P921
projecteuler.net

Golden Recurrence

ℹ️Published on Saturday, 14th December 2024, 01:00 pm; Solved by 182;
Difficulty rating: 45%

Consider the following recurrence relation: \begin{align} 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{align}

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 $n > 2$.

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

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



Soluzione

Last modified: May 01, 2025. Website built with Franklin.jl and the lovely Julia programming language.