One of the most important generalization of the classical Fibonacci numbers is the Fibonacci $p$-step numbers that is defined as follows

\begin{equation}\label{cp26} F_n^{(p)}=F_{n-1}^{(p)}+F_{n-2}^{(p)}+\cdots+F_{n-p}^{(p)}\, . \end{equation}

With boundary conditions

$$F_{0}^{(p)}=0\quad , \quad F_{1}^{(p)}=0\quad ,\, \cdots\, ,\quad F_{p-2}^{(p)}=0\quad , \quad F_{p-1}^{(p)}=1\, .$$

We can get the limit value of Fibonacci $p$-step numbers by inverse of solution of equation $x^{p+1}-2\, x+1=0$ in the interval $(0,1)$. We denote the limit value of Fibonacci $p$-step numbers with $\alpha_p$. In fact, $\alpha_p$ is defined in the following form $$ \alpha_p=\displaystyle{\lim_{n\rightarrow\infty}}\quad \frac{F^{(p)}_{n+1}}{F^{(p)}_{n}}~. $$

We denote the length of $ F_n^{(p)}$ number with $L_n^{(p)}$. For example, $ F_{10}^{(2)}=55$ therefore $L_{10}^{(2)}=2$.
With simulation, I found the following connection between the length of Fibonacci $p$-step numbers and its limit values
$$
\mid \frac{F^{(p)}_{n+1}}{F^{(p)}_{n}}-\alpha_p\mid \, \approx \, 10^{-{\displaystyle{L_n^{(p)}(\frac{p}{p-1})}}}~.
$$

I make some examples to clarify what i mean. For $p=2$, the Fibonacci $2$-step number is the classical Fibonacci number.
The limit value of the Fibonacci $2$-step number is inverse of solution of equation $x^3-2\, x+1=0$ in
the interval $(0,1)$.
so, for $p=2$, the limit value is

\begin{eqnarray} \alpha_2= \frac{1}{0.61803398874989484820458683436563811772030917980580}& & \\ & & \\ = 1.6180339887498948482045868343656381177203091798058~.& & \end{eqnarray}

The length of $F_{72}^{(2)}$ and $ F_{73}^{(2)}$ are $15$. With calculation, we have the following relation

$$
\mid \frac{F^{(2)}_{73}}{F^{(2)}_{72}}-\alpha_2\mid=
\mid \frac{ 806515533049393}{ 498454011879264}-\alpha_2\mid
\, \approx \, 10^{-30}~.
$$

For $p=3$, the Fibonacci $3$-step number is called Tribonacci number.
The limit value of the Fibonacci $3$-step number is inverse of solution of equation $x^4-2\, x+1=0$ in
the interval $(0,1)$.

\begin{eqnarray} \alpha_3= &1.8392867552141611325518525646532866004241787460975 & \\ & 92246778758639404203222081966425738435419428307014& ~. \end{eqnarray}

The length of $F_{113}^{(3)}$ and $ F_{114}^{(3)}$ are $30$. With calculation, we have the following relation

$$ \mid \frac{F^{(3)}_{114}}{F^{(3)}_{113}}-\alpha_3\mid= \mid \frac{270409000010299937423541789777}{ 147018402238651639664609196920}-\alpha_3\mid \, \approx \, 10^{-45}~. $$ Is there a analytical method that we can proof the mentioned formula. I would greatly appreciate for any suggestions.