# sequence, not easy but not hard

• Nov 11th 2007, 07:47 AM
Ununuquantium
sequence, not easy but not hard
A prime number $p$ is given. A sequence of positive integers $a_{1}, a_{2}, a_{3},...$ is determined by this conditon:
$a_{n+1}=a_{n}+p \lfloor \sqrt[p]{a_{n}} \rfloor$
Prove that there is a term in this sequence, which is a $p$-power of an integer number.
• Dec 2nd 2007, 12:07 PM
Ununuquantium
Anybody can help me ?? Is this so hard?
• Dec 4th 2007, 09:48 PM
$a_1$ is a 1-power of an integer :).

or have I misunderstood?
• Dec 9th 2007, 04:05 AM
Ununuquantium
1 is not prime number. It is harder question.

• Dec 9th 2007, 07:24 AM
topsquark
Quote:

Originally Posted by Ununuquantium
A prime number $p$ is given. A sequence of positive integers $a_{1}, a_{2}, a_{3},...$ is determined by this conditon:
$a_{n+1}=a_{n}+p \lfloor \sqrt[p]{a_{n}} \rfloor$
Prove that there is a term in this sequence, which is a $p$-power of an integer number.

What do you mean by "p-power?" For example, choose p = 2 (to keep things simple.) Then given an $a_1 = 2$ (for instance) we get
$a_2 = a_1 + 2\sqrt{a_1} = 2 + 2\sqrt{2}$

$a_3 = a_2 + 2\sqrt{a_2} = 2 + \sqrt{2} + 2\sqrt{2 + \sqrt{2}}$

etc.

Are you saying that for some n of this sequence that $a_n = k^p$ where k is some integer?

-Dan
• Dec 9th 2007, 12:08 PM
Ununuquantium
Yes exactly!!
$p$ power of an integer means
$a_{n}=k^{p}$ k integer, p prime