# [seqfan] Re: A012022, A065140 and sqrt(2)

Gerald McGarvey Gerald.McGarvey at comcast.net
Thu Dec 24 02:19:40 CET 2009

```The following appears to be true:

abs(A012022(n)) = abs(A117972(n)) * A126963(n)

A065140(n) = A117972(n) * A088802(n)

<http://www.research.att.com/%7Enjas/sequences/A117972>A117972
Numerator of Zeta'[ -2n].

If the above statements are true then

abs(A012022(k))/A065140(k) = abs(A126963(n)/A088802(n)).

In the entry for A126963 the following is stated:

f(n) -> sqrt(2) as n -> oo.

Denominators are in
<http://www.research.att.com/%7Enjas/sequences/A088802>A088802.

<http://www.research.att.com/%7Enjas/sequences/A126963>A126963
Numerators of sequence defined by f(0)=1, f(1)=5/4; f(n) =
(6n-1)*f(n-1)/(4n) - (2n-1)*f(n-2)/(4n).
<http://www.research.att.com/%7Enjas/sequences/A088802>A088802
Denominators of coefficients of powers of n^(-1) in the Romanovsky
series expansion of the mean of the standard deviation from a normal
population.

Hopefully this can be the starting point of a proof.

Regards,
Gerald

At 09:22 AM 12/23/2009, Jaume Oliver i Lafont wrote:
>Hello all,
>
>It appears that lim{k->oo} abs(A012022(k))/A065140(k) = sqrt(2)
>
>Is this easy to prove?
>
>http://www.research.att.com/~njas/sequences/?q=id:A065140|id:A012022
>
>Jaume
>
>
>_______________________________________________
>
>Seqfan Mailing list - http://list.seqfan.eu/

```