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## Proposition: Sum of Möbius Function Over Divisors

For any natural number $n\ge 1$, the sum of the Möbius function over the divisors of $n$ equals $0$ unless $n=1$. Only for this special case the sum equals $1.$ Using the Iverson notation for sums, this can be written as

$$\sum_{d\mid n}\mu(d)=[n=1].$$

| | | | | created: 2019-04-06 08:44:27 | modified: 2019-04-06 18:30:28 | by: bookofproofs | references: [701], [1272]

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