[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)

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 

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).
Denominators of coefficients of powers of n^(-1) in the Romanovsky 
series expansion of the mean of the standard deviation from a normal 

Hopefully this can be the starting point of a proof.


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?
>Seqfan Mailing list - http://list.seqfan.eu/

More information about the SeqFan mailing list