request for help with obscrure sequence

[A.N.W.Hone at kent.ac.uk]

Hi seqfans,  

That looks like the right definition of the sequence, which perhaps the author (Roger Bagula) 
can confirm. 

I'd be very surprised if there is not a simple exact formuala for the degrees - let me have a look 
at it. 

I'll try to restore the broken link to my preprint this week, but I see that there is a cached copy. 
Maybe I'll just put it on the arXiv, if it's not there already. 

Andy 


----- Original Message -----
From: Richard Mathar <mathar at strw.leidenuniv.nl>
Date: Friday, July 11, 2008 8:09 pm
Subject: Re: request for help with obscrure sequence
To: seqfan at ext.jussieu.fr

> 
> njas> From seqfan-owner at ext.jussieu.fr  Fri Jul 11 14:48:15 2008
> njas> Date: Fri, 11 Jul 2008 08:46:52 -0400
> njas> From: "N. J. A. Sloane" <njas at research.att.com>
> njas> To: seqfan at ext.jussieu.fr
> njas> Subject: request for help with obscrure sequence
> njas> Cc: jerry_metzger at und.nodak.edu, njas at research.att.com
> njas> 
> njas> Dear Seqfans,  can someone figure out what the 
> definition means?
> njas> The link is broken, and A N W Hone has not responded to
> njas> my asking for a new URL for the article.
> njas> Neil
> njas> ...
> njas> %S A122046 
> 0,0,1,3,6,10,16,24,34,46,61,79,100,124,152,184,220,260,305,355,410njas> %N A122046 a(n)=(x^(n - 1)a(n - 1)a(n - 4) + a(n - 2)*a(n - 3))/a(n - 5).
> njas> %F A122046 Comment from Jerry Metzger 
> (jerry_metzger(AT)und.nodak.edu), Jul 09 2008: It appears that 
> every 4th term is given by Sum_{m=1...n-1} (floor{m(n-2)/n})^2 
> and the intermediate terms by adding an easy "adjustment" term 
> to that sum.
> njas> ...
> 
> The definition is
> Degree of the polynomial P(n,x)= [x^(n-1)*P(n-1,x)*P(n-4,x)+P(n-
> 2,x)*P(n-3,x)]/P(n-5,x) with P(1,x)=P(0,x)=P(-1,x)=P(-2,x)=P(-3,x)=1.
> 
> The sequence continues
> 0, 0, 1, 3, 6, 10, 16, 24, 34, 46, 61, 79, 100, 124, 152, 184, 
> 220, 260, 305,
> 355, 410, 470, 536, 608, 686, 770, 861, 959, 1064, 1176, 1296, 
> 1424, 1560, 1704,
> 1857, 2019, 2190, 2370, 2560
> 
> found with a Maple program
> 
> P := proc(n) option remember ;
>         if n <= 1 then
>                 RETURN(1) ;
>         else
>                 (P(n-1)*P(n-4)*q^(n-1)+P(n-2)*P(n-3))/P(n-5) ;
>                 expand(%) ;
>                 factor(%) ;
>         fi ;
> end:
> for n from 0 to 80 do
>         bag := P(n) ;
>         printf("%d 
> %d\n",n,degree(bag,q)) ;
> od:
> 
> It does not satisfy a linear recurrence with 6 or less 
> coefficients (as
> one may have hoped spying at A014125).
> 
> Jerry's formula sum_{m=1..n-1} floor[m(n-2)/2]^2 produces the sequence
> 
> 0, 0
> 1, 0
> 2, 0
> 3, 1
> 4, 14
> 5, 62
> 6, 220
> 7, 547
> 8, 1260
> 9, 2444
> 
> generated with
> 
> for n from 0 to 15 do
>         add( floor(m*(n-
> 2)/2)^2,m=1..n-1) ;
>         print(n,%) ;
> od:
> 
> which are hardly related to A122046, because only "1" and "220"
> match terms in the sequence (?).
> 
> Richard Mathar http://www.strw.leidenuniv.nl/~mathar
> 
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