To solve this problem, we need to determine the number of ways to write a given number ( n ) as a sum of consecutive positive integers. The key insight here is that this number of ways is equivalent to the number of odd positive divrs of ( n ).

Approach Explained

The problem can be reduced to finding the number of odd divrs of ( n ) due to the following reasoning:

1. Any sum of consecutive integers can be represented using the formula: ( n = k \cdot a + \frac{k(k-1)}{2} ), where ( k ) is the number of terms and ( a ) is the starting integer. Rearranging gives ( 2n = k \cdot (2a + k - \1) ).
   • Here, ( k ) and ( (２a + k -1) ) are factors of ( 2n ) with opposite parity (one even, one odd) and ( k < (2a + k -1) ).
2. The number of such valid pairs ( (k, 2a + k -1) ) is exactly the number of odd divrs of ( n ). This is because the odd divrs of ( n ) correspond to valid values of ( k ) (or the other factor) in the equation.

Solution Code

    # Remove all factors of 2 from n
    while n % 2 == 0:
        n = n // 2

    count = 1
    i = 3
    # Check for odd divrs up to sqrt(n)
    while i * i <= n:
        exponent = 0
        while n % i == 0:
            exponent += 1
            n = n // i
        if exponent > 0:
            count *= (exponent + 1)
        i += 2

    # If remaining n is a prime number (greater than 1)
    if n > 1:
        count *= 2

    return count

# Example usage:
# print(count_consecutive_sums(9))  # Output:3
# print(count_consecutive_sums(15)) # Output:4
# print(count_consecutive_sums(6))  # Output:2

Explanation

1. Remove Even Factors: We first eliminate all factors of 2 from ( n ) since even factors do not contribute to the count of odd divrs.
2. Count Odd Divrs: For the remaining odd number, we compute the number of divrs by checking each odd number up to the square root of ( n ). For each divr, we count its exponent and multiply the result by ( (exponent +1) ) (since each exponent can range from 0 to its value in the factorization).
3. Prime Check: If the remaining number after processing all factors up to the square root is greater than 1, it means it's a prime number, so we multiply the count by 2 (since it contributes one more divr pair: itself and 1).

This approach efficiently computes the number of ways to write ( n ) as a sum of consecutive positive integers, even for large values of ( n ), as it runs in ( O(\sqrt{n}) ) time complexity (with a focus on odd factors only). This makes it suitable for most practical purposes.

作者声明:本文包含人工智能生成内容。