit: Clearly $S(m,1)$ is contained in the set of integers from $1$ to $m^2$.  Hence the above tells you $S(m,n)$ is contained in the set of integers from $n$ to $nm^2$.  This gives an upper bound $V(m,n) \le n(m^2-1)$, and your calculations suggest that the actual values have the same order of growth (in $n$).  This is supported by the idea that $V(m,1)$ is approximately the number of way to choose two numbers (with possible repetition) from $1$ to $m$, i.e., $V(m,1) \approx \frac{m(m+1)}2$.  (This approximation should be exact precisely for $m \le 3$.)  Perhaps for sufficiently large $n$, $S(m,n)$ contains all numbers between $n$ and $nm^2$ "of a certain form" so that it grows exactly linearly, with slope $m^2-1$, after a certain point.
