Coefficient of the generating function $G(z)=\frac{1}{1-z-z^2-z^3-z^4}$

The Fibonacci sequence given by $F_{n}=F_{n-1}+F_{n-2}$ has $\dfrac{z}{1-z-z^2}$ as generating function.

By analogy, $\dfrac{1}{1-z-z^2-z^3-z^4}$ is related to the recurrence $ a_{n}=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4} $.

A closed expression for this recurrence involves the roots of ${1+z+z^2+z^3=z^4}$.

The exact expression will depend on the initial values $a_0, a_1, a_2, a_3$.

Let $\left\{z_i\right\}_{i=1}^4$ be the roots of the polynomial in the denominator. Note that: \begin{eqnarray} \frac{1}{1-z-z^2-z^3-z^4} = \frac{-1}{\prod\limits_{i=1}^4 (z-z_i)} = \sum\limits_{p=1}^4 \frac{-1}{z-z_p} \cdot \left(\prod\limits_{q\neq p} \frac{1}{z_p-z_q}\right) \end{eqnarray} Now clearly : \begin{equation} a_k = \frac{1}{k!} \frac{d^k}{d z^k}\left. \left( \frac{1}{1-\sum\limits_{j=1}^4 z^j} \right) \right|_{z=0} = \sum\limits_{p=1}^4 \frac{1}{z_p^{k+1}} \cdot \left(\prod\limits_{q\neq p} \frac{1}{z_p-z_q}\right) = \left(1,1,2,4,8,15,29,56,108,208,401,\cdots\right) \end{equation}

Following your comments above: once you have the recurrence relation $a_k = a_{k-1}+a_{k-2}+a_{k-3}+a_{k-4}$ and the initial condition $(a_0,a_1,a_2,a_3)=(1,1,2,4)$, then you can solve the system by writing $$X_k = \begin{pmatrix} a_{k}\\ a_{k+1}\\ a_{k+2}\\ a_{k+3} \end{pmatrix} $$ and $$A = \begin{pmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&1&1&1\\ \end{pmatrix} $$ so that it amounts to solving $$ X_{k+1} = A X_k $$ with the initial condition $$ X_0 = \begin{pmatrix} 1\\ 1\\ 2\\ 4 \end{pmatrix} $$ To do that, first you can diagonalize $A$, to get $A=P^{-1}\Delta P$ for some diagonal matrix $\Delta$; leading to $X_k = P^{-1} \Delta^k P X_0$, which once written out gives a closed-form expression for $X_k$ (and thus $a_k$).