asking for help in connection with a new sequence
Paul D. Hanna
pauldhanna at juno.com
Tue Jul 18 09:44:41 CEST 2006
Emeric (and Max, Seqfans):
Perhaps you have noticed that the central terms of your triangle T
are:
T(2*n,n) = (2n+1)*A006605(n) where
A006605 Number of modes of connections of 2n points.
1,1,3,11,46,207,979,4797,24138,123998,647615,3428493,...
Formulas for g.f. of A006605, GF006605(x), are:
(1) GF006605(x) = (1/x)*series_reversion(x/GF001006(x))
where GF001006(x) is g.f. of A001006 (Motzkin numbers);
(2) GF006605(x)^2 = (1/x)*series_reversion( x/(1+x+x^2)^2 ).
Further, a formula for term T(2n+m-1,n), m>=0, (by inspection) is:
T(2n+3m),n+m) = (2n+3m+1)/(3m+1) * [x^n] GF006605(x)^(3m+1)
Examples:
m=1: the terms T(2n+3,n+1) begin:
[1,6,36,210,1221,7098,41328,241128,1409895,8260934,48497064,...]
compare to [x^n] GF006605(x)^4 :
[1,4,18,84,407,2028,10332,53584,281979,1501988,8082844,...]
to see:
T(2n+3,n+1) = (2n+4)/4 * [x^n] GF006605(x)^4
m=2: the terms T(2*n+6,n+2) begin:
[1,9,66,442,2835,17748,109497,669294,4065963,24597925,148381740,...]
compare to [x^n] GF006605(x)^7 :
[1,7,42,238,1323,7308,40341,223098,1237467,6887419,38469340,...]
to see:
T(2n+6,n+2) = (2n+7)/7 * [x^n] GF006605(x)^7
I am sure that there are other formulas involving powers of GF006605.
> > G(t,z)=1/[1-3tz-3t(1-t)z^2-t(1-t)^2*z^3],
> > which, by expansion, yields the sequence.
>
> Could you please elaborate a little bit on how the expansion yields
> the sequence?
A simpler form of G() is:
G(t,z) = (1-t)/(1 - t*(1 + (1-t)*z )^3 )
This generates the triangle as the coefficients:
T(n,k) = [z^n*t^k] G(t,z)
Using PARI:
T(n,k)=polcoeff(polcoeff((1-t)/(1-t*(1+(1-1*t)*z+t*O(t^n) )^3),n,z),k,t)
> >
> > (*) T(n,k)=coefficient of x^(3n-3k+2) in (1+x+x^2)^(n+1),
> >
> > i.e. the trinomial coefficient [n+1,3; 3n-3k+2]=[n+1,3; 3k-n].
This is a neat result. I do not have success showing this yet.
> T(n,k) = coefficient of x^(n-k) in [1,1,1] * M(x)^k * [1,0,0]^T
> where the matrix M(x) is defined as
> [ x^2+2*x+1 x^2+2*x x^2 + x ]
> [ x+1 x+1 x ]
> [ 1 1 1 ]
This is an interesting form, Max. Still musing over it ...
Well, that's my 2-cents worth for now.
Paul
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