Sum of Divisors

If a positive integer n is written in prime factorized form as p_1^a_1 p_2^a_2 \cdots p_k^a_k, then every divisor of n is built by choosing an exponent for each prime between 0 and a_i. Summing all divisors therefore splits into independent geometric sums for each prime factor, which is why the total equals the product of \((p_i^{a_i+1}-1)/(p_i-1)\). This multiplicative structure comes from the fact that divisors combine by multiplying the chosen prime powers.

Use this when you need the sum of all positive divisors of a number from its prime factorization, especially for classifying numbers like abundant, deficient, or perfect numbers. It is also useful in divisor-function problems in contest math and number theory proofs.

• Using the prime factors themselves instead of their exponents in the formula
• Forgetting that the result includes the number n as one of its divisors
• Adding the prime-power contributions instead of multiplying the separate geometric sums