Dirichlet Convolution - Dirichlet Inverse

Dirichlet Inverse

Given an arithmetic function ƒ its Dirichlet inverse g = ƒ−1 may be calculated recursively (i.e. the value of g(n) is in terms of g(m) for m < n) from the definition of Dirichlet inverse.

For n = 1:

(ƒ * g) (1) = ƒ(1) g(1) = (1) = 1, so
g(1) = 1/ƒ(1). This implies that ƒ does not have a Dirichlet inverse if ƒ(1) = 0.

For n = 2

(ƒ * g) (2) = ƒ(1) g(2) + ƒ(2) g(1) = (2) = 0,
g(2) = −1/ƒ(1) (ƒ(2) g(1)),

For n = 3

(ƒ * g) (3) = ƒ(1) g(3) + ƒ(3) g(1) = (3) = 0,
g(3) = −1/ƒ(1) (ƒ(3) g(1)),

For n = 4

(ƒ * g) (4) = ƒ(1) g(4) + ƒ(2) g(2) + ƒ(4) g(1) = (4) = 0,
g(4) = −1/ƒ(1) (ƒ(4) g(1) + ƒ(2) g(2)),

and in general for n > 1,

g(n) = \frac {-1}{f(1)} \sum_\stackrel{d\,\mid \,n} {d < n} f\left(\frac{n}{d}\right) g(d).

Since the only division is by ƒ(1) this shows that ƒ has a Dirichlet inverse if and only if ƒ(1) ≠ 0.

Read more about this topic:  Dirichlet Convolution

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