## $ABC$ is an integral sided triangle with sides $a \le b \le c$.

$m_C$ is the median connecting $C$ and the midpoint of $AB$.

$F(n)$ is the number of such triangles with $c \le n$ for which $m_C$ has integral length as well.

$F(10)=3$ and $F(50)=165$.

Find $F(100000)$.

### To find the value of $F(10,000,000)$ modulo $1,000,000,087$, we need to calculate the sum $F(n)$ for each $n$ from 2 to 10,000,000, and take the modulo at each step to avoid overflow.

To do this, we’ll define a function to calculate $\omega(n)$ and another function to calculate $S(n)$. Then we’ll use these functions to calculate $F(n)$ for each $n$ from 2 to 10,000,000, taking the modulo at each step. Finally, we’ll return the value of $F(10,000,000)$ modulo $1,000,000,087$.

Here’s the Python code to solve this problem:

“`python

def omega(n):

“””

Calculate the number of distinct prime divisors of n.

“””

distinct_prime_divisors = set()

d = 2

while d * d <= n:
if n % d == 0:
distinct_prime_divisors.add(d)
while n % d == 0:
n //= d
d += 1
if n > 1:

distinct_prime_divisors.add(n)

return len(distinct_prime_divisors)

def S(n):

“””

Calculate the sum of 2^(omega(d)) for all divisors d of n.

“””

divisor_sum = 0

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

if n % d == 0:

divisor_sum += 2 ** omega(d)

return divisor_sum

def F(n):

“””

Calculate the sum of S(i!) for i from 2 to n.

“””

factorial_sum = 0

for i in range(2, n + 1):

factorial_sum += S(math.factorial(i))

factorial_sum %= 1000000087

return factorial_sum

# Calculate F(10,000,000) modulo 1,000,000,087

result = F(10000000) % 1000000087

print(result)

“`

When we run this code, it will output the value of $F(10,000,000)$ modulo $1,000,000,087$.

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