Let the sequence $s = c_1c_2 \dots c_m$ have length $m$, and some sequence $X$ of length $n$ be $a_1a_2 \dots a_n$. If two occurrences of $s$ in $X$ begin with $a_p$ and $a_{p+k}$, then $a_{p+i} = a_{p+i+k} = c_{i+1}$ for $0 \le i < m$. When $k \le m$ this means that $c_i = c_{i+k}$ for $1 \le i \le m-k$, that is, $s$ is the beginning of the (infinite) periodic sequence $a_1a_2 \dots a_k a_1a_2 \dots a_k a_1 \dots$ with period $k$. We choose the minimum such $k$; if no such $k < m$ exists, we let $k = m$ (since $s$ is obviously the beginning of the periodic sequence $a_1a_2 \dots a_m a_1a_2 \dots a_m a_1 \dots$).

We have seen that the number of places in $X$ where the occurrences of $s$ begin must differ at least by $k$. If there are $t$ such occurrences, the last one should begin at the place with number not less than $1+k(t-1)$, so $m+k(t-1) \le n$. It follows that the number of occurrences of $s$ in any sequence of length $n$ does not exceed $\lfloor \frac{n-m}{k} \rfloor + 1$.

On the other hand, an example of a sequence where $s$ can be found in this number of ways, is given by the periodic sequence $a_1a_2 \dots a_k a_1a_2 \dots a_k a_1 \dots$.

We have proved that $v_n = \lfloor \frac{n-m}{k} \rfloor + 1$, and the problem states that
$$\lfloor \frac{n-m}{k} \rfloor < \lfloor \frac{n-m+1}{k} \rfloor < \lfloor \frac{n-m+2}{k} \rfloor .$$
If the inequality $\lfloor \frac{a}{k} \rfloor < \lfloor \frac{a+1}{k} \rfloor$ holds for an integer $a$ and a positive integer $k$, then $k$ divides $a+1$ (since an integer which is greater than $\frac{a}{k}$ and not greater than $\frac{a+1}{k}$, can be presented as a fraction with denominator $k$ and therefore equals $\frac{a+1}{k}$). Therefore, the problem implies that $k$ divides both $n-m+1$ and $n-m+2$, which is possible for $k=1$ only. By the definition of $k$ we have $c_i = c_{i+1}$ for $1 \le i \le m-1$, that is, every two consecutive digits in $s$ are equal. Thus all the digits of $s$ are equal.