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

[Ron Hardin]

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

A quick side calculation shows that the cumulative sequences have the same
recurrence as the individual ones (with a different series, of course).

Here's for left exceeds right by one or more, out to the 30th term
1 37
2 73
3 142
4 276
5 539
6 1059
7 2092
8 4150
9 8257
10 16461
11 32858
12 65640
13 131191
14 262279
15 524440
16 1048746
17 2097341
18 4194513
19 8388838
20 16777468
21 33554707
22 67109163
23 134218052
24 268435806
25 536871289
26 1073742229
27 2147484082
28 4294967760
29 8589935087
30 17179869711
Empirical fit from a(1..8), then matches a(9..30): 
a(n)=5*a(n-1)-9*a(n-2)+7*a(n-3)-2*a(n-4)

same as before.


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



----- Original Message ----
> From: Ron Hardin <rhhardin at att.net>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Sat, July 17, 2010 8:21:09 PM
> Subject: [seqfan] Re: Excess permutation left moves over right moves
> 
> 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
> > 
> >  _______________________________________________
> > 
> > Seqfan   Mailing list - http://list.seqfan.eu/
> > 
> 
> 
> 
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