MathDB

5

Part of 2021 Czech-Austrian-Polish-Slovak Match

Problems(1)

Interesting sequence with \sqrt{2}

Source: 2021 Czech-Polish-Slovak Match, P5

8/3/2021
The sequence a1,a2,a3,…a_1, a_2, a_3, \ldots satisfies a1=1a_1=1, and for all n≥2n \ge 2, it holds that an={an−1+3  if n−1∈{a1,a2,…,,an−1};an−1+2  otherwise. a_n= \begin{cases} a_{n-1}+3 ~~ \text{if} ~ n-1 \in \{ a_1,a_2,\ldots,,a_{n-1} \} ; \\ a_{n-1}+2 ~~ \text{otherwise}. \end{cases} Prove that for all positive integers n, we have an<n⋅(1+2). a_n < n \cdot (1 + \sqrt{2}).
Dominik Burek (Poland) (also known as [url=https://artofproblemsolving.com/community/user/100466]Burii)