Re: Lozanić's triangle (A034851) : Is there easier definition?

[Christian G.Bower]

> 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.

Christian