if μ⁢(n)=(-1)^{ω⁢(n)} then τ⁢(n)=2^{ω⁢(n)}

if

μ

⁢

(

n

)

=

(

-

1

)

ω

⁢

(

n

)

then

τ

⁢

(

n

)

=

2

ω

⁢

(

n

)

Corollary. If

n

is a

squarefree

positive integer

with a given number of

prime factors

, then the total count of its

divisors

is a

power of two

, specifically, two raised to the number of prime factors. With

μ

being the Möbius

function

,

ω

being the

number of distinct prime factors function

, and

τ

the

divisor function

, we can write that if

μ

⁢

(

n

)

=

(

-

1

)

ω

⁢

(

n

)

then

τ

⁢

(

n

)

=

2

ω

⁢

(

n

)

.

Since it is stipulated that

n

is squarefree, the number of (nondistinct) prime factors function (

http://planetmath.org/NumberOfNondistinctPrimeFactorsFunction

)

Ω

can be used instead of

ω

.

Note that for the sake of simplicity in this particular application, 1 is not considered a

prime number

.

Proof.

n

has

ω

⁢

(

n

)

prime factors, which here are labelled

p

1

,

p

2

,

…

,

p

ω

⁢

(

n

)

(they don’t necessarily have to be sorted, but it doesn’t hurt). This accounts for the

prime divisors

of

n

, therefore

τ

⁢

(

n

)

must be at least

ω

⁢

(

n

)

. The

semiprime

divisors of

n

are

p

1

⁢

p

2

,

p

1

⁢

p

3

,

…

,

p

1

⁢

p

ω

⁢

(

n

)

,

p

2

⁢

p

3

,

…

,

p

2

⁢

p

ω

⁢

(

n

)

,

…

Basically, the number of ways to choose two prime factors from among

ω

⁢

(

n

)

choices, which is the

binomial coefficient

(

ω

⁢

(

n

)

2

)

.

Likewise, the first determination, though quite obvious, was

(

ω

⁢

(

n

)

1

)

=

ω

⁢

(

n

)

.

The binomial coefficients can be used to determine the count of divisors with 3 prime factors, 4 prime factors, etc., all the way up to

ω

⁢

(

n

)

factors, which is again obvious:

(

ω

⁢

(

n

)

ω

⁢

(

n

)

)

=

1

.

This should be familiar, as these values can be looked up in Pascal’s

triangle

.

So far we’ve only looked at the divisors of

n

which are the

product

of one or more prime numbers. But every positive integer has 1 as a divisor, and since 1 has zero prime factors, we can also use Pascal’s triangle to count 1 as a divisor thus: the number of ways to choose zero prime factors from among

ω

⁢

(

n

)

choices, and that’s the 1 in the leftmost column of row

ω

⁢

(

n

)

in Pascal’s triangle. So to count the number of divisors of a squarefree number

n

with

ω

⁢

(

n

)

prime factors, it is a simple matter of adding up row

ω

⁢

(

n

)

of Pascal’s triangle:

∑

i

=

0

ω

⁢

(

n

)

(

ω

⁢

(

n

)

i

)

=

∑

i

=

0

ω

⁢

(

n

)

(

ω

⁢

(

n

)

i

)

⁢

1

ω

⁢

(

n

)

-

i

⁢

1

i

=

(

1

+

1

)

ω

⁢

(

n

)

=

2

ω

⁢

(

n

)

,

exactly as the

theorem

stated.
∎

The

consequences

of the theorem are that the number 1 has one divisor: 1; a prime

p

1

has two divisors:

1

,

p

1

; a semiprime has four divisors:

1

,

p

1

,

p

2

,

p

1

⁢

p

2

; a

sphenic number

has eight divisors:

1

,

p

1

,

p

2

,

p

3

,

p

1

⁢

p

2

,

p

2

⁢

p

3

,

p

1

⁢

p

2

⁢

p

3

; a number with four prime factors has sixteen divisors; and so on and so forth.

Title

if

μ

⁢

(

n

)

=

(

-

1

)

ω

⁢

(

n

)

then

τ

⁢

(

n

)

=

2

ω

⁢

(

n

)

Canonical name

Ifmun1omeganThentaun2omegan

Date of creation

2013-03-22 18:18:41

Last modified on

2013-03-22 18:18:41

Owner

1and2and4 (20899)

Last modified by

1and2and4 (20899)

Numerical id

5

Author

1and2and4 (20899)

Entry type

Corollary

Classification

msc 11A25

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