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Problem 921

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Golden Recurrence

Consider the following recurrence relation:
$$\begin{array}{rl}{a}_0& =\frac{\sqrt{5}+1}{2}\\ a_{n+1}& ={\displaystyle \frac{a_n(a_n^4+10a_n^2+5)}{5a_n^4+10a_n^2+1}}\end{array}$$

Note that $a_0$ is the golden ratio.

$a_n$ can always be written in the form ${\displaystyle \frac{p_n\sqrt{5}+1}{q_n}}$, where $p_n$ and $q_n$ are positive integers.

Let $s(n)=p_n^5+q_n^5$. So, $s(0)=1^5+2^5=33$.

The Fibonacci sequence is defined as: $F_1=1$, $F_2=1$, $F_n=F_{n-1}+F_{n-2}$ for $n>2$.

Define ${\displaystyle S(m)=\sum _{i=2}^{m}s(F_i)}$.

Find $S(1618034)$. Submit your answer modulo $398874989$.

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黄金递推

考虑以下递推关系：
$$\begin{array}{rl}{a}_0& =\frac{\sqrt{5}+1}{2}\\ a_{n+1}& ={\displaystyle \frac{a_n(a_n^4+10a_n^2+5)}{5a_n^4+10a_n^2+1}}\end{array}$$

注意到，$a_0$恰好是黄金比例。

$a_n$总可以写成${\displaystyle \frac{p_n\sqrt{5}+1}{q_n}}$的形式，其中$p_n$和$q_n$是正整数。

令$s(n)=p_n^5+q_n^5$。因此，$s(0)=1^5+2^5=33$。

斐波那契数列的定义是：$F_1=1$，$F_2=1$，对于$n>2$有$F_n=F_{n-1}+F_{n-2}$。

定义${\displaystyle S(m)=\sum _{i=2}^{m}s(F_i)}$。

求$S(1618034)$，并对$398874989$取余作为你的答案。

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