![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
,
is an homoemorphisme too.
,
denote the greatest prime factor of 
and 
such that there is no integer 
.
Y by pb0975
Find the maximal value of 
such that for any 
, and for any 
real numbers 
there exists real numbers 
such that 
for at least 
indices 
. Here, 
and 
denotes the fractional part of 
.
Proposed by Wong Jer Ren
such that for any 
, and for any 
real numbers 
there exists real numbers 
such that 
for at least 
indices 
. Here, 
and 
denotes the fractional part of 
.Proposed by Wong Jer Ren
This post has been edited 3 times. Last edited by navi_09220114, Aug 28, 2023, 12:54 AM
.
.
stones at 
for each 
as a copy of 
on the stones, but each edge is replaced with two directed edges in each direction. Then we take our sequence as the numbers in some Hamiltonian cycle of 
chosen, we may check that there are at most 
occurences of the inequality by considering the numbers by considering what intervals 
lie in. As such, in the limiting case![\[
c \le \frac{(N+1)(N+2)}{2(2024N)(2024N-1)} \to \frac{1}{2 \cdot 2024^2}
\]](http://latex.artofproblemsolving.com/0/a/a/0aa7b69175c2473f3402f0213e9144c1d81de280.png)
as desired.
into 
ways to connect two intervals. Furthermore, for a pair 
intervals, we can define a set of 
neighbors 
for which 
.
where 
is the expected number of edges lying in one set, which is equal to 
.![\[
c \ge \frac{(N-1)N}{2(2024N)(2024N-1)} \to \frac{2}{2024^2} \to \frac{1}{2 \cdot 2024^2}
\]](http://latex.artofproblemsolving.com/4/f/d/4fd219c1dc48a2843d355a20b6467d123f286469.png)
in the limiting case. This means that for any sequence of 
indices for any 
.
so we are done.
