Check if the given number is Ore number or not in Python

An ore number (or harmonic divisor number) is a positive integer where the harmonic mean of its divisors is an integer. The harmonic mean of divisors is calculated as the number of divisors divided by the sum of reciprocals of all divisors.

For example, 28 has divisors [1, 2, 4, 7, 14, 28]. The harmonic mean is 6 ÷ (1/1 + 1/2 + 1/4 + 1/7 + 1/14 + 1/28) = 3, which is an integer.

Finding All Divisors

First, we need a function to find all divisors efficiently by checking only up to the square root ?

import math

def get_all_divisors(n):
    divisors = []
    for i in range(1, int(math.sqrt(n)) + 1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:  # Avoid duplicate for perfect squares
                divisors.append(n // i)
    return sorted(divisors)

# Test with 28
divisors = get_all_divisors(28)
print("Divisors of 28:", divisors)

Divisors of 28: [1, 2, 4, 7, 14, 28]

Calculating Harmonic Mean

The harmonic mean of divisors is: number_of_divisors ÷ sum_of_reciprocals ?

import math

def get_all_divisors(n):
    divisors = []
    for i in range(1, int(math.sqrt(n)) + 1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n // i)
    return sorted(divisors)

def get_harmonic_mean(n):
    divisors = get_all_divisors(n)
    sum_of_reciprocals = sum(1/d for d in divisors)
    return len(divisors) / sum_of_reciprocals

# Test with 28
harmonic_mean = get_harmonic_mean(28)
print(f"Harmonic mean of divisors of 28: {harmonic_mean}")

Harmonic mean of divisors of 28: 3.0

Complete Ore Number Checker

Now we combine everything to check if a number is an ore number ?

import math

def get_all_divisors(n):
    divisors = []
    for i in range(1, int(math.sqrt(n)) + 1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n // i)
    return sorted(divisors)

def is_ore_number(n):
    if n <= 0:
        return False
    
    divisors = get_all_divisors(n)
    sum_of_reciprocals = sum(1/d for d in divisors)
    harmonic_mean = len(divisors) / sum_of_reciprocals
    
    # Check if harmonic mean is an integer
    return harmonic_mean == int(harmonic_mean)

# Test with different numbers
test_numbers = [1, 6, 28, 140, 270, 496, 672, 30]

for num in test_numbers:
    result = is_ore_number(num)
    divisors = get_all_divisors(num)
    harmonic_mean = len(divisors) / sum(1/d for d in divisors)
    print(f"{num}: {result} (harmonic mean = {harmonic_mean:.3f})")

1: True (harmonic mean = 1.000)
6: True (harmonic mean = 2.000)
28: True (harmonic mean = 3.000)
140: True (harmonic mean = 5.000)
270: True (harmonic mean = 4.000)
496: True (harmonic mean = 5.000)
672: True (harmonic mean = 6.000)
30: False (harmonic mean = 2.333)

Key Properties

• All perfect numbers (like 6, 28, 496) are ore numbers
• The smallest ore numbers are: 1, 6, 28, 140, 270, 496, 672, 1638, 2970, 6200
• Ore numbers are relatively rare − there are only 44 ore numbers less than 10,000

Conclusion

An ore number has a harmonic mean of divisors that is an integer. Use the formula: harmonic_mean = count_of_divisors ÷ sum_of_reciprocals_of_divisors. The algorithm efficiently finds divisors by checking only up to the square root.