An integer of the form $p^q q^p$ with prime numbers $p \neq q$ is called a hybrid-integer.
For example, $800 = 2^5 5^2$ is a hybrid-integer.
We define $C(n)$ to be the number of hybrid-integers less than or equal to $n$.
You are given $C(800) = 2$ and $C(800^{800}) = 10790$.
Find $C(800800^{800800})$.
To find the value of $C(800800^{800800})$, we need to count the number of hybrid-integers less than or equal to $800800^{800800}$.
We can approach this problem by iterating through all values of $p$ and $q$ and checking if the resulting number is less than or equal to $800800^{800800}$.
Let’s write the code to compute $C(800800^{800800})$ using Python:
“`python
import math
def is_prime(n):
if n <= 1:
return False
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return False
return True
def count_hybrid_integers(n):
count = 0
for p in range(2, int(math.sqrt(n)) + 1):
if is_prime(p):
for q in range(2, int(math.log(n, p)) + 1):
if is_prime(q) and p != q:
number = p ** q * q ** p
if number <= n:
count += 1
return count
n = 800800 ** 800800
result = count_hybrid_integers(n)
print("C(800800^{800800}) =", result)
```
In this code, we define a helper function `is_prime` to check if a number is prime. Then, we define the main function `count_hybrid_integers` that iterates through all possible values of $p$ and $q$ within a certain range. For each combination of $p$ and $q$, we calculate the hybrid-integer and check if it is less than or equal to the given $n$. If it is, we increment the count. Finally, we print the result.
Note: This code may take a long time to run due to the large value of $n$. You can consider optimizing the code if needed.
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