Difference between revisions of "2017 IMO Problems/Problem 1"
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| + | ==Problem== | ||
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For each integer <math>a_0 > 1</math>, define the sequence <math>a_0, a_1, a_2, \ldots</math> for <math>n \geq 0</math> as | For each integer <math>a_0 > 1</math>, define the sequence <math>a_0, a_1, a_2, \ldots</math> for <math>n \geq 0</math> as | ||
<cmath>a_{n+1} = | <cmath>a_{n+1} = | ||
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\end{cases} | \end{cases} | ||
</cmath>Determine all values of <math>a_0</math> such that there exists a number <math>A</math> such that <math>a_n = A</math> for infinitely many values of <math>n</math>. | </cmath>Determine all values of <math>a_0</math> such that there exists a number <math>A</math> such that <math>a_n = A</math> for infinitely many values of <math>n</math>. | ||
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| + | ==Solution== | ||
| + | {{solution}} | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{IMO box|year=2017|before=First Problem|num-a=2}} | ||
Revision as of 00:38, 19 November 2023
Problem
For each integer 
, define the sequence 
for 
as
![\[a_{n+1} = \begin{cases} \sqrt{a_n} & \text{if } \sqrt{a_n} \text{ is an integer,} \\ a_n + 3 & \text{otherwise.} \end{cases}\]](http://latex.artofproblemsolving.com/7/e/5/7e5005fa83196dcdb7e5270b0dc20e1f13e22bd4.png)
Determine all values of 
such that there exists a number 
such that 
for infinitely many values of 
.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See Also
| 2017 IMO (Problems) • Resources | ||
| Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
| All IMO Problems and Solutions | ||
