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Problem 921

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Golden Recurrence

Consider the following recurrence relation:
$$
\begin{aligned}
a_0 &= \frac{\sqrt 5 + 1}{2}\\
a_{n+1} &= \dfrac{a_n(a_n^4 + 10a_n^2 + 5)}{5a_n^4 + 10a_n^2 + 1}
\end{aligned}
$$

Note that $a_0$ is the golden ratio.

$a_n$ can always be written in the form $\dfrac{p_n\sqrt{5}+1}{q_n}$, 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 $n \gt 2$.

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：
$$
\begin{aligned}
a_0 &= \frac{\sqrt 5 + 1}{2}\\
a_{n+1} &= \dfrac{a_n(a_n^4 + 10a_n^2 + 5)}{5a_n^4 + 10a_n^2 + 1}
\end{aligned}
$$

注意到，$a_0$恰好是黄金比例。

$a_n$总可以写成$\dfrac{p_n\sqrt{5}+1}{q_n}$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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