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

Paul D. Hanna pauldhanna at juno.com
Thu Mar 1 09:45:38 CET 2007

```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(""))
>

```