20171216, 15:54  #23 
May 2017
ITALY
522_{10} Posts 
I thought
6^X where X is N^(1/X)=1 
20171216, 16:13  #24 
May 2017
ITALY
2·3^{2}·29 Posts 
I am not practical in mathematics.
It should be about 6 ^ [log_2 (log_10 (N))] 
20171216, 16:59  #25  
Feb 2017
Nowhere
2·3^{3}·5·23 Posts 
Let's see here: Subject says O([log_9(N)]^3)
Post #5 says Quote:
Quote:
Post #24 says Quote:
O(log(N)^(log(6)/log(2))). Since log(6)/log(2) < 3, that's smaller than your original claim. I can hardly wait to see what's next Last fiddled with by Dr Sardonicus on 20171216 at 17:04 

20171216, 17:04  #26  
May 2017
ITALY
2·3^{2}·29 Posts 
Quote:


20171217, 18:44  #27 
Aug 2006
5,987 Posts 

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