## Let $G(a, b)$ be the smallest non-negative integer $n$ for which $\operatorname{\mathbf{gcd}}$Greatest common divisor$(n^3 + b, (n + a)^3 + b)$ is maximized.

For example, $G(1, 1) = 5$ because $\gcd(n^3 + 1, (n + 1)^3 + 1)$ reaches its maximum value of $7$ for $n = 5$, and is smaller for $0 \le n \lt 5$.

Let $H(m, n) = \sum G(a, b)$ for $1 \le a \le m$, $1 \le b \le n$.

You are given $H(5, 5) = 128878$ and $H(10, 10) = 32936544$.

Find $H(18, 1900)$.

### To solve this problem, we need to find the expected area of the convex polygons for each value of n, and then calculate the cumulative sum of these expected values up to n = 50.

Let’s begin by writing a function that calculates the area of a convex polygon given the lengths of its sides using Python’s math library and the Heron’s formula:

“`python

import math

def calculate_area(sides):

a, b, c = sides

s = (a + b + c) / 2 # semi-perimeter

area = math.sqrt(s * (s – a) * (s – b) * (s – c))

return area

“`

Next, we will create a function to generate all possible splits of a line segment into n segments of integer length. We can do this using recursion and keeping track of the remaining length and the current split:

“`python

def generate_splits(n, remaining_length, current_split):

if n == 0:

return [current_split]

splits = []

for length in range(1, remaining_length – (n – 1) + 1):

splits += generate_splits(n-1, remaining_length-length, current_split + [length])

return splits

“`

Now, we can write a function to calculate the expected area for a given n. This function will generate all possible splits, calculate the area for each split, and then return the average area:

“`python

def calculate_expected_area(n):

splits = generate_splits(n, 2*n-3, [])

total_area = 0

for split in splits:

areas = []

for i in range(n):

sides = [split[i], split[(i+1)%n], split[(i+2)%n]]

areas.append(calculate_area(sides))

total_area += max(areas)

expected_area = total_area / len(splits)

return expected_area

“`

Finally, we can write a function to calculate the cumulative sum of the expected areas up to n = 50:

“`python

def calculate_cumulative_sum(n):

cumulative_sum = 0

for i in range(3, n+1):

cumulative_sum += calculate_expected_area(i)

return cumulative_sum

“`

Now we can call the `calculate_cumulative_sum()` function with n = 50 and round the result to 6 decimal places to find S(50):

“`python

result = calculate_cumulative_sum(50)

rounded_result = round(result, 6)

print(rounded_result)

“`

Running this code will give us the value of S(50) rounded to 6 decimal places.

##### More Answers:

Maximum Number of DivisorsSums of Power Sums

Unbalanced Nim