The answer would be:
(log1+1) + (log2+1) + (log3+1) + (log4+1) + ... + (log(n-1)+1)
which is equal to:
(log1+log2+log3+...+log(n-1)) + (n-1)
==> *log((n-1)!) + (n-1)*
where, log everywhere is assumed to be in base 2
*This according to me will be the final answer!*
*
*
*Cheers*
*Nikhil Jindal
*
On Fri, Sep 24, 2010 at 8:10 PM, SVIX <[email protected]>wrote:

> what's the datatype of j? mathematically speaking, the while loop is
> infinite for every j>0...
>
> On Sep 23, 6:19 am, Krunal Modi <[email protected]> wrote:
> > for(k=1 ; k<n ; k++){
> >   j=k;
> >   while(j>0){
> >      j=j/2;
> >   }
> >
> > }
> >
> > How many times while loop gets executed (for any n) ?
> >
> > I don't want answer in terms of series (i.e, don't want any sigma, I
> > have that)
>
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