The Inverse of a Finite Series and a Third-order Recurrent Sequence

It is apparently not well-known that g(n) = f(n) + f(n− 1) + f(n− 2) if and only if f(n) = n ∑ k=0 g(n− k) [ k2 ] ∑ j=0 (−1)k−j ( k − j j ) , where we suppose that f(n) = 0 for n < 0. This may also be expressed as f(n) = n ∑ k=0 (−1)g(n− k) 2 ( (−1) k3 ] + (−1) k+1 3 ] ) . We show how to solve for f(n) in the general case g(n) = r ∑ k=0 f(n− k), where f(n) = 0 for n < 0, with 1 ≤ r ≤ n. We shall also see that the values at which g is evaluated in forming the inverse satisfy a third-order recurrence relation of the form an = an−1 + an−2 − an−3.