gistfile1.txt

from math import floor, sqrt

def find_next_prime(n):
     """ We use is_prime passed through a while loop to check if a given value (starting at n+1) is_prime
     and exit if we've found our number. I think this too can be accomplished potentially using itertools to 
     greater effect and would love to hear criticism. I know that while True conditionals are generally frowned upon
     in python."""
    i = n+1
    while True:
        if is_prime(i):
            return i
        i+=1       
        
def is_prime(n):    
    """ The sqrt of an integer n is the highest point of a possible paired factor for n (e.g. n**1/2 * n**1/2),
    so we start at the integer 2 and work our way up to the floor of the square root. If the modulo of n by any number
    i in this range is = to 0 we know the number cannot be prime. There are more efficient ways of doing this calculation
    including vectorisation in numpy, but I am unfortunately not so mathematically gifted :)."""
    for i in xrange(2,int(floor(sqrt(n)))+1):
        if not n % i:
            return False
    return True

[(x, is_prime(x)) for x in xrange(2,12)]
[find_next_prime(6),
find_next_prime(10),
find_next_prime(11),
find_next_prime(200),
find_next_prime(2000),
find_next_prime(200000)]