# Discrete generalization of Fibonacci golden ratios?

I have found the connection of :

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

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

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

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

How can I prove it ?

Update nr 1

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

• Have you tried diagonalising?
– Zima
May 27 at 14:14
• It follows from the general theory of linear recurrence relations. May 27 at 14:50

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 $$i \neq j$$ $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\sum_{i=1}^k c_i \lambda_i^{n+1}}{\sum_{i=1}^k c_i \lambda_i^{n}} = \lambda_1.$$