What is the general continuous form of discrete generalization of Fibonacci golden ratios? What is the general continuous form?
The general form for discrete numbers is:

$$
a_n = a_{n-1} + a_{n-2}
$$
$$
A = \lim_{n \to \infty} \biggl( \frac{a_n}{a_{n-1}} \biggr)
$$
$$a_0 = 1$$
$$a_1 = 1$$

$$
b_n = b_{n-1} + b_{n-2} + b_{n-3}
$$
$$
B = \lim_{n \to \infty} \biggl( \frac{b_n}{b_{n-1}} \biggr)
$$
$$b_0 = 1$$
$$b_1 = 1$$
$$b_2 = 1$$

$$
c_n = c_{n-1} + c_{n-2} + c_{n-3} + c_{n-4}
$$
$$
C = \lim_{n \to \infty} \biggl( \frac{c_n}{c_{n-1}} \biggr)
$$
$$c_0 = 1$$
$$c_1 = 1$$
$$c_2 = 1$$
$$c_3 = 1$$

$$
\dots
$$

 A: All these (homogeneous) recurrence relations are linear with constant coefficients, such that you can use the characteristic method.
The $m^\mathrm{th}$ "step" corresponds to the recurrence relation $a_n = a_{n-1} + a_{n-2} + \ldots + a_{n-m}$, which leads to the following characteristic polynomial :
$$
\lambda^m - \lambda^{m-1} - \lambda^{m-2} - \ldots - 1 = \lambda^m - \sum_{k=0}^{m-1}\lambda^k = \lambda^m - \frac{\lambda^m-1}{\lambda-1} = \frac{\lambda^{m+1}-2\lambda^m+1}{\lambda-1}
$$
Its roots $\{\lambda_1,\ldots,\lambda_m\}$ are those of the $m$-degree polynomial $\lambda^{m+1}-2\lambda^m+1$, except $\lambda_0=1$ because of the denominator. Unfortunately, I'm not aware of a closed form.
Nevertheless, the general solution will be given by the linear combination $a_n = c_1\lambda_1^n + \ldots + c_m\lambda_m^n$, hence
$$
A_m := \lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{c_1\lambda_1^{n+1}+\ldots+c_m\lambda_m^{n+1}}{c_1\lambda_1^n+\ldots+c_m\lambda_m^n} = \max_{1\le i\le m} \lambda_i
$$

N.B. : I wouldn't call your problem a continuous generalization of the Fibonacci sequence; indeed, it would imply a continuous variable $n$, renamed $x$, hence the relation $f_m(x) = f_m(x-1) + f_m(x-2) + \ldots + f_m(x-m)$ at the $m^\mathrm{th}$ step, which will be usually treated as a functional equation rather than a recurrence relation.
