# Elliptic Convolution Identity

Paul D. Hanna pauldhanna at juno.com
Thu Mar 11 05:49:28 CET 2004

```Seqfans,
I would like to share with you a surprising identity
involving a special product which I call the
"elliptic convolution" of a power series.

The elliptic convolution is a type of self-convolution
of a power series, but where the power series is put to
varying QUADRATIC powers, while still mapping integer series
to integer series.  I have not yet been able to define the
elliptic convolution between two different power series
such that integer series are mapped to integer series.

After stating the conjecture I give examples and PARI code.
This result has not been proved, but empirical results
strongly suggest the veracity of the conjecture.

Would someone venture to prove the conjecture?

Thanks,
Paul
==========================================================

CONJECTURE.
----------------------------------------------------------
Define the elliptic convolution of a power series F(x) at power n
with parameter M to be the power series G(x,n,M) given by:

(1) [x^k]G(x,n,M) =

sum_{j=0..k} [x^j]F(x)^(n/2+sqrt(k^2+M*(k/2-j)^2)) *
[x^(k-j)]F(x)^(n/2-sqrt(k^2+M*(k/2-j)^2))

where [x^k]f(x) denotes the coefficient of x^k in f(x).

Then, amazingly,

(2) G(x,n,M) = F(x)^n /(1 - M*(x*F'(x)/F(x))^2 )

for all M and n, where F'(x) is the derivative of F(x) wrt x.

Thus, the elliptic convolution maps integer series to integer series
for all integer parameter M.
==========================================================

SPECIAL CASE: M=0.
At M=0, we have the identity:

(3) [x^k]F(x)^n = sum_{j=0..k} [x^j]F(x)^(n/2+k) * [x^(k-j)]F(x)^(n/2-k)

which holds for all n and integer k>=0.

The simplest example of (3) is the binomial indentity:

(4) C(n,k) = sum_{j=0..k} C(n/2+k,j)*C(n/2-k,k-j)

which is true.
==========================================================

EXAMPLE:
The elliptic convolution of 1/(1-x-x^2) with parameter M
at power n=1 equals:
(1-x-x^2)/(1-2*x-(M+1)*x^2-(4*M-2)*x^3-(4*M-1)*x^4)

= (1-x-x^2)/[(1-x-x^2)^2 - M*x^2*(1-2*x)^2]
= (1/F(x))/[(1/F(x))^2 - M*x^2*(d/dx 1/F(x) )^2]
= F(x)/(1 - M*( x*F'(x)/F(x) )^2 ).

M: elliptic convolution of 1/(1-x-x^2) with parameter M, power 1
-- -------------------------------------------------------
0: (1-x-x^2)/(1-2*x-1*x^2+2*x^3+1*x^4) = 1/(1-x-x^2)
1: (1-x-x^2)/(1-2*x-2*x^2-2*x^3-3*x^4)
2: (1-x-x^2)/(1-2*x-3*x^2-6*x^3-7*x^4)
3: (1-x-x^2)/(1-2*x-4*x^2-10*x^3-11*x^4)
4: (1-x-x^2)/(1-2*x-5*x^2-14*x^3-15*x^4)

==========================================================
PARI CODE in examples below.
==========================================================
EXAMPLE 1.
Demonstrate that the conjecture holds for F(x)=1/(1-x-x^2).

/* Using definition of elliptic convolution: */
{F(x)=1/(1-x-x^2)}
{P(n,k)=polcoeff((F(x)+O(x^(k+1)))^n,k)}
{T(n,k,M)=sum(j=0,k,P(n/2+sqrt(k^2+M*(k/2-j)^2),j)*P(n/2-sqrt(k^2+M*(k/2-
j)^2),k-j))}
{for(k=0,20,print1(round(T(1,k,1)),","))}

RESULTS:
1,1,3,10,31,91,273,820,2461,7381,22143,66430,199291,597871,1793613,
5380840,16142521,48427561,145282683,435848050,1307544151,

/* Compare to results using the formula:       */
/* G(x,n,M) = F(x)^n/(1 - M*(x*F'(x)/F(x))^2 ) */
{F(x)=1/(1-x-x^2)}
{G(x,n,M)=F(x)^n/(1 - M*( x*deriv(F(x))/F(x) )^2 )}
{T(n,k,M)=polcoeff(G(x,n,M)+O(x^(k+1)),k)}
{for(k=0,20,print1(T(1,k,1),","))}

RESULTS MATCH:
1,1,3,10,31,91,273,820,2461,7381,22143,66430,199291,597871,1793613,
5380840,16142521,48427561,145282683,435848050,1307544151,

==========================================================
EXAMPLE 2.
Demonstrate that the conjecture holds for F(x)=exp(x).

/* Using definition of elliptic convolution: */
{F(x)=exp(x)}
{P(n,k)=polcoeff((F(x)+O(x^(k+1)))^n,k)}
{T(n,k,M)=sum(j=0,k,P(n/2+sqrt(k^2+M*(k/2-j)^2),j)*P(n/2-sqrt(k^2+M*(k/2-
j)^2),k-j))}
{for(k=0,5,print(T(1,k,1)*1.0,","))}

RESULTS:
1.000000000000000000000000000,
1.000000000000000000000000000,
1.500000000000000000000000000,
1.166666666666666666666666667,
1.541666666666666666666666664,
1.175000000000000000000000010,

/* Compare to results using the formula:       */
/* G(x,n,M) = F(x)^n/(1 - M*(x*F'(x)/F(x))^2 ) */
{F(x)=exp(x)}
{G(x,n,M)=F(x)^n/(1 - M*( x*deriv(F(x))/F(x) )^2 )}
{T(n,k,M)=polcoeff(G(x,n,M)+O(x^(k+1)),k)}
{for(k=0,5,print(T(1,k,1)*1.0,","))}

RESULTS MATCH:
1.000000000000000000000000000,
1.000000000000000000000000000,
1.500000000000000000000000000,
1.166666666666666666666666667,
1.541666666666666666666666667,
1.175000000000000000000000000,
==========================================================
END.

```