Skip to main content
MathOverflow

Return to Question

replaced http://mathoverflow.net/ with https://mathoverflow.net/
Source Link
Community Bot
Community Bot
  • 1
  • 2
  • 3

It is proved HEREHERE that there is a natural number $N$ such that for any $n > N$ it is possible to place numbers $1,2,\cdots, n^2$ is an $n\times n$ square such that the numbers in any two adjacent square (having a mutual edge) are coprime. How can one estimate $N$ !? My guess is that the least such $N$ is equal to $1$.

It is proved HERE that there is a natural number $N$ such that for any $n > N$ it is possible to place numbers $1,2,\cdots, n^2$ is an $n\times n$ square such that the numbers in any two adjacent square (having a mutual edge) are coprime. How can one estimate $N$ !? My guess is that the least such $N$ is equal to $1$.

It is proved HERE that there is a natural number $N$ such that for any $n > N$ it is possible to place numbers $1,2,\cdots, n^2$ is an $n\times n$ square such that the numbers in any two adjacent square (having a mutual edge) are coprime. How can one estimate $N$ !? My guess is that the least such $N$ is equal to $1$.

Source Link
Hue2012
Hue2012
  • 41
  • 2

Placing numbers $1,2,\ldots,n^2$ in a square so that numbers of any two adjacent unit subcube are coprime

It is proved HERE that there is a natural number $N$ such that for any $n > N$ it is possible to place numbers $1,2,\cdots, n^2$ is an $n\times n$ square such that the numbers in any two adjacent square (having a mutual edge) are coprime. How can one estimate $N$ !? My guess is that the least such $N$ is equal to $1$.