[seqfan] BIPRODUCT of Two Sequences

[Paul D. Hanna]

     Given two sequences with known g.f., what is the g.f. of the
sequence 
resulting from the term-by-term product of the sequences?
 
Suppose we call this the "BIPRODUCT" of the two sequences 
(is there a better term for this?).
 
Then, so far I have derived: 
(1)  
BIPRODUCT( 1/(1-ax-bx^2), 1/(1-cx-dx^2) ) =
(1 - bd*x^2)/(1 - ac*x - (da^2 + bc^2 + 2bd)*x^2 - abcd*x^3 +
b^2*d^2*x^4)
 
EXAMPLE:
the biproduct of Fibonacci and Pell is A001582:
{1*1, 1*2, 2*5, 3*12, 5*29, 8*70, 13*169, ...} =
{1,2,10,36,145,560,2197,8568,33490,130790,510949,1995840,7796413,30454814
,...}
where the G.f. is given by: 
 
BIPRODUCT( 1/(1-x-x^2), 1/(1-2*x-x^2) ) = 
    (1-x^2)/(1-2*x -7*x^2 -2*x^3 +x^4)

Q:  Can someone extend this to higher orders --
i.e., what is biproduct( 1/(1-ax-bx^2-cx^3), 1/(1-dx-ex^2-fx^3) ) = ?
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(2)  Catalan biproducts with 2nd order recurrence.
Given the recurrence: 
     S(n) = b*S(n-1) + c*S(n-2), 
then the g.f. for the term-by-term product: 
     a(n) = S(n)*Catalan(n) 
is given by: 
     sum_{n>=0} S(n)*Catalan(n)*x^n 
 
=  sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.
 
= (1/x)* series reversion of  x*( sqrt(1-4*c*x^2) - b*x )

EXAMPLE:
Biproduct of Catalan and Fibonacci (A098614):
{1,1,4,15,70,336,1716,9009,48620,267410,1494844,8465184,...}
has g.f.:
BIPRODUCT(  G000108(x), G000045(x) ) =
    sqrt( (1-2*x - sqrt(1-4*x-16*x^2))/10 )/x.

So the BIPRODUCT of two sequences retains some of the properties of both.
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Would this interest anyone to come up with a more general formula?
 
Thanks,
       Paul
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