[seqfan] Re: Asymptotic formula
Bernard Cloitre
benoit7848c at gmail.com
Sun May 1 20:42:21 CEST 2011
Thanks to Jean-Paul hint and more thought, it seems we have in any case
a(2^n+1)=-r^n (the rule of sign for a(n) is somewhat related to Thue Morse
sequence). So the asymptotic would be a(N)<<N^(log(r)/log(2)). I guess it
remains true for the more general:
a(n)=f(n)+r*(a(ceil(n/2))-a(floor(n/2))) with f(n)-->0.
Benoit
2011/4/30 <franktaw at netscape.net>
> 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
>
>
> -----Original Message-----
> From: Olivier Gerard <olivier.gerard at gmail.com>
>
> Benoit's message has suffered a transmission problem,
> Here is a more legible version:
> ==============================================
>
> Dear seqfans,
>
>
> I came across this sequence defined by the recursion:
>
> a(1)=1, a(2)=1, n>=3, a(n)=r*(a(ceil(n/2))-a(floor(n/2))) where r>1.
>
> I'm interested with the asymptotic behaviour of a(n). For instance I guess
> u(n)<<n if r=2 and u(n)<<n^(1/2)L(n) if r=3/2 where L is a slowly varying
> function. Does anyone know a reference for the general case or confirm this
> fact for r=2 and r=3/2?
>
> Thanks
> BC
>
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