# [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:

a(n)=if(n<3,1,r*(a(ceil(n/2))-a(floor(n/2))))

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

1,0,-2,0,-4,4,0,0,-8,8,8,-8,0,0,0,0,-16,16,16,-16,16,-16,-16,16,0,0,0,0,0,0,0,0,-32,32,32,-32,32,-32,-32,32,32,-32,-32,32,-32,32,32,-32,

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:

a(n)=if(n<3,1,f(n)+r*(a(ceil(n/2))-a(floor(n/2))))

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

Benoit

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?
>