Problem 710: One Million Members

Mathematical Foundation

Theorem 1 (Structure of Palindromic Partitions). A palindromic partition of $n$ is equivalent to choosing:

1. A “center” part $c$ (which may be 0, meaning no center), and
2. A partition of $(n - c)/2$ into unrestricted parts (each appearing with even multiplicity in the original, counted once here).

The center $c$ must satisfy $c \le n$, $c \equiv n \pmod{2}$, and $(n - c)/2 \ge 0$.

Proof. In a palindromic partition, each part except possibly one center part appears an even number of times. Removing the center part $c$ (or setting $c = 0$ if all parts have even multiplicity) leaves a multiset where every part has even multiplicity. Halving each multiplicity gives an unrestricted partition of $(n - c)/2$. This map is a bijection. $\square$

Theorem 2 (Generating Function for Even-Part Palindromic Partitions). When restricted to even parts, the center $c$ must be even (or 0), and the “half-partition” uses even parts only. Let $p_e(m)$ denote the number of partitions of $m$ into even parts. Then

Proof. Direct from Theorem 1 with the constraint that all parts are even. The center $c \in \{0, 2, 4, \ldots\}$ is even (or absent), and the remaining parts form an even-part partition. $\square$

Lemma 1 (Partition into Even Parts). $p_e(m) = p(\lfloor m/2 \rfloor)$ where $p(k)$ is the standard partition function, since dividing each even part by 2 gives a bijection with unrestricted partitions.

Proof. A partition of $m$ into even parts $\{2a_1, 2a_2, \ldots\}$ bijects with a partition of $m/2$ into positive integers $\{a_1, a_2, \ldots\}$ (when $m$ is even; when $m$ is odd, $p_e(m) = 0$). $\square$

Theorem 3 (Recurrence for $t(n)$). The values $t(n)$ satisfy a computable recurrence derived from the partition function recurrence (Euler’s pentagonal number theorem):

with $p(0) = 1$ and $p(n) = 0$ for $n < 0$. Using this to build a table of $p(n)$ values, $t(n)$ is computed via the sum in Theorem 2.

Proof. Euler’s pentagonal number theorem gives $\prod_{k=1}^{\infty}(1-x^k) = \sum_{k=-\infty}^{\infty}(-1)^k x^{k(3k-1)/2}$. Inverting yields the recurrence for $p(n)$. $\square$

Editorial

We build partition function table using pentagonal recurrence. Finally, compute t(n) for each n and check divisibility by 10^6. We enumerate the admissible parameter range, discard candidates that violate the derived bounds or arithmetic constraints, and update the final set or total whenever a candidate passes the acceptance test.

Pseudocode

Build partition function table using pentagonal recurrence
Compute t(n) for each n and check divisibility by 10^6
if m is even
else p_e(m) = 0

Complexity Analysis

• Time: $O(n_{\max} \sqrt{n_{\max}})$ for building the partition table (each entry uses $O(\sqrt{n})$ pentagonal terms). Computing $t(n)$ for each candidate takes $O(n)$ per candidate, but can be optimized.
• Space: $O(n_{\max})$ for the partition function table.

Answer

#include <bits/stdc++.h>
using namespace std;
int main() {
    int N = 1000;
    vector<long long> dp(N + 1, 0);
    dp[0] = 1;
    for (int k = 1; k <= N; k++)
        for (int j = N; j >= 2*k; j--)
            dp[j] += dp[j - 2*k];

    for (int n = 1; n <= N; n++) {
        long long t = dp[n];
        for (int c = 1; c <= n; c++)
            t += dp[n - c];
        if (t > 1000000) {
            cout << n << endl;
            return 0;
        }
    }
    return 0;
}

"""
Problem 710: One Million Members
"""

def solve():
    # Count palindromic partitions
    # Using DP approach
    N = 1000
    # Partitions where each part appears even number of times
    dp = [0] * (N + 1)
    dp[0] = 1
    for k in range(1, N + 1):
        for j in range(N, 2*k - 1, -1):
            dp[j] += dp[j - 2*k]

    # Add center element options
    t = [0] * (N + 1)
    for n in range(N + 1):
        t[n] = dp[n]  # no center
        for c in range(1, n + 1):
            if n - c >= 0:
                t[n] += dp[n - c]

    # Find first n where t(n) > 10**6
    for n in range(1, N + 1):
        if t[n] > 10**6:
            print(n)
            return n
    return -1

answer = solve()
print(answer)