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Mar 11, 2012 at 23:03 history closed Will Jagy
Felipe Voloch
Chandan Singh Dalawat
Andy Putman
Suvrit
too localized
Mar 11, 2012 at 14:32 answer added Kevin O'Bryant timeline score: 1
Mar 11, 2012 at 14:29 answer added Brendan McKay timeline score: 2
Mar 11, 2012 at 13:10 answer added iensen timeline score: 1
Mar 10, 2012 at 23:33 answer added David Harris timeline score: 0
Mar 10, 2012 at 22:34 comment added iensen >It will be bounded from below by logn/loglogn Yes,i have got this by replacing LOGN with LOG N_F,where N_F is N which was an argument of the first call. Do you have any ideas for the upper bound?
Mar 10, 2012 at 22:30 comment added David Harris It will be bounded from below by $\log n/\log \log n$.
Mar 10, 2012 at 22:29 comment added iensen This equation (like most questions on math.stackexchange.com,but i didn't find exactly this one) originate from Cormen's book about algorithms. It is not stated clearly anywhere but it can be composed on the base of one of the book's chapters.
Mar 10, 2012 at 22:26 comment added iensen I believe the approximation log(n/logn) is not correct. T(1000 000)= 7,T(1000 000 000 )=9...,it grows much slower. I wonder if there is a method for solving such equations
Mar 10, 2012 at 22:24 comment added Yemon Choi This looks vaguely (but probably not exactly) like questions asked on math.stackexchange.com - where does this recurrence relation originate?
Mar 10, 2012 at 22:18 comment added Gerhard Paseman T(n) will be bounded above by log(n) with the log base being 2. It is likely to be approximated for large n by something like log(n/log(n)), but I have not verified this. Gerhard "Ask Me About System Design" Paseman, 2012.03.10
Mar 10, 2012 at 22:15 history edited iensen CC BY-SA 3.0
added 55 characters in body; added 1 characters in body
Mar 10, 2012 at 22:14 comment added iensen Oh sorry,the log is "binary" log with base 2.
Mar 10, 2012 at 22:11 comment added Woett Well, there is no such function. Check N = e, for example.
Mar 10, 2012 at 22:00 history asked iensen CC BY-SA 3.0