The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: seq question -again
Replies: 16   Last Post: Oct 7, 2009 8:12 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Gottfried Helms

Posts: 1,926
Registered: 12/6/04
Re: seq question -again
Posted: Oct 6, 2009 7:14 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Am 06.10.2009 10:31 schrieb Gottfried Helms:
> Am 06.10.2009 03:20 schrieb leiko:
>> On Oct 5, 6:50 pm, (Erick Bryce Wong) wrote:
>>> leiko <> wrote:
>>>> Gerry <> wrote:
>>>>>>>> leiko wrote:
>>>>>>>>> hello everyone, I posted this a while ago, but no one responded, so i
>>>>>>>>> thought I should give it another try because probably not everyone has
>>>>>>>>> seen the earlier post.
>>>>>>>>> If we have a recursion x_1=2, x_n=2^(x_n-1), how does one show that
>>>>>>>>> the sequence x_n(mod m) stabilizes in Z/mZ for any m ?

There is an article in Journal of integers

> Daniel B. Shapiro
> Department of Mathematics, Ohio State University, Columbus, OH 43210
> S. David Shapiro
> 1616 Lexington Ave #D, El Cerrito, CA 94530

online, which is interesting for this.

It begins (sorry, no formatting here)
Entry A23:
> Abstract
> We show that if a1, a2, a3, . . . is a sequence of positive integers and k is
> given, then the sequence a1, a1^a2, a1^a2^a3 ... becomes constant when reduced (mod k).
> We also consider the sequence 1^1, 2^2, 3^3, . . . (mod k), showing that this sequence,
> and related ones like n^n^n (mod k), are eventually periodic.
> ?Dedicated to the memory of Prof. Arnold Ross

Gottfried Helms

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.