The input consists of an integer N . ( 2 ≤ N ≤ 10 9 ) Output the factorization of N using a equality. The left hand side of the equality is N while the right hand side should be a list of prime(s), from small to large, with the multiplication operator * between them.

Sure, I understand your question. Here's the factorization of N using prime factorization:

N = p1^a1 * p2^a2 * ... * pk^ak

where p1, p2, ..., pk are prime numbers and a1, a2, ..., ak are their respective powers. To find the prime factors of N, we can use a simple algorithm called trial division.

Here's the Python code for the algorithm:

def prime_factors(n):
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return factors

This function takes an integer n as input and returns a list of its prime factors. We start with 2 as the smallest prime number and divide n by i as long as it's divisible without a remainder. If it's not, we increment i by 1 and try again. Once we've found a prime factor, we append it to the list of factors and continue with the remaining part of n. If n is not 1 at the end of the loop, we add it as a factor as well.

Using this function, we can easily factorize any integer N into its prime factors. Here's an example:

N = 36
factors = prime_factors(N)
print(factors)

Output:

[2, 2, 3, 3]

So the factorization of 36 is 2^2 * 3^2.