Difference between revisions of "1995 OIM Problems/Problem 6"
| Line 2: | Line 2: | ||
A function <math>f: N \to N</math> is circular if for every <math>p</math> in <math>N</math> there exists <math>n</math> in <math>N</math> with <math>n \le p</math> such that | A function <math>f: N \to N</math> is circular if for every <math>p</math> in <math>N</math> there exists <math>n</math> in <math>N</math> with <math>n \le p</math> such that | ||
| − | <cmath>f^n(p) = f( f( \cdots n | + | <cmath>f^n(p) = f( f( \cdots \text{n times} \cdots f(p))) = p</cmath> |
The function f has degree of repulsion <math>k</math>, <math>0 < k<1</math>, if for each <math>p</math> in <math>N</math>, <math>f^i(p) \ne p</math> for <math>i=1, 2, \cdots , \left\lfloor k.p \right\rfloor</math> '''(*)'''. | The function f has degree of repulsion <math>k</math>, <math>0 < k<1</math>, if for each <math>p</math> in <math>N</math>, <math>f^i(p) \ne p</math> for <math>i=1, 2, \cdots , \left\lfloor k.p \right\rfloor</math> '''(*)'''. | ||
Revision as of 13:57, 13 December 2023
Problem
A function 
is circular if for every 
in 
there exists 
in with 
such that
![\[f^n(p) = f( f( \cdots \text{n times} \cdots f(p))) = p\]](http://latex.artofproblemsolving.com/1/2/3/1230098fb871a4281112095eca3993e1d06896b8.png)
The function f has degree of repulsion 
, 
, if for each in , 
for 
(*).
Find the greatest degree of repulsion that a circular function can have.
Note (*): 
indicates the largest integer less than or equal to 
.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
