Can there exist a recurrence relation for "sequential search"?
Sure.

The cost of searching a list of length zero is zero; the worst-case cost of searching a list of length&nbsp;$n$ is one greater than the cost of searching a list of length $n-1$. So
$$\begin{align*}T(0)&amp;=0 \\ T(n)&amp;=T(n-1)+1\,,\end{align*}$$
with solution being, as you'd expect, $T(n)=n$.