[seqfan] Re: LRS formulas for n>=...?

[rhhardin at att.net]

A163002-A163154 are the same computation without the <=2 adjacent
condition.  As I recall these had lots of recurrences but I didn't
mention them because I hadn't tested that software very much.

These new ones were submitted as A163684-A163741 (a couple added
as the last of the laptops finished overnight), though they may not get those
numbers in the end of course.


Hamming: The purpose of computation is understanding, not numbers.
--
rhhardin at mindspring.com
rhhardin at att.net (either)
  
-------------- Original message ----------------------
From: Hagen von Eitzen <math at von-eitzen.de>
>
> rhhardin at att.net schrieb:
> > Yes, that looks good.
> >
> > In fact just proving the existence of a recursion is enough to prove
> > an empirical recursion, with some decision about how many terms
> > it might have at most.
> >
> > Enumeration of enough terms fades fast as m increases in nXm however.
> >
> >   
> BTW, I'm missing sequences without any "path form here to there" 
> condition, i.e.
> let F(n,m) be the number of nxm binary arrays with all 1's connected and 
> no 1 having more than two 1s adjacent.
> Then e.g. your A000032(n) = F(n,4) - 2*F(n-1,4) + F(n-2,4) and 
> A000040(n) = F(n,4) - 2*F(n,3) + F(n,2).
> 
> Hagen
> 
> 
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