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Omni-MATH / Let a sequence {a_(n)}_(n=0)^(∞) be defined by a₀=√(2),a₁=2, and a_(n+1)=a_(n)a_(n−1)² for n≥1. The sequence of remainders when a₀,a₁,a₂,⋯ are divided by 2014 is eventually periodi…

Problem

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problem

Let a sequence {an}n=0∞\left\{a_{n}\right\}_{n=0}^{\infty} be defined by a0=2,a1=2a_{0}=\sqrt{2}, a_{1}=2, and an+1=anan−12a_{n+1}=a_{n} a_{n-1}^{2} for n≥1n \geq 1. The sequence of remainders when a0,a1,a2,⋯a_{0}, a_{1}, a_{2}, \cdots are divided by 2014 is eventually periodic with some minimal period pp (meaning that am=am+pa_{m}=a_{m+p} for all sufficiently large integers mm, and pp is the smallest such positive integer). Find pp.
Plain-text mathematical notation (without MathML)
Let a sequence {a_(n)}_(n=0)^(∞) be defined by a₀=√(2),a₁=2, and a_(n+1)=a_(n)a_(n−1)² for n≥1. The sequence of remainders when a₀,a₁,a₂,⋯ are divided by 2014 is eventually periodic with some minimal period p (meaning that a_(m)=a_(m+p) for all sufficiently large integers m, and p is the smallest such positive integer). Find p.
Original LaTeX notation
Let a sequence $\left\{a_{n}\right\}_{n=0}^{\infty}$ be defined by $a_{0}=\sqrt{2}, a_{1}=2$, and $a_{n+1}=a_{n} a_{n-1}^{2}$ for $n \geq 1$. The sequence of remainders when $a_{0}, a_{1}, a_{2}, \cdots$ are divided by 2014 is eventually periodic with some minimal period $p$ (meaning that $a_{m}=a_{m+p}$ for all sufficiently large integers $m$, and $p$ is the smallest such positive integer). Find $p$.

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Source and history

task_id
6ccaffa4-cb0e-5cec-a6ff-f1918c0e0728
task_key
test--6ccaffa4-cb0e-5cec-a6ff-f1918c0e0728
task_revision_id
1
config / split
/ test

Official source

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