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Divisor Sum Calculator – σ(n) Function

Quick answer

σ(n) sums all divisors: σ(12) = 1+2+3+4+6+12 = 28, σ(6) = 12.

  • Perfect iff σ = 2n — plus aliquot sum σ−n and prime factor shortcut.

Figures are estimates from this calculator’s standard formula — adjust the inputs in the tool below for your exact number.

How the Divisor Sum Calculator Works

σ(n) = Σ d over all divisors | σ(pᵃ) = (pᵃ⁺¹−1)/(p−1)

The divisor function σ(n) adds every positive divisor of n. For 12, the divisors 1, 2, 3, 4, 6, 12 sum to σ(12) = 28. The prime-factorization shortcut computes it without listing: 12 = 2²×3 gives σ = (1+2+4)(1+3) = 7×4 = 28, using the geometric-series rule σ(pᵃ) = 1 + p + … + pᵃ = (pᵃ⁺¹−1)/(p−1). Multiplicativity across coprime parts lets σ(10) = σ(2)·σ(5) = 3×6 = 18. The calculator shows the divisor list with running sum, the factorized computation, and the derived quantities: the aliquot sum σ(n) − n (16 for 12, flagging it abundant), and the perfect-number test σ = 2n (σ(28) = 56 ✓, σ(6) = 12 ✓). It also reports τ(n), the divisor count — τ(12) = 6 — and notes squares have odd τ.

  1. 1

    Enter n

    Any positive integer

  2. 2

    List the divisors

    All d | n with a running total

  3. 3

    Factor shortcut

    (1+2+4)(1+3) = 28 recomputes σ(12)

  4. 4

    Classify

    Perfect (σ = 2n), abundant (σ > 2n), or deficient (σ < 2n)

Use Cases

Perfect Number Hunt

σ(28) = 56 = 2×28 certifies perfection

Number Theory Homework

Verify multiplicativity: σ(10) = σ(2)σ(5) = 18

Aliquot Sequences

Iterate σ(n) − n to explore amicable chains

Tips

  • 1

    σ(pᵃ) = (pᵃ⁺¹−1)/(p−1) — the geometric series rule

  • 2

    Perfect ⇔ σ(n) = 2n; aliquot sum = σ(n) − n

  • 3

    τ(n) counts divisors; squares have odd τ

Common Mistakes

  • Omitting n itself from the divisor list

  • Multiplying σ values for non-coprime parts (must factor first)

  • Confusing σ with the aliquot sum σ − n

  • Missing the factor (p⁰ = 1) in the geometric series

FAQs

What is σ(12)?

1+2+3+4+6+12 = 28; shortcut (1+2+4)(1+3) = 28 agrees.

Why is 28 perfect?

σ(28) = 1+2+4+7+14+28 = 56 = 2×28.

What is the aliquot sum of 12?

σ(12) − 12 = 16, making 12 abundant.

How do primes behave?

σ(p) = p+1: σ(7) = 8, deficient like all primes.

How many divisors does 12 have?

Six — τ(12) = 6 from (2+1)(1+1) in exponent form.