some help is needed Re: SEQ+# A127157 FROM Emeric Deutsch (fwd)

[Paul D. Hanna]

Emeric (and SeqFans), 
    We can finish this with just simple algebraic manipulation. 
Show that (7) and (8) are equivalent: 
(7) t*z/(1-z)^2 = A*( 1 - z/(1-z)*A )^2 
where F = 1 + x*F^3 is the g.f. of A001764; 
(8) A(z,t) = t*z/(1-z)^2 * F( t*z^2/(1-z)^3 )^2 .
 
Substitute the A of (8) into the first A of the right-side of (7): 
t*z/(1-z)^2 = { t*z/(1-z)^2 * F( t*z^2/(1-z)^3 )^2 } * ( 1 - z/(1-z)*A
)^2
 
Eliminate common factors to obtain: 
F( t*z^2/(1-z)^3 ) = 1/( 1 - z/(1-z)*A )
 
Substitute A by (8) again:
F( t*z^2/(1-z)^3 ) = 1/( 1 - t*z^2/(1-z)^3 * F( t*z^2/(1-z)^3 )^2 )
 
or
F( t*z^2/(1-z)^3 )*( 1 - t*z^2/(1-z)^3 * F( t*z^2/(1-z)^3 )^2 ) = 1
 
which is equivalent to: 
F( t*z^2/(1-z)^3 ) = 1  +  t*z^2/(1-z)^3 * F( t*z^2/(1-z)^3 )^3 
 
Therefore 
F(x) = 1 + x*F(x)^3  where  x=t*z^2/(1-z)^3 
 
and F(x) is the g.f. of A001764 
(here we can define F easily by Lagrange Inversion). 
 
Therefore, your g.f. (modified) in (3) and your formula in (2) agree. 
 
   
Perhaps the simplest expressions for the g.f. are: 
 
G.f. A(z,t) satisfies: t*z - A*(1-z - z*A)^2 = 0.
 
and 
 
G.f.: A(z,t) = t*z/(1-z)^2 * F( t*z^2/(1-z)^3 )^2 
where F = 1 + x*F^3 is the g.f. of A001764 (ternary trees).
 
I will submit these as a COMMENT soon. 
-- Paul 
============================================
On Thu, 1 Mar 2007 01:40:27 -0500 "Paul D. Hanna" <pauldhanna at juno.com>
writes:
> Emeric (and Seqfans), 
>     Could it be that your g.f.: 
> (1) z^2*G^3-z(z+2)G^2+(1+2z)G-t^2*z-1=0 
> has a slight typo (should be t^1 instead of t^2)? 
>  
> I believe that the g.f. should be: 
> (3) z^2*G^3 - z(z+2)G^2 + (1+2z)G - t*z - 1 = 0
>  
> which can be rewritten as:
> 1+t*z - (1+2*z)*G + z*(z+2)*G^2 - z^2*G^3 = 0 
>  
> But wait, this G has a constant term: G(0,0) = 1 
> that your triangle does not include.
>  
> To remove the constant, substitute G = A+1  to get: 
>  
> (4) 1+t*z - (1+2*z)*(A+1) + z*(z+2)*(A+1)^2 - z^2*(A+1)^3 = 0 
>  
> Here, A = A(z,t) is the correct g.f. of your triangle. 
>  
> Putting (4) into powers of A: 
> (5) t*z - (1-z)^2*A + 2*z*(1-z)*A^2 - z^2*A^3 = 0
>  
> which is equivalent to: 
> (6) t*z - A*(1-z - z*A)^2 = 0
>  
> Restating: 
> (7) t*z/(1-z)^2 = A*( 1 - z/(1-z)*A )^2 
>   
>  
> Now, from your nice formula for [t^k z^n] A(z,t): 
> (2) [t^k z^n] = 2*binomial(3k-1,2k)*binomial(n-1+k,3k-2)/(3k-1) 
> we can easily derive A to be: 
>  
> (8) A(z,t) = t*z/(1-z)^2 * F( t*z^2/(1-z)^3 )^2
>  
> where F = 1 + x*F^3 is the g.f. of A001764 = [1,1,3,12,55,273,...]. 
> 
>   
>  
> So the problem now is to obtain (8) from (7). 
>   
> I have verified that (7) and (8) do in fact agree 
> (I give my PARI code below) 
>  
> but can we apply Lagrange Inversion to (7) to obtain (8)? 
>   
> --Paul
>  
> /* Using Equation (8) where F = 1 + x*F^3: */
> F=1+x;for(i=0,12,F=1+x*F^3+O(x^13));Vec(F)
> A=t*z/(1-z +O(z^13))^2*subst(F^2,x,t*z^2/(1-z +O(z^13))^3);
> T(n,k)=polcoeff(polcoeff(A,n,z),k,t)
> for(n=0,12,for(k=0,(n+1)\2,print1(T(n,k),","));print("")) 
>  
> /* Using Equation (7): */
> A=z+t;for(i=0,12,A=A + t*z - A*(1-z - z*A)^2 +O(z^13) +O(t^13))
> T(n,k)=polcoeff(polcoeff(A,n,z),k,t)
> for(n=0,12,for(k=0,(n+1)\2,print1(T(n,k),","));print("")) 
>