Python - count the number of prime divisors without using range

I have to write a function that counts the total number of prime divisors of a given positive integer n. Since I have just started learning python a week ago, I am not allowed to use for loop with range. Below is what I have so far:

def count_prime_divisors(n):
    return num_of_divisors(n, 1)

def num_of_divisors(n, i):
    if i > n:
        return 0
    if n % i == 0 and num_of_divisors(i, 1) == 2:
        return 1 + num_of_divisors(n, i+1)
    else:
        return num_of_divisors(n, i+1)

So I know that the exceeding maximum recursion depth error occurs at this line if n % i == 0 and num_of_divisors(i, 1) == 2 but I don't know how to check if the divisor is prime so that the function can work appropriately. Maybe I should write another helper function? Can someone help me with this? Any help is much appreciated :( Thank you!

You can try this.

def prime_fac(n,count=0,prime_num=2):
    if n==1:
        print('Number of prime factors',count)
        return
    elif n%prime_num==0:
        print(prime_num)
        return prime_fac(n//prime_num,count+1,prime_num)
    else:
        return prime_fac(n,count,prime_num+1)

prime_fac(100)
prime_fac(12)
prime_fac(999)

---

output

2
2
5
5
Number of prime factors 4
2
2
3
Number of prime factors 3
3
3
3
37
Number of prime factors 4

---

Use this if want to store the count of prime factors and unique prime factors.

def prime_fac(n,count=0,prime_num=2,prime_set=set()):
    if n==1:
        print('Number of prime factors',count)
        return count,prime_set
    elif n%prime_num==0:
        print(prime_num)
        prime_set=prime_set|{prime_num}  #set union operation
        return prime_fac(n//prime_num,count+1,prime_num,prime_set)
    else:
        return prime_fac(n,count,prime_num+1,prime_set)

a=prime_fac(100)
print(a)

---

output

2
2
5
5
Number of prime factors 4

(4, {2, 5}) #a looks like this where 4 is prime factors count and {2,5} is unique prime factors.