[seqfan] Re: Zeros in A172390 and A172391

Paul D Hanna pauldhanna at juno.com
Mon Mar 22 22:24:35 CET 2010

Joerg (and SeqFans), 
    Thanks for the comments and example using the hypergeometric expression. 
My response to your question is inserted below. 
>>       Is it known/trivial that: 
>> (5) [x^(4n+2)] EllipticK(4x)^(-4n) = 0 for n>=1.  
> > If (5) is true, it would imply that   A172390(2n+1) = 0  for n>=1. 
> I do not see this equivalence, is it easy to point out?
In general, if 
A(x) = B(x/A(x)) and 
B(x) = A(x*B(x)) 
A(x) = Sum_{n>=0} a(n)*x^n and 
B(x) = Sum_{n>=0} b(n)*x^n 
then by Lagrange Inversion formula: 
b(n) = [x^n] A(x)^(n+1)/(n+1), 
a(n) = [x^n] B(x)^(1-n)/(1-n) for n>1. 
A(x) = x/Series_Reversion(x*B(x)), and 
B(x) = (1/x)*Series_Reversion(x/A(x)). 
As applied in this case, A(x) and G(x) satisfy: 
A(x) = G(x/A(x))^2 and 
G(x)^2 = A(x*G(x)^2) 
where A(x) is the g.f. of A172390, and 
G(x) = Sum_{n>0} C(2n,n)^2*x^n or, equivalently, 
G(x) = 1/AGM(1,sqrt(1-16x)) and also 
G(x) = (2/Pi)*EllipticK(4*sqrt(x)). 
So it follows that 
if a(2n+1) = 0 for n>=1 
then [x^(2n+1)] G(x)^(-4n) = 0 for n>=1, 
also [x^(2n+1)] AGM(1,sqrt(1-16x))^(4n) = 0 for n>=1, 
thus [x^(4n+2)] EllipticK(4x)^(-4n) = 0 for n>=1. 
> IMHO a comment in the seq should mention that
> G(x)==hypergeom([1/2,1/2],[1],16*x,N)
> == ellipticK(4*x) (mod that factor Pi/2)
Joerg, would you mind sending in the hypergeometric formula and code? 

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