Some recurrences "build themselves" if you
- consider cases and
- read the terms out loud.
An example will clarify what I mean.
Let $a(n)$ be the number of binary strings of length $n$ that do not have two consecutive 0's. There are two cases: either the string begins with a 0 or it begins with a 1.
If the string begins with a 0, we know something more about the string: the next digit cannot be a 0. So, the first two digits are "01", and the rest of the string can be any properly formatted (meaning without two consecutive 0's) string of length $n-2$. There are exactly $a(n-2)$ of these, because reading this term out loud means "the number of properly formatted strings of length $n-2$".
If, instead, the entire string begins with a 1, we know nothing more about it except that the remaining $n-1$ digits must be properly formatted. There are exactly $a(n-1)$ of these, because reading this term out loud means "the number of properly formatted strings of length $n-1$".
Since the two cases were disjoint and account for all possible properly formatted strings, we can add them together to get the recurrence
a(n) = a(n-2) + a(n-1).