Problem 627: Counting Products

Mathematical Analysis

Recurrence

The number of ordered factorizations satisfies:

The “+1” accounts for the trivial factorization $n$ itself. Equivalently:

where the sum over proper divisors includes $f(1) = 1$ as the base case (corresponding to “use $n$ as a single factor”).

Dirichlet Series

The Dirichlet generating function is:

This follows because $f = f * \mathbf{1}_{>1}$ in Dirichlet convolution: the DGF $F(s)$ satisfies $F(s) = F(s)(\zeta(s) - 1) + 1$, giving $F(s)(2 - \zeta(s)) = 1$.

Multiplicative Structure

While $f$ is not multiplicative, it satisfies useful identities:

• $f(p) = 1$ for prime $p$
• $f(p^2) = 2$ (namely $p^2$ and $p \cdot p$)
• $f(p^k) = 2^{k-1}$ (compositions of $k$)
• $f(pq) = 3$ for distinct primes $p, q$ (namely $pq$, $p \cdot q$, $q \cdot p$)

Concrete Values

Derivation

Algorithm: Sieve-Like DP

Compute $f(n)$ for $n = 1, \ldots, N$ using a divisor-sieve approach:

1. Initialize $f(1) = 1$ and $f(n) = 0$ for $n \ge 2$.
2. For each $d$ from 1 to $N$, for each multiple $m = 2d, 3d, \ldots, N$:
   • $f(m) \mathrel{+}= f(d)$ (this adds the contribution of factorizations starting with factor $m/d$).

This runs in $O(N \log N)$ time (harmonic sum).

Alternative: Recursive with Memoization

For each $n$, recursively sum $f(d)$ over proper divisors $d$ of $n$.

Proof of Correctness

Theorem. The sieve computes $f(n) = \sum_{d \mid n, d < n} f(d)$ for all $n \ge 2$.

Proof. Each proper divisor $d$ of $n$ contributes $f(d)$ to $f(n)$ when $d$ is processed and its contribution propagated to multiple $n$. Since we iterate over all $d$ and all multiples, every divisor pair is covered exactly once. $\square$

Corollary. $F(s) = 1/(2 - \zeta(s))$ gives the analytic continuation, confirming the asymptotic $F(N) \sim CN$ for a constant $C$ determined by the pole structure.

Complexity Analysis

• Sieve DP: $O(N \log N)$ time, $O(N)$ space.
• Recursive: $O(N \sigma_0(N))$ worst case, where $\sigma_0$ is the divisor count function.

Answer

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

/*
 * Problem 627: Counting Products
 *
 * f(n) = ordered factorizations of n into factors >= 2.
 * Recurrence: f(n) = sum_{d|n, d<n} f(d), f(1) = 1.
 * DGF: F(s) = 1/(2 - zeta(s)).
 *
 * Method 1: Sieve-like DP in O(N log N)
 * Method 2: Recursive over divisors (verification)
 */

const int MAXN = 1000001;
ll f[MAXN];

void solve_sieve(int N) {
    memset(f, 0, sizeof(f));
    f[1] = 1;
    for (int d = 1; d <= N; d++) {
        for (int m = 2 * d; m <= N; m += d) {
            f[m] += f[d];
        }
    }
}

int main() {
    int N = 1000000;
    solve_sieve(N);

    // Verify small values
    assert(f[1] == 1);
    assert(f[2] == 1);
    assert(f[4] == 2);
    assert(f[6] == 3);
    assert(f[8] == 4);
    assert(f[12] == 8);

    // Verify f(p^k) = 2^{k-1}
    for (int k = 1; k <= 19; k++) {
        int pk = 1 << k;
        if (pk > N) break;
        assert(f[pk] == (1 << (k - 1)));
    }

    ll total = 0;
    for (int n = 2; n <= N; n++) total += f[n];
    cout << "F(" << N << ") = " << total << endl;

    return 0;
}

"""
Problem 627: Counting Products

f(n) = number of ordered factorizations of n into factors >= 2.
F(N) = sum_{n=2}^{N} f(n).

Dirichlet series: sum f(n)/n^s = 1/(2 - zeta(s)).
Recurrence: f(n) = sum_{d|n, d<n} f(d), f(1) = 1.

Method 1: Sieve-like DP in O(N log N)
Method 2: Recursive with memoization (verification)
"""

# --- Method 1: Sieve DP ---
def solve_sieve(N: int) -> list:
    """Compute f(n) for n=1..N using sieve-like DP."""
    f = [0] * (N + 1)
    f[1] = 1
    for d in range(1, N + 1):
        for m in range(2 * d, N + 1, d):
            f[m] += f[d]
    return f

# --- Method 2: Recursive ---
def solve_recursive(N: int) -> list:
    """Compute f(n) recursively."""
    f = [0] * (N + 1)
    f[1] = 1
    for n in range(2, N + 1):
        total = 0
        d = 1
        while d * d <= n:
            if n % d == 0:
                if d < n:
                    total += f[d]
                if d != n // d and n // d < n:
                    total += f[n // d]
            d += 1
        f[n] = total
    return f

# Verify
N = 1000
f_sieve = solve_sieve(N)
f_rec = solve_recursive(N)

for n in range(1, N + 1):
    assert f_sieve[n] == f_rec[n], f"Mismatch at n={n}"

# Known values
assert f_sieve[1] == 1
assert f_sieve[2] == 1
assert f_sieve[4] == 2
assert f_sieve[6] == 3
assert f_sieve[8] == 4
assert f_sieve[12] == 8
assert f_sieve[16] == 8

# f(p^k) = 2^{k-1}
for k in range(1, 10):
    assert f_sieve[2**k] == 2**(k-1), f"f(2^{k}) = {f_sieve[2**k]}, expected {2**(k-1)}"

F_N = sum(f_sieve[2:])
print(f"F({N}) = {F_N}")
print("All verifications passed.")