Esempio n. 1
0
def find_consecutive(howmany):
    latest = [] # factors of the previous consecutive numbers
    for n in itertools.count(2):
        facn = factors.factor_count(n).items()
        if len(facn) != howmany:
            latest = []
            continue
        latest.append(set(facn))
        latest = latest[-howmany:]

        ## are they distinct?
        all_factors = functools.reduce(set.union, latest)
        if len(all_factors) == howmany**2:
            return n - howmany + 1
Esempio n. 2
0
def count_divisors(n):
    fdict = factors.factor_count(n)
    if not fdict:
        return 1
    return functools.reduce(operator.mul, (c+1 for c in fdict.values()))