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mathmateBest ResponseYou've already chosen the best response.2
Does that have to do with log(n)/n and the RiemannZeta function?
 2 years ago

acrossBest ResponseYou've already chosen the best response.0
With this question, I am seeking to clash a mathematical approach with a computerized one. Although it would be interesting to see in how many different ways we can tackle this exercise.
 2 years ago

mathmateBest ResponseYou've already chosen the best response.2
So that makes 6692.7, say 6693. Is there a better estimate using RZ function, I don't really understand it.
 2 years ago

acrossBest ResponseYou've already chosen the best response.0
mathmate, your approximation is fairly close!
 2 years ago

acrossBest ResponseYou've already chosen the best response.0
For corroboration, I wrote this: http://ideone.com/qoDbe I am now trying an analytical approach.
 2 years ago

mathmateBest ResponseYou've already chosen the best response.2
Can you elaborate on your analytical approach, or is it proprietary?
 2 years ago

mathmateBest ResponseYou've already chosen the best response.2
6693 wasn't even close, you were being polite! Using the "offset logarithmic integral" \[Li(n) = \int\limits_{2}^{n}\ \frac{dt}{\log(t)}\] I get 7212.99 > 7213. Using your code, I get 7216. So now it's getting close. Also, your code included 1 as a prime, which it is not. This does not change the counts over 1 though.
 2 years ago

acrossBest ResponseYou've already chosen the best response.0
That made me want to go back and recheck the code. ^^
 2 years ago

acrossBest ResponseYou've already chosen the best response.0
Interesting. Using the logarithmic integral from 2,000,000 to 2,100,000 you get 6881, whereas the (fixed) code chunks out 6871. The error is now negative.
 2 years ago

GTBest ResponseYou've already chosen the best response.0
Code looks good except comparison to 1.
 2 years ago

mathmateBest ResponseYou've already chosen the best response.2
I get the same answers, except that I get 6872 for the actual count, both using your code and my code.
 2 years ago
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