Is_prime check in Python: can this be more pythonic? I have the following function that checks if a number is prime:

def is_prime(n):
    """ Is n prime? """

    if n &lt; 2:
        return False
    if n in (2, 3):
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False

    limit   = int(math.sqrt(n))
    divisor = 5

    while divisor &lt;= limit:
        if n % divisor == 0 or n % (divisor + 2) == 0:
            return False
        divisor += 6

    return True


Most of it are checks and calculations to save up time, to make it faster than a simple loop through a range.

Can it be written in a more pythonic way? Can it be improved in some way?

Any other feedback is welcome too.
Your code looks nice, seems to follow PEP 8 and has docstrings. I don't have much to say about it but here are a few changes to make it look more pythonic.

The while loop can be rewritten using a for loop with range (or xrange).

for divisor in xrange(5, 1+int(math.sqrt(n)), 6):
    if n % divisor == 0 or n % (divisor + 2) == 0:
        return False
return True


Then, this looks like a scenario where we could use all/any. The obvious way to convert the code above is :

return not any(n % divisor == 0 or n % (divisor + 2) == 0 for divisor in xrange(5, int(math.sqrt(n))+1, 6))


but it looks better (and makes more sense) to use all with reversed conditions :

return all(n % divisor != 0 and n % (divisor + 2) != 0 for divisor in xrange(5, int(math.sqrt(n))+1, 6))


then, you can use the fact that non-zero numbers evaluates to True in a boolean context (and zero evaluates to False) :

return all(n % divisor and n % (divisor + 2) for divisor in xrange(5, int(math.sqrt(n))+1, 6))


Then, you could use the same kind of idea in the previous if then merge duplicated code :

return n % 2 and n % 3 and all(n % divisor and n % (divisor + 2) for divisor in xrange(5, int(math.sqrt(n))+1, 6))


This is pretty much all I have to say about the code. Now, I think it might be worth commenting the algorithm a bit more.

def is_prime(n):
    """ Check if n (assumed to be an int) is a prime number. """
    # Handle values up to 3
    if n &lt; 2:
        return False
    if n in (2, 3):
        return True
    # Handles bigger values : divisors for n have to be
    # * multiples of 2
    # * multiples of 3
    # * of the form 5 + 6*i or 5 + 6*i + 2 (other cases already handled)  
    return n % 2 and n % 3 and all(n % divisor and n % (divisor + 2) for divisor in xrange(5, int(math.sqrt(n))+1, 6))