Double Pandigital Number Divisible by $11$

We call a positive integer double pandigital if it uses all the digits $0$ to $9$ exactly twice (with no leading zero). For example, $40561817703823564929$ is one such number.
How many double pandigital numbers are divisible by $11$?

To solve this problem, we need to determine the function `P(k)` that counts the number of visible peaks from the `k`th mountain. We can then sum up the values of `P(k)` for `k` ranging from 1 to 2,500,000.

To determine the visibility of the peaks, we need to iterate through the mountains and check if the current mountain is higher than any previous mountain.

Here’s a possible approach to solving this problem in Python:

Step 1: Generate a list of prime numbers up to a certain limit. We can use the sieve of Eratosthenes algorithm to efficiently generate the prime numbers.

“`python
def sieve_of_eratosthenes(n):
primes = [True] * (n + 1)
primes[0] = primes[1] = False

for p in range(2, int(n**0.5) + 1):
if primes[p]:
for i in range(p * p, n + 1, p):
primes[i] = False

return [p for p, is_prime in enumerate(primes) if is_prime]
“`

Step 2: Generate a list of mountains using the prime numbers. Each mountain is represented as a tuple of its up-slope and downslope heights.

“`python
def generate_mountains(primes):
mountains = []
for i in range(len(primes) // 2):
up_slope = primes[2 * i]
down_slope = primes[2 * i + 1]
mountains.append((up_slope, down_slope))
return mountains
“`

Step 3: Implement the `P(k)` function to count the number of visible peaks.

“`python
def count_visible_peaks(mountains, k):
visible_peaks = 0
max_height = 0

for i in range(k):
up_slope, down_slope = mountains[i]

if up_slope > max_height:
visible_peaks += 1
max_height = up_slope

if down_slope > max_height:
visible_peaks += 1
max_height = down_slope

return visible_peaks
“`

Step 4: Compute the sum of `P(k)` for `k` ranging from 1 to 2,500,000.

“`python
primes = sieve_of_eratosthenes(5000000) # Generate prime numbers up to a higher limit
mountains = generate_mountains(primes)

total_visible_peaks = sum([count_visible_peaks(mountains, k) for k in range(1, 2500001)])
print(total_visible_peaks)
“`

This code should output the sum of `P(k)` for `k` ranging from 1 to 2,500,000.

Note: This code may take some time to execute since it involves calculating prime numbers and performing a large number of iterations. You can optimize the code further if needed by using dynamic programming to store intermediate results of `P(k)` to avoid redundant calculations.

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