Proposition: Calculating the Sum of Divisors

Let $n > 1$ and let $n=\prod_{p\mid n}p^l$ be the factorization of $n.$ Then the sum of divisors of $F(n)$ can be calculated using the formula $$F(n)=\prod_{p\mid n}\frac{p^{l+1}-1}{p-1}.$$

Example

If $n=28$ then $n=2^2\cdot 7^1$ is its factorization. Therefore, $$F(28)=\frac{2^{2+1}-1}{2-1}\cdot \frac{7^{1+1}-1}{7-1}=\frac{7}{1}\cdot \frac{48}{6}=7\cdot 8=56.$$

Indeed, the divisors of $28$ are $\{1,2,4,7,14,28\}$ and their sum is $$1+2+4+7+14+28=56.$$

Proofs: 1

Proofs: 1


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References

Bibliography

  1. Landau, Edmund: "Vorlesungen ├╝ber Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927