Question

Could someone take a look at my code for ps1b? Thanks.

Answer 1

Here is the solution I came up with. It works, but for numbers above 20,000 or so, it can take a little while to compute the answer. I was just wondering if there's a better/faster way to do it?

from math import *

n = input('Pick a number: ')
sum = 0
for number in range(2,n):
divisor = 2
while (divisor < (number/2)) and (number % divisor !=0):
divisor = divisor + 1
if number % divisor != 0:
prime_number = number
sum = sum + log(prime_number)

print sum + log(2)
print n
print (sum/n)

Answer 2

Here is all the code, highlighted with comments, if you need it:
http://dpaste.com/hold/450061/

Answer 3

Last line, use this :

print ((sum+log(2))/n)

If not, you are not counting the log of 2 in your ratio and your theorem fails