=Lozanic's triangle A0348512 -- Matrix Log Holds Key

[Paul D. Hanna]

Dear Christian (and Seqfans)
         Very interesting triangle ... when one takes the matrix log, 
the pattern is obvious and significant.  
This pattern accounts for the unusual sign pattern in the matrix inverse:
log(T) = 
0;
1, 0;
1, 0, 0;
0, 1, 1, 0;
0, 0, 2, 0, 0;
0, 0, 0, 2, 1, 0;
0, 0, 0, 0, 3, 0, 0;
0, 0, 0, 0, 0, 3, 1, 0; ...
 
This pattern also says something about the e.g.f of your original
triangle. 
  
Paul

On Fri, 03 Mar 2006 18:49:15 -0800 "Christian G.Bower" <bowerc at usa.net>
writes:
> > The definition of most positions in Lozanic's triangle are
> > understoodby most people (even politicians who think pi is 3), but 
> the
> > "red"entries require subtracting C(n/2-1,(k-1)/2). This triangle
> > "deservesto be better known" (Neil Sloane, in "Classic Sequences") 
> but
> > the lackof a simple definition for the "red" entries stands in the 
> way
> > of thishappening. I've been looking for ways to calculate those
> > entries basedon previous rows of the triangle but I've come up 
> empty
> > every time.Anyone have any ideas?  Alonso
> 
> I don't have an answer to your question, but it got me to thinking
> about why the triangle is important which led to a discovery.
> 
> I normally think of Lozanic's triangle position n,k as the number of
> strings of 2 colored beads that can be reversed: n total beads k of
> which are white.  Pascal's triangle is the same except you can't
> reverse the string.  So obviously to calculate it, just take Pascal,
> subtract all the strings that are the same when reversed, divide by 
> 2,
> then add back all those reversed strings.  Okay, not exactly any
> easier but it gets the concept across.  If the triangle of self
> reversing strings is easy to calculate, it may have some value.  So 
> I
> calculated that triangle:
> 
> 1 
> 1 1 
> 1 0 1 
> 1 1 1 1 
> 1 0 2 0 1 
> 1 1 2 2 1 1 
> 1 0 3 0 3 0 1 
> 1 1 3 3 3 3 1 1 
> ...
> 
> I'm not convinced it's EIS worthy, but maybe someone can change my
> mind.  Then I calculated its matrix inverse (because I always
> calculate the inverse of triangles that have all 1's on the
> diagonal.):
> 
> 1 
> -1 1 
> -1 0 1 
> 1 -1 -1 1 
> 1 0 -2 0 1 
> -1 1 2 -2 -1 1 
> -1 0 3 0 -3 0 1 
> 1 -1 -3 3 3 -3 -1 1 
> 
> I've seen many triangles whose matrix inverse is the same except
> for the signs (e.g. Pascal's), but this is the first time I've
> seen one where the sign pattern was not a simple alternation
> scheme.
> 
> 
> 
> 
> 
> 
> 
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