[seqfan] Re: Zeros in A172390 and A172391

[Joerg Arndt]

Thanks for the explanation.

I worked today on this problem, but no success.
I guess there is some (nontrivial) modular magic
involved, Conway would possibly know ;-)

My (simple minded pari/gp) code for the hypergeometric
series is at  http://www.jjj.de/pari/
Do you need more than this?

The part about elliptic integerals, theta functions, etc of
  http://www.jjj.de/fxt/#fxtbook
may be of interest here.

I'll keep thinking about this little wonder.


* Paul D Hanna <pauldhanna at juno.com> [Mar 23. 2010 07:52]:
> 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)) 
> where 
> 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. 
> Incidentally, 
> 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? 
> Thanks, 
>       Paul  
> 
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