6 examples of 'how to find factors of a number in python' in Python

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def factors(number):
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    """
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    Find all of the factors of a number and return it as a list
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    number:
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    The number to find the factors for
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    """
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    factors = []
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    for i in range(1, number + 1):
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        if number % i == 0:
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            factors.append(i)
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    return factors

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def get_factors(n):
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    """[summary]
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    Arguments:
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        n {[int]} -- [to analysed number]
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    Returns:
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        [list of lists] -- [all factors of the number n]
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    """
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    def factor(n, i, combi, res):
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        """[summary]
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        helper function
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        Arguments:
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            n {[int]} -- [number]
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            i {[int]} -- [to tested divisor]
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            combi {[list]} -- [catch divisors]
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            res {[list]} -- [all factors of the number n]
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        Returns:
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            [list] -- [res]
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        """
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        while i * i <= n:
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            if n % i == 0:
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                res += combi + [i, int(n/i)],
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                factor(n/i, i, combi+[i], res)
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            i += 1
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        return res
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    return factor(n, 2, [], [])

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def inverse_factorial(number, round_up=True):
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    '''
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    Get the integer that the factorial of would be `number`.
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    If `number` isn't a factorial of an integer, the result will be rounded. By
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    default it'll be rounded up, but you can specify `round_up=False` to have
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    it be rounded down.
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    Examples:
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        >>> inverse_factorial(100)
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        5
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        >>> inverse_factorial(100, round_up=False)
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        4
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    '''
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    assert number >= 0
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    if number == 0:
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        return 0
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    elif number < 1:
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        return int(round_up) # Heh.
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    elif number == 1:
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        return 1
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    else:
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        current_number = 1
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        for multiplier in itertools.count(2):
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            current_number *= multiplier
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            if current_number == number:
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                return multiplier
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            elif current_number > number:
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                return multiplier if round_up else (multiplier - 1)

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def closest_square_factors(integer, larger_first=True):
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	factors = []
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	root = ceil(sqrt(integer))
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	for i in range(root, integer+1):
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		if integer % i == 0:
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			return (i, int(integer / i)) if  larger_first else (int(integer / i), i)
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	return (1, integer)

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def factor_modulus(n, d, e):
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    """
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    Efficiently recover non-trivial factors of n
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    See: Handbook of Applied Cryptography
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    8.2.2 Security of RSA -> (i) Relation to factoring (p.287)
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    http://www.cacr.math.uwaterloo.ca/hac/
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    """
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    t = (e * d - 1)
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    s = 0
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    while True:
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        quotient, remainder = divmod(t, 2)
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        if remainder != 0:
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            break
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        s += 1
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        t = quotient
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    found = False
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    while not found:
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        i = 1
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        a = random.randint(1,n-1)
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        while i <= s and not found:
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            c1 = pow(a, pow(2, i-1, n) * t, n)
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            c2 = pow(a, pow(2, i, n) * t, n)
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            found = c1 != 1 and c1 != (-1 % n) and c2 == 1
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            i += 1
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    p = fractions.gcd(c1-1, n)
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    q = n // p
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    return p, q

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def prime_number_factorisation(n):
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    if n < 2:
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        return [n]
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    i = 2
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    factors = []
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    while i * i <= n:
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        if n % i:
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            i += 1
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        else:
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            n //= i
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            factors.append(i)
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    if n > 1:
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        factors.append(n)
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    return factors