@air1. Consider the sequence,
f(n) = 2.f(n -1)
= f(n - 1) + f( n - 1)
Now, if we expand $2^{nd}$ f(n - 1), then we get previous terms. If we repeat the procedure of expanding only $2^{nd}$ terms, then we get
f(n) = f( n - 1 ) + f( n - 2 ) + f( n - 3 ) + f( n - 4 ) .........2.f(0).
Even though this is indirect expansion, I think we can have such recurrances. I havent studied higher mathematics :) .