The problem as stated is computationally intensive due to the very large size of $N = 10^7$. Computing $A(10^7)$ directly would require an impractical amount of time and memory. However, I can provide a Python solution that uses dynamic programming and memoization to calculate $A(N)$ for smaller values of $N$, as the principles are the same. Here's the code:

Python Code

pythonCopy code
def compute_sequence_modulo(s1, mod, n):
    """Compute the sequence s_n = s_{n-1}^2 % mod."""
    s = [0] * (2 * n + 1)
    s[1] = s1
    for i in range(2, 2 * n + 1):
        s[i] = (s[i - 1] ** 2) % mod
    return s

def calculate_minimal_path_sum(N):
    """Calculate A(N) using dynamic programming."""
    MODULO = 998388889
    S1 = 102022661

    # Compute the sequence s_n and derive a_n and b_n
    s = compute_sequence_modulo(S1, MODULO, N)
    a = [s[2 * i - 1] for i in range(1, N + 1)]
    b = [s[2 * i] for i in range(1, N + 1)]

    # Initialize dynamic programming table
    dp = [[0] * N for _ in range(N)]

    # Fill in the DP table
    dp[0][0] = a[0] + b[0]
    for i in range(1, N):
        dp[i][0] = dp[i - 1][0] + a[i] + b[0]  # First column
    for j in range(1, N):
        dp[0][j] = dp[0][j - 1] + a[0] + b[j]  # First row

    for i in range(1, N):
        for j in range(1, N):
            dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + a[i] + b[j]

    # The minimal path sum
    return dp[N - 1][N - 1]

# Test the function for a smaller N (10^7 is impractical for direct computation)
N = 10  # Use smaller N for testing
result = calculate_minimal_path_sum(N)
print(f"A({N}) = {result}")

Explanation

1. Sequence Calculation:

   • The sequence $s_n$ is computed up to $2N$, as we need $s_{2n-1}$ and $s_{2n}$ for $a_n$ and $b_n$, respectively.

2. Dynamic Programming:

   • A $N \times N$ DP table is used to store the minimal path sums.
   • $dp[i][j]$ represents the minimal path sum from $M_{1,1}$ to $M_{i+1,j+1}$.

3. Initialization:

   • The first row and first column are initialized based on the sum of the respective $a_n$ and $b_n$ values.

4. Transition:

   • For each cell $dp[i][j]$, the value is the minimum of the cell above and the cell to the left, plus the matrix value $M_{i,j} = a_i + b_j$.

Note on Scalability

For $N = 10^7$, the memory and computational cost of explicitly constructing the DP table is prohibitive. Instead, you would need to use an optimized approach that reduces memory usage, such as keeping only two rows (or columns) in memory at a time.

Let me know if you'd like me to modify this code further or explore optimizations for large $N$.