FW: another Motzkin triangle, no shortage of'm

[wouter meeussen]

hi Ralf & seqfanners,

the procedure
GF(w)={Prepend[0*Range[w-1],1]}.Inverse[IdentityMatrix[w]-x*A[w]].Table[{1},{w}]
gives the GF for each size w.
The trick is to compare them, and come up with one GF(w) that 'binds them all':

A011973 is a give-away (Author=njas):
define f[n_] := Sum[(-1)^k Binomial[n-k,k] x^k, {k,0,Floor[n/2]}]

GF(w) equals f[w]/f[w+2]

Even better would be to be able to prove that lim(w->inf) GF(w) equals the
GF for A001006 (Motzkin numbers)=(1 -x-(1-2*x-3*x^2)^(1/2))/(2*x^2).
but Mma  *don'wanna*

A combinatorial equivalence between the matrix

> 0 1 0 0 0 ...
> 1 0 1 0 0
> 1 1 0 1 0
> 1 1 1 0 1
> 1 1 1 1 0
> ...

and the Motzkin path-counting would also be nice
NE-E-SE

          1
       /    \
     1 -- 2 -- 5
  /    /\   /\   \
1 -- 1 -- 2 -- 4 -- 9



Wouter.

(ps, typo in my substitution game:
replace "2->1" by "2->13"     )





----- Original Message -----
From: "Ralf Stephan" <ralf at ark.in-berlin.de>
To: "Meeussen Wouter (bkarnd)" <wouter.meeussen at vandemoortele.com>
Cc: "Seqfan (E-mail)" <seqfan at ext.jussieu.fr>
Sent: Saturday, April 10, 2004 9:01 AM
Subject: Re: FW: another Motzkin triangle, no shortage of'm


Wouter Meeussen:
> > {1, 1, 2, 4, 9, 21, 51, 126, 315, 792, 1998, 5049, 12771, 32319, 81810, 207117, 524394}

This seems to be 1/3*(A062882(n)+A062882(n+1)) or 1/9 the partial
sums of A062882. Working out a(n+2)-a(n+1)-a(n) gives A090400/3 so
the g.f. should have as denominator (1-x-x^2)*(1-3x+3x^3).

> > {1, 1, 2, 4, 9, 21, 51, 127, 322, 826, 2135, 5545, 14445, 37701, 98514, 257608, 673933}

This one appears to be A033192(n)+A033192(n+1) = C(F(n)+1,2)+C(F(n+1)+1,2).
Or 1/2*(A059727(n)+A059727(n+1)), so the g.f. is
[1-2x-x^2+x^3]/[(1-3x+x^2)(1-x-x^2)] but you have to mul. with x and add 1.

As these are all rational gfs, superseeker should be able to give you
better answers.

ralf