[seqfan] BIPRODUCT of Two Sequences
Paul D. Hanna
pauldhanna at juno.com
Sun Nov 7 21:41:02 CET 2004
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) ) = ?
----------------------------------------------------------------------
(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.
------------------------------------------------------------------------
Would this interest anyone to come up with a more general formula?
Thanks,
Paul
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