To solve this problem using Dynamic Programming (DP), we need to compute the minimal path sum $A(N)$ in the matrix $M$. Here's a detailed breakdown of how we can model this problem:

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Restating the Problem

1. The sequence $s_n$ is generated recursively:

   • $s_1 = 102022661$
   • $s_n = s_{n-1}^2 \mod 998388889$ for $n > 1$.

2. Two subsequences are defined:

   • $a_n = s_{2n-1}$ (odd-indexed terms of $s_n$)
   • $b_n = s_{2n}$ (even-indexed terms of $s_n$).

3. Matrix $M$ of size $N \times N$ is defined as:

   • $M_{i,j} = a_i + b_j$.

4. $A(N)$ is the minimal path sum from $M_{1,1}$ (top-left) to $M_{N,N}$ (bottom-right), with only right or down moves allowed.

Plan to Solve Using DP

To find $A(N)$, we use a DP table $dp[i][j]$ where:

• $dp[i][j]$ is the minimal path sum to reach $M_{i,j}$.

The recurrence relation for $dp[i][j]$ is: [ dp[i][j] = M_{i,j} + \min(dp[i-1][j], dp[i][j-1]) ] with the base case: [ dp[1][1] = M_{1,1}. ]

To compute $M_{i,j}$, we generate $a_i$ and $b_j$ efficiently using modular arithmetic.

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Algorithm

1. Precompute $s_n$: Generate the sequence $s_n$ modulo $998388889$ up to the required size $2N$ (since we need both $a_n$ and $b_n$).

2. Extract $a_n$ and $b_n$:

   • $a_n = s_{2n-1}$
   • $b_n = s_{2n}$.

3. Construct Matrix $M$:

   • Compute $M_{i,j} = a_i + b_j$ for $1 \leq i, j \leq N$.

4. Compute DP Table:

   • Use the recurrence relation to compute $dp[i][j]$ iteratively.

5. Return $dp[N][N]$: This gives the minimal path sum $A(N)$.

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Python Implementation

Here's the Python code to solve the problem for any $N$:

pythonCopy code
def solve_A(N):
    MOD = 998388889
    s = [0] * (2 * N + 1)
    s[1] = 102022661
    
    # Step 1: Precompute s_n
    for i in range(2, 2 * N + 1):
        s[i] = (s[i-1] * s[i-1]) % MOD
    
    # Step 2: Compute a_n and b_n
    a = [s[2 * i - 1] for i in range(1, N + 1)]
    b = [s[2 * i] for i in range(1, N + 1)]
    
    # Step 3: Initialize M matrix and dp table
    dp = [[0] * (N + 1) for _ in range(N + 1)]
    
    # Step 4: Fill dp table
    for i in range(1, N + 1):
        for j in range(1, N + 1):
            M_ij = a[i-1] + b[j-1]  # M[i][j] = a_i + b_j
            if i == 1 and j == 1:
                dp[i][j] = M_ij
            elif i == 1:
                dp[i][j] = dp[i][j-1] + M_ij
            elif j == 1:
                dp[i][j] = dp[i-1][j] + M_ij
            else:
                dp[i][j] = M_ij + min(dp[i-1][j], dp[i][j-1])
    
    # Step 5: Return result
    return dp[N][N]

# Example usage
N = 10**7  # Replace with smaller values for testing
print(solve_A(N))

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Explanation of Complexity

1. Precomputing $s_n$: $O(2N)$
2. Matrix Construction: $O(N^2)$
3. DP Computation: $O(N^2)$

For $N = 10^7$, direct computation may be infeasible due to memory and time constraints. Optimization techniques (e.g., space reduction in DP) or numerical simulations may be required for large $N$.

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Next Steps

Let me know if you want:

1. An optimization for handling very large $N$ (e.g., space-efficient DP).
2. A detailed explanation of the modular arithmetic in $s_n$.
3. A breakdown of memory usage for $N = 10^7$.
4. Insights on how to compute paths for smaller submatrices.
5. Guidance on how to verify results for smaller $N$.

Tip: Start testing the code with smaller values of $N$ (like 100 or 1000) to verify correctness before scaling up.