Closed form for $F(n) = \sum_{i=1}^k F(n-i)$ If we have $F(n) = F(n-1) + F(n-2)$ and $F(0) = 0$ and $F(1) =1$ then we get the Fibonacci sequence.  A closed form solution is:
$$
F_n = \left[\frac{\left(\frac{1+\sqrt{5}}{\sqrt{5}}\right)^n}{\sqrt{5}}\right]
,$$
where $[]$ rounds to the nearest integer.
What do we get as a closed form solution for $F(n) = \sum_{i=1}^3 F(n-i)$ with $F(0) = 1, F(1) = 2, F(2) = 4$?

In general, if we set $F(i) = a_i$ for $i \in
 \{0,\dots,k-1\}$, $a_i \geq 0$ and constant $k >1$, what does the closed
  form solution for $F(n) = \sum_{i=1}^k F(n-i)$ look like for large $n$?

 A: Example:
$F(n)=F(n-1)+F(n-2)\enspace $ with given $\enspace F(0)\enspace $ and $\enspace F(1)$ 
$x^n=x^{n-1}+x^{n-2}\enspace$ means $\enspace x^2=x^1+x^0\enspace$ => $\enspace x\in\{x_1,x_2\}$   
$F(n):=ax_1^n+bx_2^n$
Linear Equation System:   $\enspace ax_1^0+bx_2^0=F(0) \enspace $   and  $\enspace ax_1^1+bx_2^1=F(1)$ 
It follows $\,a\,$ and $\,b\,$ and therefore $\,F(n)$ .  
Note: Now you know how to solve for the general $k\,$ (above: $\,k=2$) .
A: The solution has an increasing algebraic difficulty because a closed form is linked to the roots of an equation of degree $n$ just as in Fibonacci sequence you must solve $x^2-x-1=0$ for the your question $F(n) = \sum_{i=1}^3 F(n-i)$ you need to solve $ z^3-z^2-z-1=0$. Two roots are complex numbers and the general solution is quite ugly
For $n>4$, as you know,  in general there is no way to find roots by radicals  but the aspect is always similar to Fibonacci closed form. Roots of polynomial equations raised to $n$ and added together
An approximated form which works great is
$f(n)\approx -(0.0687258\, -0.123522 i) (-0.419643+0.606291 i)^n+(-0.0687258-0.123522 i) (-0.419643-0.606291 i)^n+1.13745\cdot 1.83929^n$
which gives the first terms
$1,\;2,\;4,\;7,\;13,\;24,\;44,\;81,\;149,\;274,\;504,\;927,\;1705,\;3136,\;5768,\;10609,\;19513,\;35890,\;66012,\;121415,\;223317,\ldots$
