nth root calculation proof of concept

sqrtn.py

def sqrtn(n, p, z):
    print("sqrtn {} with {} precision is:".format(n, p))
    if len(n) %z != 0:
        n = "0"*(z-len(n)%z) +  n 
    splited_n = [n[i:i+z] for i in range(0,len(n),z)]
    A,B,Remainer,Result = 0,0,0,""
    def get_binomial_coefficient(n,k):
        if k == 0:
            return 1
        if k == n:
            return 1
        return get_binomial_coefficient(n-1, k-1) + get_binomial_coefficient(n-1, k)
    def binomial_coefficients(n):
        return [get_binomial_coefficient(n, k) for k in range(0,n+1)]
    def findb(N, a, b, remainer, result):
        for j in range(0, 11):
            c = remainer*pow(10, z) + int(N)
            b_cs = binomial_coefficients(z)
            from functools import reduce
            import operator
            if reduce(operator.add,
                        [ pow(10, z-idx-1)*x*y for idx,(x,y) in 
                            enumerate(zip(b_cs[1:],
                                          [ pow(a, z-i)*pow(j, i) for i in range(1,z+1)]
                    ))]) > c:
                b = j-1
                remainer = c - reduce(operator.add,
                [ pow(10, z-idx-1)*x*y for idx,(x,y) in 
                    enumerate(zip(b_cs[1:],
                              [ pow(a, z-i)*pow(b, i) for i in range(1,z+1)]))])
                a = 10*a + b
                result = "{}{}".format(result, b)
                break
        return a,b,remainer,result
    for i in splited_n:
        A,B,Remainer,Result = findb(i, A, B, Remainer, Result)
    if Remainer != 0:
        Result = "{}.".format(Result)
        for i in ["000" for i in range(p)]:
            A,B,Remainer,Result = findb(i, A, B, Remainer, Result)
    return Result,A,Remainer

> sqrtn("13310", 10, 5) # 5th root of 13310 with 10 precision
sqrtn 13310 with 10 precision is:
('6.6809098427', 66809098427, 47871672829096003837148308169945579226879093)