Define a sequence $ a_n = n^n + (n - 1)^{n+1}$ when $n$ is a positive integer. Define all those positive integer $m$ , for which this sequence of numbers is eventually periodic modulo $m$, e.g. there are such positive integers $K$ and $s$ such that $a_k \equiv a_{k+s}$ ($mod \,m$), where $k$ is an integer with $k \ge K$.
Problem
Source: Finland 2019, p4
Tags: Periodic sequence, Sequence, periodic, bir tinga qimmat masala
NikoIsLife
09.09.2019 13:18
parmenides51 wrote: Define a sequence $ a_n = n^n + (n - 1)^{n+1}$ when $n$ is a positive integer. Define all those positive integer modules $m$, where this sequence of numbers is eventually periodic, e.g. there are such positive integers $K$ and $s$ such that $a_k \equiv a_{k+s}$ ($mod \,m$), when $k \ge K$ is an integer. What is the problem?
parmenides51
09.09.2019 14:17
find all the positive integers $m$, for which this sequence above mod $m$, is eventually periodic I hope it makes sense now, if you could suggest a better way to rewrite the wording, it is welcome (my translation's attempt comes with the help of Google Translate, but it is not that good with Finnish)
bora_olmez
19.09.2020 14:17
parmenides51 wrote: Define a sequence $ a_n = n^n + (n - 1)^{n+1}$ when $n$ is a positive integer. Define all those positive integer $m$ , for which this sequence of numbers is eventually periodic modulo $m$, e.g. there are such positive integers $K$ and $s$ such that $a_k \equiv a_{k+s}$ ($mod \,m$), where $k$ is an integer with $k \ge K$. Any solutions? Anyone?
Sprites
28.07.2021 14:24
$a_x \equiv a_{x+p(p-1)} \pmod p $ where $m=p$
Just expand it$\pmod p$ and they will be identical.
$a_x \equiv a_{x+p \cdot \prod {r_i}}$ where $r_i=ord_m(x_i)$
Note:-Suppose we are considering some divisor of $m$ where $m|d^2$ then this case we won't be able to use orders,however since $x+pQ(x) \equiv x \pmod p$ so again you just have to expand $(d_r-1)^{d_r+1}$ where $d_r|M$ and then get the answer.
Just expand it$\pmod p$ and they will be identical.
Hence,we have showed that it is periodic$\pmod m$ for all $m$
laikhanhhoang_3011
28.07.2021 16:57
https://laikhanhhoangchuyenndu.blogspot.com/2021/07/2019-finland-mo-p4.html Am i wrong? Sorry but i dont know how to use latex