Recurring Styles 13 - Game of Inequalities!

Algebra Level 5

a n + 1 = a n n + n a n , a 1 = 1 \Large{ a_{n+1} = \dfrac{a_n}{n} + \dfrac{n}{a_n} \quad, \quad a_1 = 1}

Let the sequence < a n > \left< a_n \right> be defined as above for all positive integers n n . Evaluate ⌊ a 2015 ⌋ \large{\left \lfloor a_{2015} \right \rfloor }


The answer is 44.

Satyajit Mohanty by Satyajit Mohanty , 24, India

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2 solutions

Aareyan Manzoor
Sep 20, 2015

easy python3:

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Does anyone have a math solution?

Alan Yan - 5 years, 8 months ago
Karan Siwach
Sep 17, 2015 
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#include <stdio.h>


#include <stdlib.h>


float series(int n)


{


if (n==1){return 1;}else


 {


float t= series(n-1);


float f=t/(float)(n-1)+(float)(n-1)/t;


return f;


}


}



int main()


{


float f;


f=series(2015);


printf("%f", f);


return 0;


}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Is there a mathematical method??? @Pi Han Goh @Brian Charlesworth ??

Aaghaz Mahajan - 1 year ago

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I think yes, ⌊ a n ⌋ \lfloor a_n \rfloor is closest integer value to n \sqrt n ... maybe?

Pi Han Goh - 1 year ago

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