Multiplicative Functions
We consider here arithmetic functions which are defined on the natural numbers
1, 2, 3, 4, · · · .
The divisor function. We define d(n) to be the number of positive divisors of n.
Thus for example d(24) = 8 since the divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Also
d(1) = 1, d(2) = 2, d(3) = 2, d(4) = 3.
Practice Problems
1. Compute d(n) for 5 ≤ n ≤ 25.
2. Show d(n) = 2 if and only if n is a prime.
3. What can you say about n if d(n) = 3?
4. What can you say about n if d(n) = 4?
5. If n = p1 p2 p3 · · · pk , where all the pi ’s are distinct from each other, find a formula
for d(n). (Such n are called squarefree.)
6. In general if n = p1 a1 p2 a2 p3 a3 · · · pk ak , find a formula for d(n).
7. Compute d(100), d(1000), d(1, 000, 000), d(1099 ).
Multiplicative functions An arithmetic function f is multiplicative if
f (mn) = f (m)f (n),
where (m, n) = 1;
here (m, n) means the greatest common divisor (the gcd) of m and n. If (m, n) = 1
we say m and n are relatively prime.
Practice Problems
1. Show a multiplicative function must have f (1) = 1 or else be identically zero:
f (n) = 0 for all n.
2. Show d(n) is multiplicative.
3. Show σ(n) the sum of the divisor of n is multiplicative.
4. Show ω(n) the number of distinct prime factors of n is multiplicative.
5. Show δ(n) defined
to be 1 if n = 1 and zero otherwise is multiplicative.
X
6. Let g(n) =
f (d), where f (n) is multiplicative. Show g(n) is multiplicative
d|n
The Möbius function We define the Möbius function µ(n) by: µ(1) = 1, µ(n) =
(−1)k if n = p1 p2 p3 · · · pk where the pi ’s are distinct (n is squarefree), and µ(n) = 0
if n is not squarefree. Thus µ(3) = −1, µ(4) = 0, and µ(6) = 1.
Practice Problems
1. Compute µ(n) for 1 ≤ n ≤ 25.
2. Show µ(n)
X is multiplicative.
3. Show
µ(d) = δ(n), where δ(n) is the function above where δ(1) = 1 and
d|n
δ(n) = 0 if n 6= 1.
1
4. Show
X
k|n
n
µ(k)d( ) = 1
k
Dirichlet Convolution We define
f ∗ g(n) =
X
d|n
n
f (d)g( ),
d
and call this the Dirichlet convolution of f and g.
Practice Problems
1. Show f ∗ g = g ∗ f . (It is also true that (f ∗ g) ∗ h = f ∗ (g ∗ h) and f ∗ (g + h)) =
f ∗ g + f ∗ h)
2. Show d(n) = 1(n) ∗ 1(n) (or d(n) = 1 ∗ 1), where 1(n) = 1 for all n.
3. Show δ = 1 ∗ µ
4. Show f = δ ∗ f
5. Show if g = 1 ∗ f then f = µ ∗ g. This is called the Möbius inversion formula.