A very Recursive Sequence

Number Theory Level 4

If a recursive sequence is defined by a 1 = 1 a_{1}=1 , a 2 = 1 2 a_{2}=\dfrac{1}{2} , and a n = 1 − a n − 1 2 a n − 2 a_{n}=\dfrac{1-a_{n-1}}{2a_{n-2}} for all n ≥ 3 n \geq {3} , then a 2019 a_{2019} can be written as p q \dfrac{p}{q} , where p p and q q are relatively prime positive integers. What is p + q p+q ?


The answer is 2021.

Aidan Poor by Aidan Poor , 18, USA

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1 solution

Mark Hennings
Mar 20, 2019

A simple induction shows that x 3 n = n 2 n + 2 x 3 n + 1 = n + 2 2 n + 2 x 3 n + 2 = 1 2 x_{3n} \; = \; \frac{n}{2n+2} \hspace{1cm} x_{3n+1} \; = \; \frac{n+2}{2n+2} \hspace{1cm} x_{3n+2} \; = \; \frac12 for all n ≥ 0 n \ge 0 . Thus x 2019 = 673 2 × 673 + 2 = 673 1348 x_{2019} \; = \; \frac{673}{2\times673+2} = \frac{673}{1348} and hence p + q = 673 + 1348 = 2021 p+q = 673 + 1348 = \boxed{2021} .

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