[seqfan] Re: Asymptotic formula

[Aai]

Ignoring the 0's of the sequence and the first 1, the sign pattern of the 
numbers is as follows:
(using Haskell)

*Main> take 4 $ iterate (ap (++) (map negate)) [-1]
[[-1],[-1,1],[-1,1,1,-1],[-1,1,1,-1,1,-1,-1,1]]

This is: every next sub sequence is itself extended with its negation:
   [-1,1] -->   [-1, 1,1,-1]


Using this, the non-zero elements are simply obtained iteratively.
e.g. for r=2:

an r = (1:). concat
   $ zipWith (map. (*)) (map (r^)[1..]) (iterate (ap (++) (map negate)) [-1])

*Main> take 16 $ an (2)
[1,-2,-4,4,-8,8,8,-8,-16,16,16,-16,16,-16,-16,16]




> 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
>
-- 
Met vriendelijke groet,
=@@i