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Return all divisors of n sorted from 1..n by default.
If generator is True an unordered generator is returned.

The number of divisors of n can be quite large if there are many
prime factors (counting repeated factors). If only the number of
factors is desired use divisor_count(n).

Examples
========
(more...)

        def divisors(n, generator=False):
    r"""
    Return all divisors of n sorted from 1..n by default.
    If generator is True an unordered generator is returned.

    The number of divisors of n can be quite large if there are many
    prime factors (counting repeated factors). If only the number of
    factors is desired use divisor_count(n).

    Examples
    ========

    >>> from sympy import divisors, divisor_count
    >>> divisors(24)
    [1, 2, 3, 4, 6, 8, 12, 24]
    >>> divisor_count(24)
    8

    >>> list(divisors(120, generator=True))
    [1, 2, 4, 8, 3, 6, 12, 24, 5, 10, 20, 40, 15, 30, 60, 120]

    This is a slightly modified version of Tim Peters referenced at:
    http://stackoverflow.com/questions/1010381/python-factorization

    See Also
    ========

    primefactors, factorint, divisor_count
    """

    n = int(abs(n))
    if isprime(n):
        return [1, n]
    if n == 1:
        return [1]
    if n == 0:
        return []
    rv = _divisors(n)
    if not generator:
        return sorted(rv)
    return rv
        


src/s/y/sympy-HEAD/sympy/polys/numberfields.py   sympy(Download)
from sympy.simplify.simplify import _mexpand, _is_sum_surds
from sympy.ntheory import sieve
from sympy.ntheory.factor_ import divisors
from sympy.mpmath import pslq, mp
 
 
            # x**(2*q) = product(factors)
            factors = [cyclotomic_poly(i, x) for i in divisors(2*q)]
            mp = _choose_factor(factors, x, ex)
            return mp