FW: another Motzkin triangle, no shortage of'm

[FRANCISCO SALINAS]

Wouter Meeussen wrote:

>should have been:
>mat[w_Integer] := Table[If[i==j-1 || i>j,1,0], {i,w}, {j,w}]
>
>GF[w_]:= (1+x)g[w]/g[w+2]
>
>with g[w_] := Det[IdentityMatrix[w] - x*mat[w]]
>
>without beautiful 'closed form' for g(w).

I think it hasn't an ugly one:

g(w)= x * D_(w-1)(1+x, x*(1+x)) + (1-2*x) * E_(w-1)(1+x, x*(1+x))

D_n (x, a) denotes the Dickson polynomial of the first kind of degree n, 
where D_n is defined by sum_{i=0..floor(n/2)} (n/(n-i)) * binomial(n-i,i) * 
(-a)^i * x^(n-2*i) and has this ogf in y: (2-x*y)/(1-x*y+a*y^2). E_n (x, a) 
is called the Dickson polynomial of the second kind of degree n, where E_n 
is defined by sum_{i=0..floor(n/2)} binomial(n-i,i)*(-a)^i*x^(n-2*i) with 
ogf: 1/(1-x*y+a*y^2).

So, we have:

g(w+1)= sum_{i=0..floor(w/2)} (1-x*((w-2*i)/(w-i))) * binomial(w-i,i) * 
(-x)^i * (1+x)^(w-i) with the following ogf: 
(1-x*(1+x)*y)/(1-(1+x)*y+x*(1+x)*y^2) = sum_{w=0..inf} g(w) * y^w

The last generating function can also be easily deduced from the recurrence:

g(w)=g(w-1)-(g(w-2)*x^2+g(w-3)*x^3+..+g(1)*x^(w-1)+x^w), w>1

which is obtained when we evaluate the determinant of IdentityMatrix[w] - 
x*mat[w] through repeated expansion by minors using the two left-most 
columns at each stage.

PS: According to your original definition of GF(w) (*), it should be: 
GF[w_]:= (1+x)g[w-2]/g[w] (w>2), and not GF[w_]:= (1+x)g[w]/g[w+2] as you 
stated.


(*) GF(w)={Prepend[0 Range[w-1],1]}.Inverse[IdentityMatrix[w] -x  A[w] ] .
Table[{1},{w}]

_________________________________________________________________
MSN Amor: busca tu ½ naranja http://latino.msn.com/autos/