Problem 926: Binomial Coefficient Divisibility

Mathematical Foundation

Theorem 1 (Lucas’ Theorem, 1878). Let $p$ be a prime and let $m, n$ be non-negative integers with base-$p$ representations $m = \sum_{i=0}^{s} m_i p^i$ and $n = \sum_{i=0}^{s} n_i p^i$. Then

Proof. We use the Frobenius endomorphism: in $\mathbb{F}_p[x]$, $(1+x)^p = 1 + x^p$. By induction, $(1+x)^{p^i} = 1 + x^{p^i}$. Writing $m = \sum m_i p^i$:

The coefficient of $x^n$ on the left is $\binom{m}{n}$. On the right, writing $n = \sum n_i p^i$ and using the uniqueness of base-$p$ representation, the coefficient of $x^n$ is $\prod_{i} \binom{m_i}{n_i}$. (Cross-terms from different $p^i$-positions cannot combine to give the same power of $x$ because of the uniqueness of base-$p$ digits.) $\square$

Theorem 2 (Non-divisibility Criterion). $\binom{m}{n} \not\equiv 0 \pmod{p}$ if and only if $n_i \leq m_i$ for every digit position $i$ in base $p$.

Proof. From Theorem 1, $\binom{m}{n} \equiv \prod_i \binom{m_i}{n_i} \pmod{p}$. Since $0 \leq n_i, m_i \leq p-1$, each factor $\binom{m_i}{n_i}$ is nonzero mod $p$ iff $n_i \leq m_i$ (as $\binom{a}{b} = 0$ for $b > a$ and $\binom{a}{b} \in \{1, \ldots, p-1\}$ for $0 \leq b \leq a < p$). The product is nonzero iff every factor is nonzero. $\square$

Theorem 3 (Per-Row Count). The number of entries $\binom{n}{k}$ ($0 \leq k \leq n$) not divisible by $p$ equals

where $n = d_s d_{s-1} \cdots d_0$ in base $p$.

Proof. By Theorem 2, we need to count $(k_0, k_1, \ldots, k_s)$ with $0 \leq k_i \leq d_i$ for each $i$. Since the digits are independent, the count is $\prod_i (d_i + 1)$. $\square$

Lemma 1 (Specialization to $p = 2$). For $p = 2$, each digit $d_i \in \{0, 1\}$, so $d_i + 1 \in \{1, 2\}$, and the number of odd entries in row $n$ is $2^{s_2(n)}$, where $s_2(n) = \mathrm{popcount}(n)$ is the number of 1-bits.

Proof. When $d_i = 0$, the factor is 1; when $d_i = 1$, the factor is 2. The product is $2^{(\text{number of 1-digits})} = 2^{s_2(n)}$. $\square$

Theorem 4 (Kummer’s Theorem, 1852). The largest power of $p$ dividing $\binom{m+n}{m}$ equals the number of carries when adding $m$ and $n$ in base $p$.

Proof. By Legendre’s formula, $v_p(k!) = \sum_{j \geq 1} \lfloor k/p^j \rfloor$. Then

Each term $\lfloor(m+n)/p^j\rfloor - \lfloor m/p^j \rfloor - \lfloor n/p^j \rfloor$ equals the carry into position $j$ when adding $m$ and $n$ in base $p$. $\square$

Editorial

Optimized digit DP for general $p$. We use dynamic programming over the state space implied by the derivation, apply each admissible transition, and read the answer from the final table entry.

Pseudocode

Compute base-p digits of N
S(p^k - 1) = ((p+1)/2)^k for p=2: number of non-zero entries in rows 0..p^k-1
General: product_{digit positions} sum over digit values of (d+1)
Contribution from rows with digit_i < d at position i
and arbitrary lower digits (within full range 0..p-1)

Complexity Analysis

• Time (brute force): $O(N \log N)$ — iterating over rows and computing popcount.
• Time (digit DP): $O(\log_p^2 N)$.
• Space: $O(\log_p N)$ for digit representation.

Answer

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

/*
 * Problem 926: Binomial Coefficient Divisibility
 *
 * Count entries in Pascal's triangle (rows 0-1000) NOT divisible by p=2.
 *
 * By Lucas' theorem: binom(n,k) != 0 mod p iff each base-p digit
 * of k <= corresponding digit of n.
 * Non-zero entries in row n = prod(d_i + 1) over base-p digits d_i.
 * For p=2: = 2^popcount(n).
 *
 * Kummer's theorem: v_p(binom(m+n,m)) = # carries adding m,n in base p.
 *
 * Complexity: O(N * log_p(N)).
 */

int main() {
    int N = 1000, p = 2;
    long long total = 0;
    for (int n = 0; n <= N; n++) {
        long long prod = 1;
        int nn = n;
        if (nn == 0) { total += 1; continue; }
        while (nn > 0) {
            prod *= (nn % p + 1);
            nn /= p;
        }
        total += prod;
    }
    cout << total << endl;

    // Verify: row 0 has 1 entry, row 1 has 2, row 3 has 4
    assert(1 == 1);  // row 0
    // row 3 = 11 in binary, popcount=2, so 2^2=4 odd entries
    int r3 = 1;
    int nn = 3;
    while (nn > 0) { r3 *= (nn%2+1); nn /= 2; }
    assert(r3 == 4);

    return 0;
}

"""
Problem 926: Binomial Coefficient Divisibility

Count non-zero entries mod p in Pascal's triangle rows 0..N. For each row n,
the number of entries C(n,k) not divisible by p equals the product of (d_i + 1)
where d_i are the base-p digits of n (Kummer's theorem / Lucas' theorem).

Key ideas:
    - By Lucas' theorem, C(n,k) != 0 mod p iff each base-p digit of k <= digit of n.
    - Number of non-zero entries in row n = product of (d_i + 1) for base-p digits d_i.
    - For p=2, this equals 2^(popcount(n)), giving a self-similar fractal structure.
    - Sierpinski triangle pattern emerges in Pascal's triangle mod 2.

Methods:
    1. Digit-product formula (Lucas' theorem application)
    2. Direct binomial computation for small cases (verification)
    3. Fractal dimension analysis
    4. Comparison across different primes
"""

import numpy as np
from math import comb

def count_nonzero_row(n, p):
    """Count entries in row n of Pascal's triangle not divisible by p."""
    prod = 1
    while n > 0:
        prod *= (n % p + 1)
        n //= p
    return prod

def solve(N=1000, p=2):
    """Sum of non-zero-mod-p entries across rows 0..N."""
    total = 0
    for n in range(N + 1):
        total += count_nonzero_row(n, p)
    return total

def count_nonzero_row_brute(n, p):
    """Count entries in row n not divisible by p by direct computation."""
    count = 0
    for k in range(n + 1):
        if comb(n, k) % p != 0:
            count += 1
    return count

def fractal_dimension_estimate(p, max_power=10):
    """Estimate fractal dimension: total non-zero entries up to p^k rows.

    For Pascal mod p, the fractal dimension is log(p*(p+1)/2) / log(p).
    """
    ratios = []
    for k in range(2, max_power + 1):
        N = p**k
        total = sum(count_nonzero_row(n, p) for n in range(N))
        if k > 2:
            prev_N = p**(k - 1)
            prev_total = sum(count_nonzero_row(n, p) for n in range(prev_N))
            ratios.append(np.log(total / prev_total) / np.log(p))
    return ratios

def row_counts_for_prime(N, p):
    """Return list of non-zero entry counts per row for given prime."""
    return [count_nonzero_row(n, p) for n in range(N)]

# Solve and verify
answer = solve()

# Verify digit-product formula against brute force for small cases
for n in range(50):
    assert count_nonzero_row(n, 2) == count_nonzero_row_brute(n, 2), f"Mismatch at row {n}"

# Verify for p=3
for n in range(40):
    assert count_nonzero_row(n, 3) == count_nonzero_row_brute(n, 3), f"Mismatch at row {n} for p=3"

# Known: row 2^k - 1 has all entries odd (all 2^k entries are odd)
for k in range(1, 8):
    n = 2**k - 1
    assert count_nonzero_row(n, 2) == 2**k

print(answer)