Duplicate hunting, pt. 10

Andrew Plewe aplewe at sbcglobal.net
Fri May 4 23:46:59 CEST 2007

submissions to allow for list members to take care of as many as possible).
submitted as well; I can post that on the blog but I don't think it'd be of
	-Andrew Plewe-
From: Max Alekseyev [mailto:maxale at gmail.com]
Sent: Friday, May 04, 2007 2:13 PM
To: Andrew Plewe
Cc: seqfan at ext.jussieu.fr
Subject: Re: Duplicate hunting, pt. 10
Return-Path: <maximilian.hasler at gmail.com>
X-Ids: 168
DKIM-Signature: a=rsa-sha1; c=relaxed/relaxed;
        d=gmail.com; s=beta;
DomainKey-Signature: a=rsa-sha1; c=nofws;
        d=gmail.com; s=beta;
Message-ID: <3c3af2330705041604w1560980cl5c725acf483cabcc at mail.gmail.com>
Date: Fri, 4 May 2007 19:04:09 -0400
From: "Maximilian Hasler" <maximilian.hasler at gmail.com>
To: "Andrew Plewe" <aplewe at sbcglobal.net>
Subject: Re: Duplicate hunting, pt. 10
Cc: seqfan at ext.jussieu.fr
In-Reply-To: <200705042147.l44Ll3SI032724 at shiva.jussieu.fr>
MIME-Version: 1.0
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Content-Transfer-Encoding: 7bit
Content-Disposition: inline
References: <d3dac270705041412x7b193ffdrd5434a302c479090 at mail.gmail.com>
	 <200705042147.l44Ll3SI032724 at shiva.jussieu.fr>
X-Greylist: IP, sender and recipient auto-whitelisted, not delayed by milter-greylist-3.0 (shiva.jussieu.fr []); Sat, 05 May 2007 01:04:17 +0200 (CEST)
X-Virus-Scanned: ClamAV 0.88.7/3206/Fri May  4 21:42:28 2007 on shiva.jussieu.fr
X-Virus-Status: Clean
X-j-chkmail-Score: MSGID : 463BBBF0.001 on shiva.jussieu.fr : j-chkmail score : X : 0/50 1 0.478 -> 1
X-Miltered: at shiva.jussieu.fr with ID 463BBBF0.001 by Joe's j-chkmail (http://j-chkmail.ensmp.fr)!

> For ease of setup and speed, I've created a blog:
> http://duplicate-sequences.blogspot.com/

Great idea. A wiki would be even better, aren't there some free
providers for that ?

A question close to the topic: what about sequences that just differ by 1,
i.e. a(n)=b(n)+1 for all n ?
I ask it since I just ran across the following (included below), the
comment is easily proved:
A117805(n+1) = (A117805(n)-1) A117805(n)  by definition, and
A005267(n+1) +1 = product(A005267(k), k=1..n ) by definition.
If, by assumption, the latter is equal to A117805(n),
then A117805(n+1) = A005267(n+1) product(A005267(k), k=1..n )
 =product(A005267(k), k=1..n+1 ) = A005267(n+2) +1,
i.e. the assumption is true for all following n.


http://www.research.att.com/~njas/sequences/A117805 :
A117805 		Start with 3. Square the previous term and subtract it. 	
	3, 6, 30, 870, 756030, 571580604870, 326704387862983487112030,
(list; graph; listen)
Apparently a(n)=A005267(n+1)+1. - R.J. Mathar
(mathar(AT)strw.leidenuniv.nl), Apr 22 2007

More information about the SeqFan mailing list