Discrete generalization of Fibonacci golden ratios?

Question

I have found the connection of :

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

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

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

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It seems that A, B, C is a root of :
$$A \rightarrow a^1 + a^2 = a^3$$
$$B \rightarrow a^1 + a^2 + a^3 = a^4$$
$$C \rightarrow a^1 + a^2 + a^3 + a^4 = a^5$$

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How can I prove it ?

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Update nr 1

$$ \sum_{n=1}^{m} x ^ n = \frac{x (x^m - 1)}{x - 1} = x ^ {m+1} $$ $$ \frac{(x - 2) x^m + 1}{x - 1} = 0 $$

Answer 1

For any recurrent relation $$ a_n = a_{n-1} + a_{n-2} + \ldots + a_{n-k} $$ we can find $k$ geometric progressions that will satisfy this relation. Indeed,$a_n = \lambda^n$, where $\lambda$ is a root of characteristic polynomial $\lambda^k = \lambda^{k-1} + \lambda^{k-2} + \ldots + \lambda +1$ satisfies the recurrent relation.

It can be shown that general solution is $$ a_n = \sum_{i=1}^k c_i \lambda_i^n, $$ where $\lambda_i$ are the roots of characteristic polynomial, $c_i$ are constants that can be found from the condition $$ a_1 = a_2 = \ldots = a_k = 1. $$ Without losing of generality let $c_1 \neq 0$ and $|\lambda_1| = \max_{i} |\lambda_i|$. Then assuming $\lambda_i \neq \lambda_j$ for $i \neq j$ $$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\sum_{i=1}^k c_i \lambda_i^{n+1}}{\sum_{i=1}^k c_i \lambda_i^{n}} = \lambda_1. $$