[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?
>
> Franklin T. Adams-Watters
>
>
>
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