# [seqfan] Re: Zeros in A172390 and A172391

Joerg Arndt arndt at jjj.de
Mon Mar 22 20:39:27 CET 2010

Marvelous seqs!

* Paul D Hanna <pauldhanna at juno.com> [Mar 22. 2010 19:01]:
> Seqfans,
>       Is it known/trivial that:
> (5) [x^(4n+2)] EllipticK(4x)^(-4n) = 0 for n>=1.
>

? n=3;  \\ any old n \in N
? v=Vec(hypergeom([1/2,1/2],[1],(4*x)^2,N)^(-4*n))
[1, 0, -48, 0, 816, 0, -5632, 0, 12336, 0, -768, 0,
4096, 0, 0, 0, -196560, 0, -3144960, 0, -40884480, 0, -503193600, ...]
? v[4*n+2 +1]  \\ ==0 (note offset 1)
0

Interesting.  I don't think this is known.

> This statement (5) is equivalent to (4) given in prior email.
>
> 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?

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)

>
> Thanks,
>      Paul
> ---------- Original Message ----------
> From: "Paul D Hanna" <pauldhanna at juno.com>
> To: seqfan at list.seqfan.eu
> Subject: [seqfan]  Zeros in A172390 and A172391
> Date: Sat, 20 Mar 2010 16:29:42 GMT
>
> SeqFans,
>     Sequences A172390 and A172391 record 2 surprising observations.
> Is there any reason why the following statements should be true?
> (1)  A172390(2n+1) = 0  for n>=1;
> (2)  A172391(2n+1) = 0  for n>=1.
>
> Here is a fact that may be a big clue for (1):
> (3)  Sum_{n>=0} C(2n,n)^2*x^n  =  1/AGM(1, (1-16x)^(1/2) )
> where AGM is the arithmetic-geometric mean.
>
> Statement (3) makes (1) equivalent to:
> (4)  [x^(2n+1)] AGM(1, (1-16x)^(1/2) )^(4n)  = 0 for n>=1.
>
> Can someone show that this (4) is true?
> [...]
>
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