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Topic: Average length of random 1-expansions
Replies: 9   Last Post: Nov 20, 2000 4:55 AM

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Hugo van der Sanden

Posts: 234
Registered: 12/8/04
Re: Average length of random 1-expansions
Posted: Nov 20, 2000 4:55 AM
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r3769 wrote:
> >Note that f(1/(2^k n)) = n/2 + f(1/n).
>
> I think you wanted to write f(1/2^k n))=k/2 + f(1/n), no?


Yes, my bad.

> It is kind of interesting to note that the "coin-flip 1-expansion" defined
> above can be generalized:
>
> Given any integer b>0, let amin be the smallest integer s.t.
> floor(x[i]*amin)=b and amax the largest such integer.
>
> Now let f(b,x)=the average length of all the coin-flip b-expansions of x.
>
> For example f(1,1/11)=16, f(2,1/11)=6, f(4,1/11)=4, f(11,1/11)=51, and
> f(113,1/11)=18.
>
> I suppose it's not too difficult to show the sequence <f(b,1/11)> is
> bounded, so what is it's sup?


I haven't had a chance to look at this yet, but I'll try to come back to
it.

Hugo







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