Number of ways to make $n$ cents
If we try to apply your recurrence to $n = 9$ and add the base case $F(0) = 1$, then we get $F(9) = 2$. Now this might be the answer you want, since there are indeed two possibilities: penny+nickel and nickel+penny, but usually we consider these the same.

There are two ways to proceed if we want to compute the number of ways without regard to order. The first way is to consider three different functions $F,G,H$ which correspond to using all coins ($F$), only nickels and dimes ($G$), and only dimes ($H$). The recurrences are
$$
F(n) = F(n-4) + G(n-5) + H(n-10), \\
G(n) = G(n-5) + H(n-10), \\
H(n) = H(n-10),
$$
with initial conditions $F(0) = G(0) = H(0) = 1$ and $F(m) = G(m) = H(m) = 0$ for negative $m$. This gives a joint linear recurrence for the triplet $F(n),G(n),H(n)$.

The other approach is to use generating functions, which give us the following expression:
$$
\sum_{n=0}^\infty F(n) x^n = \frac{1}{(1-x^4)(1-x^5)(1-x^{10})} = \frac{1}{1-x^4-x^5+x^9-x^{10}+x^{14}+x^{15}-x^{19}}.
$$
This shows that the sequence $F(n)$ satisfies the recurrence
$$
F(n) = F(n-4) + F(n-5) + F(n-10) \\- F(n-9) - F(n-14) - F(n-15) + F(n-19)
$$
with initial conditions $F(0) = 1$ and $F(m) = 0$ for negative $m$.

This can be seen as a kind of inclusion/exclusion principle &mdash; I'll let you work out the details.