[seqfan] Re: Excess permutation left moves over right moves

Ron Hardin rhhardin at att.net
Sun Jul 18 02:21:09 CEST 2010

I'll try that next; but am diverted by the puzzle with the A000002-A000005.

Next, actually, is redoing the columns for those jagged column ones (with more 
offsets) with a less error-prone adhoc directory structure,
to avoid slips of the finger contaminating the data.

Then on to the cumulative excess ones you suggest.  (Is it obvious that if the 
individual ones have recurrences that the cumulative must also?)

I suspect the 0 excess will once again not fit in any recurrence; you have to 
start with 1 excess.  I bet.

 rhhardin at mindspring.com
rhhardin at att.net (either)

----- Original Message ----
> From: William Keith <wjk26 at drexel.edu>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Sat, July 17, 2010 7:30:37 PM
> Subject: [seqfan] Re: Excess permutation left moves over right moves
> Suppose you take A000006 through A000010 and produce the linear recursions  
>associated to those, then form the jagged table of the signed coefficients that  
>appear in the recursions with them and the recursion in A000011 through   
>A000013, i.e. these:
> >  a(n)=5*a(n-1)-9*a(n-2)+7*a(n-3)-2*a(n-4)
> (giving 5, -9, 7,  -2)
> Perhaps there might be an interesting Pascaline rule on the  coefficients thus 
>arranged?  It would be especially interesting if it could  be proved that the 
>recurrences hold indefinitely, which it seems plausible to me  that they would.
> William  Keith
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> Seqfan  Mailing list - http://list.seqfan.eu/

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