[seqfan] Re: Asymptotic formula

Olivier Gerard olivier.gerard at gmail.com
Sat Apr 30 13:10:48 CEST 2011

Here is the answer from Benoit in legible form.


Yes here a(2n)=0. Here a pari code:


For r=2 a(2n)=0 and a(2n-1) for n>=1 begins:


So there is a structure allowing us to say a(2^n+1)=-2^(n-1) for n>=2 and
a(n)<<n is true in that case.

If 1<r<2 things are less simple. My guess is a(n)<<n^(r-1)L(n) for a slowly
varying function L.

Note I consider a more general case for a given function f satisfying
f(n)<<1/n and r>1:


so a(2n)=0 is not always the case.


On Sat, Apr 30, 2011 at 10:19, <franktaw at netscape.net> wrote:

> I think there's something wrong here. As written, a(n) = 0 for any even n >
> 2. Is this really what is intended?
> Franklin T. Adams-Watters

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