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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

**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