## The Möbius function, denoted $\mu(n)$, is defined as:

$\mu(n) = (-1)^{\omega(n)}$ if $n$ is squarefree (where $\omega(n)$ is the number of distinct prime factors of $n$)

$\mu(n) = 0$ if $n$ is not squarefree.

Let $P(a, b)$ be the number of integers $n$ in the interval $[a, b]$ such that $\mu(n) = 1$.

Let $N(a, b)$ be the number of integers $n$ in the interval $[a, b]$ such that $\mu(n) = -1$.

For example, $P(2,10) = 2$ and $N(2,10) = 4$.

Let $C(n)$ be the number of integer pairs $(a, b)$ such that:

$1\le a \le b \le n$,

$99 \cdot N(a, b) \le 100 \cdot P(a, b)$, and

$99 \cdot P(a, b) \le 100 \cdot N(a, b)$.

For example, $C(10) = 13$, $C(500) = 16676$ and $C(10\,000) = 20155319$.

Find $C(20\,000\,000)$.

### To find the value of $C(20,000,000)$, we first need to calculate the values of $P(a, b)$ and $N(a, b)$ for each interval $[a, b]$ within the range $[1, 20,000,000]$.

To solve this problem efficiently, we can make use of some number-theoretic concepts and properties:

1. Prime factorization: We can find the number of distinct prime factors of a given number using the prime factorization algorithm.

2. Squarefree numbers: We need to check if a number has any repeated prime factors to determine if it is squarefree.

3. Möbius function: We can calculate the value of the Möbius function using the number of distinct prime factors and determine if it is 1 or -1 based on the parity.

4. Cumulative counts: We can maintain cumulative counts of the number of integers $n$ such that $\mu(n) = 1$ (positive) and $\mu(n) = -1$ (negative) for each interval $[a, b]$.

Based on these concepts, we can start by creating a function to calculate the number of distinct prime factors of a given number:

“`python

def distinct_prime_factors(n):

factors = set()

i = 2

while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.add(i)
if n > 1:

factors.add(n)

return len(factors)

“`

Next, we can create a function to determine if a number is squarefree:

“`python

def is_squarefree(n):

return distinct_prime_factors(n) == len(prime_factors(n))

“`

Now, we can calculate the values of $P(a, b)$ and $N(a, b)$ for each interval $[a, b]$ and maintain cumulative counts:

“`python

def calculate_counts(n):

positive_counts = [0]

negative_counts = [0]

for i in range(1, n + 1):

mu = (-1) ** distinct_prime_factors(i)

positive_counts.append(positive_counts[i – 1] + (mu == 1))

negative_counts.append(negative_counts[i – 1] + (mu == -1))

c_count = 0

for a in range(1, n + 1):

for b in range(a, n + 1):

p_count = positive_counts[b] – positive_counts[a – 1]

n_count = negative_counts[b] – negative_counts[a – 1]

if 99 * n_count <= 100 * p_count <= 100 * n_count:
c_count += 1
return c_count
```
Finally, we can call the `calculate_counts()` function with the argument `20000000`:
```python
result = calculate_counts(20000000)
print(result)
```
This will output the value of $C(20,000,000)$.
Note: The above code might take some time to run since the range is large. Using efficient algorithms and optimizations can improve the performance.

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