Difference between revisions of "1976 IMO Problems/Problem 6"
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<cmath>\lfloor u_{n} \rfloor = 2^{\frac {(2^{n} - ( - 1)^{n})}{3}}</cmath> | <cmath>\lfloor u_{n} \rfloor = 2^{\frac {(2^{n} - ( - 1)^{n})}{3}}</cmath> | ||
| − | (where <math>\lfloor x\rfloor</math> denotes the smallest integer <math>\leq</math> x)<math>.</math> | + | (where <math>\lfloor x\rfloor</math> denotes the smallest integer <math>\leq</math> <math>x</math>)<math>.</math> |
== Solution == | == Solution == | ||
Revision as of 13:03, 16 July 2018
Problem
A sequence 
is defined by
![\[u_{0} = 2 \quad u_{1} = \frac {5}{2}, u_{n + 1} = u_{n}(u_{n - 1}^{2} - 2) - u_{1} \quad \textnormal{for} n = 1,\ldots\]](http://latex.artofproblemsolving.com/b/a/9/ba97b544a1e5ab3c6cad35f88a8e176bf4bd07db.png)
Prove that for any positive integer 
we have
![\[\lfloor u_{n} \rfloor = 2^{\frac {(2^{n} - ( - 1)^{n})}{3}}\]](http://latex.artofproblemsolving.com/8/4/c/84c80cbc75bf044a90cc4ab9e30678b82fd5b905.png)
(where 
denotes the smallest integer 
)
Solution
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See also
| 1976 IMO (Problems) • Resources | ||
| Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Final Question |
| All IMO Problems and Solutions | ||
