I'm assuming "vector of length $n$" means the number of components is $n$.  In this case, the answer is $3^{n-1}$.  I will create a bijection between vectors ($n$-tuples) of the form described and words of length $n-1$ on an alphabet of three symbols.

Suppose given a vector $v = (a_1, \ldots, a_n)$ satisfying the three conditions, and consider the sequence 
$$D(v) = (a_2 - a_1, a_3 - a_2, \ldots, a_n - a_{n-1}).$$
  By condition (2), D(v) has all entries equal to 0 or $\pm 1$. Conversely, given a sequence d, the condition
$$D(v) = d$$
determines $v$ up to addition of a vector of the form $(a, a, \ldots, a)$. But conditions (1) and (3) exactly state that the minimum component of $v$ must equal $1$; so $D(v)$ uniquely determines $v$.

Thus, 
$$v \mapsto D(v)$$
is a bijection from the set of vectors we wish to count to the set of $(n-1)$-tuples with each entry $0$ or $\pm 1$.