[seqfan] Re: Problem reproducing A103750 (forth order linear recurrence)
zak seidov
zakseidov at yahoo.com
Sat Jul 31 18:31:13 CEST 2010
My best guess is:
a[0]=1;a[1]=1;a[2]=-2;
a[n_]:=a[n]=2*a[n-3]-a[n-2]-a[n-1];
Table[a[n],{n,0,40}]
which gives:
1,1,-2,3,1,-8,13,-3,-26,55,-35,-72,217,-215,-146,
795,-1079,-8,2677,-4827,2134,8047,-19835,16056,
19873,-75599,87838,27507,-266543,414712,-93155,
-854643,1777222,-1108889,-2377619,7040952,
-6881111,-4915079,25878094,-34725237,-983015
and taking Abs (but why not to live as is and tag "sign"?)
we get the A103750.
Zak
--- On Sat, 7/31/10, Richard Mathar <mathar at strw.leidenuniv.nl> wrote:
> From: Richard Mathar <mathar at strw.leidenuniv.nl>
> Subject: [seqfan] Problem reproducing A103750 (forth order linear recurrence)
> To: seqfan at seqfan.eu
> Date: Saturday, July 31, 2010, 11:02 AM
>
> Can someone with Mma reproduce http://www.oeis.org/classic/A103750 ?
> According to the formula this should be the absolute terms
> of a
> 4th order linear recurrence a(n)=a(n-1)-a(n-3)+2*a(n-4).
> I cannot reproduce this from any starting configuration
> (1,1,1,2),
> (-1,1,1,2), (1,-1,1,2), (1,1,-1,2), (-1,-1,1,2) etc
> of the a(0) to a(3).
>
> In addition: what is the sense of referencing A073058 in
> A103750?
>
> In contrast, A103749 is consistent with the 3rd order
> recurrence provided there.
>
>
> RJM
>
>
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