I’ve been trying to find an explicit formula for the following: $$f(n)=f(n-1)+f(n-2)+\cdots+f(n-a)$$ $$f(1)=F(1), f(2)=F(2), \ldots, f(a)=F(a)$$ When $F(x)$ is the $x^\text{th}$ Fibonacci number
I can easily find that $\lim_{n\to\infty}\frac{f(n+1)}{f(n)}=r=\sum_{k=0}^{a-1}r^{-k}$. We can easily see that this has $a$ number of $r$’s which I call $r_1,r_2,r_3\ldots r_a$, and we know all these values.
For the case $a=2$, we can form the formula $f(n)=\frac{r_1^n-r_2^n}{r_1-r_2}$ and for $a=3$, we have $f(n)= \frac{r_1^{n+1}}{\left(r_1-r_2\right)\left(r_1-r_3\right)}+ \frac{r_2^{n+1}}{\left(r_2-r_1\right)\left(r_2-r_3\right)}+ \frac{r_3^{n+1}}{\left(r_3-r_1\right)\left(r_3-r_2\right)}$
But for any other $a$, I’m stuck, I don’t know what operations need to be done with all the $r$’s and what will be the denominator. However, I need a formula without the floor function or an approximation, I need such a formula where we can input any $n$ such that $n\in\mathbb{C}$ gives us a perfect answer. Can anyone please help?
Instead of this, if we look at another case when
$$f(2-a)=0,f(3-a)=0,\ldots,f(a-a)=0,f(1)=1$$ Then we can see that the explicit formula can be given as: $$f(n)=\frac{r_1^{n+a-2}}{\left(r_1-r_2\right)\left(r_1-r_3\right)\cdots\left(r_1-r_a\right)}+ \frac{r_2^{n+a-2}}{\left(r_2-r_1\right)\left(r_2-r_3\right)\cdots\left(r_2-r_a\right)}+\cdots+\frac{r_a^{n+a-2}}{\left(r_a-r_1\right)\left(r_a-r_2\right)\cdots\left(r_a-r_{a-1}\right)}$$
But my situation is a bit different, so what can I do?
Mostly I want to know about the cases $a\gt 3$