Yes, it's beautiful. Thank you very much, you're great!

On Jul 21, 7:06 pm, Ashesh <[EMAIL PROTECTED]> wrote:
> On Jul 21, 8:38 pm, zpedja <[EMAIL PROTECTED]> wrote:

> > I'm now dealing with your solution.
> > I've managed to prove that the argument in k-th recursion acts like
> > ceil(n/log(n)^k) , but I don't know how you obtain k=Θ( log(n)/
> > log(log(n)) ).
>
> Asymptotically speaking, the recurrence reaches an end when n/log(n)^k
> = ϴ(1) = c, where c is a constant. So, c.log(n)^k = n
> or, k.log(log(n)) + logc = log(n)
> getting rid of the constant, the asymptotic expression for k would
> then be: k=ϴ( log(n) / log(log(n)) ).
>
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