For My Friend Nihar Mahajan (Happy Birthday Man)

Algebra Level 5

$\large{{ x }_{ n+1 }=\left\lfloor \sqrt [ 3 ]{ { x }_{ n }^{ 3 }-\left\lfloor \sqrt [ 3 ]{ { x }_{ n } } \right\rfloor } \right\rfloor }$

Let there be a sequence defined by above rule for $n \ge 1$ . For $n \ge k$ the value of $a_{n}$ will be $0$ , for some positive integer $k$ . Find the value of $k$ , if $a_{1}=839475687$ .

The answer is 839475688.

1 solution

Gian Sanjaya
Oct 7, 2015

For $a_n$ big enough, more than 1, the RHS expression has $a_n$ cubed (while it is a positive integer) substracted by a positive integer less than $3a_n^2-3a_n+1$ , actually even less than $a_n$ , then taken the cube root, so we get a value less than $a_n$ but more than $a_n-1$ , and then floored, which means, actually, $a_{n+1}=a_n-1$ for every positive integer n such that $a_n\geq 2$ . Checking $a_n=1$ , we will find that it also works. Then, $k=1+a_1=\boxed{839475688}$

You are absolutely right nice solution upvoted.

Department 8 - 5 years, 8 months ago

For My Friend Nihar Mahajan (Happy Birthday Man)

The answer is 839475688.

1 solution

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