1 Introduction

# Sums involving the number of distinct prime factors function

Sums involving the number of distinct prime factors function \ArticleNameOn sums involving the number of distinct prime factors function

\Author

Tanay V. Wakhare

T. Wakhare

University of Maryland, College Park MD 20742 \Address Center for Nanoscale Science and Technology, National Institute of Standards and Technology, Gaithersburg MD 20899 \EmailDtwakhare@gmail.com

\ArticleDates

Received XX January 2017 in final form ????; Published online ????

\Abstract

The main object of this paper is to find closed form expressions for finite and infinite sums that are weighted by , where is the number of distinct prime factors of . We then derive general convergence criteria for these series. The approach of this paper is to use the theory of symmetric functions to derive identities for the elementary symmetric functions, then apply these identities to arbitrary primes and values of multiplicative functions evaluated at primes. This allows us to reinterpret sums over arbitrary complex numbers as divisor sums and sums over the natural numbers.

\Keywords

Symmetric Polynomials; Analytic Number Theory; Distinct Prime Factors; Dirichlet Series; Divisor Sums; Multiplicative Number Theory

\Classification

11C08;11M06;11N99;11Z05

## 1 Introduction

The number of distinct prime factors of a natural number , denoted by , has been the subject of extensive study, from Hardy and Ramanujan [1] to Erdős and Kac [2]. Letting , we have . For example, , , and . Throughout, we assume that . The function is also additive, so for any coprime and .

In this work, we address two major types of identities: divisor sums and Dirichlet series. Many of the closed form expressions available here do not seem to be present anywhere. This is because the easiest functions to work with are multiplicative, while is additive. Identities mixing both types of functions are not easy to derive using standard methods. In this paper, we use the theory of symmetric functions to find an alternative proof for many series expansions. The methods readily generalize, and can be applied to other additive functions. Our two main results are that

 ∑d|nμ(d)ω(d)f(d)=⎛⎝∑d|nμ(d)f(d)⎞⎠⎛⎝∑p|nf(p)f(p)−1⎞⎠ (1)

and

 ∑n∈Nω(n)f(n)ns=(∑n∈Nf(n)ns)(∑pap1+ap), (2)

with suitable restrictions on and , and .

For convenience, we note here that a multiplicative function satisfies for , where gives the greatest common divisor of and . A completely multiplicative function satisfies for any and . Both satisfy . An additive function satisfies for , while a completely additive function satisfies for any and .

Now, we address some notation. We let and denote sums and products over all primes , beginning with . We let and denote sums and products over the distinct primes that divide a positive integer . Finally, we let and denote sums and products over the positive divisors of a positive integer , including and . For example if , and . For the duration of this paper, the symbol will denote the set of positive integers and will be referred to as the natural numbers. We obey the convention that for , with . We also define .

We also utilize the theory of Euler products. An Euler product, described in [3, (27.4.1-2)], is the product form of a Dirichlet series. For any multiplicative function we have [5, (11.8)] , where

 ap:=∞∑m=1f(pm)pms. (3)

The left-hand side is an Dirichlet series, and the right-hand is an Euler product. Furthermore, letting , the abscissa of absolute convergence is the unique real number such that absolutely converges if and only if .

Some multiplicative functions which will be used in this paper include , which denotes the Euler totient function. This counts the number of natural numbers less than or equal to which are coprime to [3, (27.2.7)]. A multiplicative expression for it [3, (27.3.3)] is

 ϕ(n)=n∏p|n(1−1p). (4)

This is generalized by Jordan’s totient function, which is defined in [3, (27.3.4)] as

 Jk(n):=nk∏p|n(1−1pk). (5)

Using , we can also define the Möbius function, which gives the parity of the number of prime factors in a squarefree number. It is defined in [3, (27.2.12)] as

 μ(n):=⎧⎨⎩1,if n=1,0,if n is non-squarefree,(−1)ω(n),if n is squarefree. (6)

The paper is organized as follows. In Section 2, we derive identities for the elemtary symmetric functions. In Section 3, divisor sums of weighted by other functions are explored. In Section 4, Dirichlet series of weighted by other functions are explored. Finally, in Section 5 we present an extension to higher orders, and lay out other possible extensions of this work.

## 2 Factorization identities

The main results of this paper are based on the following proposition.

###### Proposition \thelemma.

Let and . Then

 (n∏i=1(1−xi))(n∑i=1xixi−1)=n∑k=1(−1)kkek, (7)

where

 ek:=∑1⩽i1
###### Proof.

Following Macdonald [9, (Chapter 1)], we have the relation

 E(t)=n∏i=1(1−txi)=n∑k=0(−1)ktkek, (9)

with . Taking a logarithmic derivative of yields

 E′(t)=E(t)n∑k=1xitxi−1. (10)

Taking completes the proof. ∎

###### Corollary \thelemma.

Let and . Then

 (n∏i=1(1+xi))(n∑i=1xixi+1)=n∑k=1kek. (11)
###### Proof.

We repeat the previous argument with but map to before taking the logarithmic derivative. ∎

## 3 Divisor sums

The following theorems are obtained by reinterpreting (7). We consider divisor sums of a multiplicative function weighted by and or . Throughout, will refer to the value of evaluated at any prime . We define as if , where is any function, not necessarily multiplicative, because has no distinct prime factors. Under this convention, the theorems in this section also hold for .

###### Theorem \thelemma.

Let be a multiplicative function, , and for any prime that divides . Then

 ∑d|nμ(d)ω(d)f(d)=⎛⎝∏p|n(1−f(p))⎞⎠⎛⎝∑p|nf(p)f(p)−1⎞⎠. (12)
###### Proof.

First take any squarefree natural number , so that by the fundamental theorem of arithmetic. We then let in (7), such that each is an arithmetic function evaluated at each distinct prime that divides . We can evaluate , yielding

 ek=∑1⩽i1

Since is multiplicative and each is coprime to the others by definition, we have

 ek=∑1⩽i1

Now we can regard as the sum of evaluated at the divisors of with prime factors. This is because each term in trivially has prime factors, and every possible product of primes that divide is included in . This is equivalent to partitioning the divisors of based on their number of distinct prime factors. Then (7) transforms into

 ⎛⎝ω(n)∏i=1(1−f(pi))⎞⎠⎛⎝ω(n)∑i=1f(pi)f(pi)−1⎞⎠=∑d|nω(d)(−1)ω(d)f(d).

Each divisor is squarefree since is squarefree, so we can replace by , where is the Möbius function which is defined by . Rewriting the product and sum over , , as a product and sum over gives (12) for squarefree numbers. However, we can immediately see that if is non-squarefree, eliminates any non-squarefree divisors on the left-hand side. Meanwhile, the right-hand side is evaluated over the distinct primes that divide so changing the multiplicities of these primes will not affect the sum in any way. Therefore (12) is valid for all , which completes the proof. ∎

###### Corollary \thelemma.

Let be a multiplicative function, , and for any prime that divides . Then

 ∑d|nμ(d)ω(d)f(d)=⎛⎝∑d|nμ(d)f(d)⎞⎠⎛⎝∑p|nf(p)f(p)−1⎞⎠. (13)
###### Proof.

A general theorem for any multiplicative function , found in [5, (2.18)], is that

 ∑d|nμ(d)f(d)=∏p|n(1−f(p)). (14)

Substituting this relation into (12) completes the proof. ∎

###### Theorem \thelemma.

Let be a multiplicative function, , and for any prime that divides . Then

 ∑d|n|μ(d)|ω(d)f(d)=⎛⎝∏p|n(1+f(p))⎞⎠⎛⎝∑p|nf(p)1+f(p)⎞⎠. (15)
###### Proof.

We follow the reasoning of the proof of (12) but substitute into (11) instead of (7). Hence, we only have to take the divisor sum over squarefree divisors without multiplying by . We do this by multiplying the divisor sums by , the characteristic function of the squarefree numbers. This completes the proof. ∎

###### Corollary \thelemma.

Let be a multiplicative function with , , and . Then

 ∑d|n|μ(d)|ω(d)f(d)=⎛⎝∑d|n|μ(d)|f(d)⎞⎠⎛⎝∑p|nf(p)1+f(p)⎞⎠. (16)
###### Proof.

We substitute into (14), where is multiplicative, ensuring that is also multiplicative. Noting that since the Möbius function only takes values of and yields

 ∑d|n|μ(d)|g(d)=∏p|n(1+g(p)).

Mapping to to maintain consistent notation and substituting into (15) completes the proof. ∎

Specializing yields a variety of new formulae involving convolutions with . Below we use a variety of functions along with (12), (13), and (15) to find new expressions for divisor sums involving .

###### Theorem \thelemma.

Let . Then

 ∑d|n|μ(d)|ω(d)=ω(n)2ω(n)−1. (17)
###### Proof.

Substituting into (15) gives Since the product and sum on the right-hand side are over the distinct primes that divide , each is evaluated times. This simplifies to

###### Theorem \thelemma.

Let . Then

 ∑d|nμ(d)ω(d)(nd)k=Jk(n)∑p|n11−pk. (18)
###### Proof.

Substituting into (12) gives

 ∑d|nμ(d)ω(d)dk=⎛⎝∏p|n(1−1pk)⎞⎠⎛⎝∑p|n11−pk⎞⎠. (19)

Using (5) to see that and substituting this relation into (19) completes the proof. ∎

###### Theorem \thelemma.

Let . Then

 ∑d|n|μ(d)|ω(d)(nd)k=J2k(n)Jk(n)∑p|n11+pk. (20)
###### Proof.

Substituting into (15) gives

 ∑d|n|μ(d)|ω(d)dk=⎛⎝∏p|n(1+1pk)⎞⎠⎛⎝∑p|n11+pk⎞⎠.

Now we complete the proof by using (5) to see that

 ∏p|n(1+1pk)=n2kn2k∏p|n(1−1p2k)∏p|n(1−1pk)=1nkJ2k(n)Jk(n).

###### Corollary \thelemma.

Let . Then

 ∑d|n|μ(d)|ω(d)(nd)=ψ(n)∑p|n11+p. (21)
###### Proof.

Substituting into (20) and noting that completes the proof. ∎

###### Theorem \thelemma.

Let with squarefree. Then

 ∑d|nω(d)dk=J2k(n)Jk(n)∑p|npk1+pk. (22)
###### Proof.

Substituting into (15) gives

 ∑d|n|μ(d)|ω(d)dk=∏p|n(1+pk)∑p|npk1+pk. (23)

Now if we let be squarefree, then . We also know that (5) transforms into

 Jk(n)=nk∏p|n(1−1pk)=nk∏p|n(pk−1)∏p|npk.

However we now have , so . Then

 J2k(n)Jk(n)=∏p|n(p2k−1)∏p|n(pk−1)=∏p|n(p2k−1)(pk−1)=∏p|n(pk+1). (24)

Letting be a squarefree natural number in (15), we can also eliminate from the sum on the left-hand side since every divisor of will already be squarefree. Substituting (24) into (23) completes the proof. ∎

## 4 Infinite sums

The propositions (7) and (11) hold for a finite number of elements . However, we can take the limit to extend this sum. This enables us to find closed form product expressions for Dirichlet series, which are described in [3, (27.4.4)], of the form for many commonly encountered multiplicative functions .

For later convenience we introduce the prime zeta function, described in Fröberg (1968) [6, (0.1)], denoted by . We define it by

 P(s):=∑p1ps,

and note that it converges for . It is an analog of the Riemann zeta function, described in [3, (25.2.1)], with the sum taken over prime numbers instead of all natural numbers. For notational convenience we also define the shifted prime zeta function as

 P(s,a):=∑p1ps+a,

such that .

{lemma}

Let and . Then P(s,a) converges absolutely if and only if , .

###### Proof.

The result follows from using a direct comparison test with to prove the divergence of , then taking an absolute value to bound it above and prove absolute convergence for . ∎

{lemma}

Let with . Then converges absolutely if and only if .

###### Proof.

The result follows from taking an absolute value and multiplying the top and bottom by , then applying Lemma 4, whether or not. ∎

###### Theorem \thelemma.

Let be a multiplicative function, , , and for any prime . If and both converge absolutely for , then

 ∑n∈Nω(n)f(n)ns=(∑n∈Nf(n)ns)(∑pap1+ap), (25)

which converges absolutely for .

###### Proof.

We let denote the th prime number. We must choose a suitable to substitute into (11), so we let , . We still retain the condition to avoid dividing by . Substituting this into (11) gives

 ∞∑k=1kek=∏p(1+ap)∑pap1+ap.

The product and sum on the right now go through every prime since each is in a one-to-one correspondence with a sum over the th prime. We also have that the product over primes is the Euler product for the Dirichlet series . We now prove that sums over every natural number that has distinct prime factors.

{lemma}

Let , , denote the th prime, denote any multiplicative function, and . Furthermore, let and

 Sk:={n∈N:ω(n)=k}, (26)

so that is the set of natural numbers with distinct prime factors. If has an abscissa of absolute convergence , then

 ek:=∑1⩽i1

which converges absolutely for .

###### Proof.

Let , , and , . We can directly evaluate as

 ek=∑1⩽i1

Here varies because it goes over every single power of which is present in . Since is a subseries of , it will also converge absolutely for and any rearrangement of its terms does not change the value of the sum. Since is multiplicative and each is coprime to the others by definition, we have

 ek=∑1⩽i1

If we take an arbitrary natural number with distinct prime factors, it will be present in the sum with the th symmetric function, . The th symmetric function contains every natural number with prime factors, since dictates the number of terms that are multiplied together to form every term in . The multiplicity also doesn’t matter, since that varies with which is independent of .

We also know that by the fundamental theorem of arithmetic, there is a bijection between the natural numbers and the products of distinct primes with any multiplicity. This means that every product of distinct primes in the expression for corresponds to a natural number . Taking it all together, it follows that goes over every natural number with distinct prime factors. Rewriting each product of primes as then gives equation (27). Since is a subseries of , it will also converge absolutely for . ∎

We then have , where is defined by (26). This means that as goes from to the sum of each from the left-hand side can be interpreted to go over every natural number except because they have been partitioned based on how many distinct prime factors they have. The series fails to sum over , which does not have any prime factors, but so this does not affect the sum in any way.

We can also see is the weight that’s represented by since we can bring it inside the inner sum as and rewrite the double sum as a sum over the natural numbers. We can also change the bottom limit from to since the term is . This gives

 ∞∑k=1k∑Skf(n)ns=∞∑k=0∑Skω(n)f(n)ns=∑n∈Nω(n)f(n)ns.

The inner sum converges absolutely for , but the sum over does not converge on this half plane in general. This rearrangement is valid if converges absolutely. However, as a Dirichlet series it is guaranteed to have an abscissa of absolute convergence and therefore the rearrangement is valid for some .

Simplifying (11) finally shows that

 ∑n∈Nω(n)f(n)ns=(∑n∈Nf(n)ns)(∑pap1+ap).

The left-hand side has the same convergence criteria as the right-hand side. Therefore if has an abscissa of absolutely convergence and converges absolutely for some , the left-hand side will converge absolutely for . This shows that weighting the terms of the Dirichlet series of any multiplicative function by multiplies the original series by a sum of over primes. ∎

###### Theorem \thelemma.

Let be a completely multiplicative function, , and . If has an abscissa of absolute convergence , then

 ∑n∈Nω(n)f(n)ns=(∑n∈Nf(n)ns)(∑pf(p)ps), (28)

which converges absolutely for .

###### Proof.

If , then converges absolutely and therefore . Then

 xi=api=∞∑m=1f(pmi)pims=∞∑m=1f(pi)mpims=11−f(pi)pis−1.

Replacing by in (25) and simplifying completes the proof. We also note that is a subseries of , so it will also converge absolutely for and we can simplify our convergence criterion. ∎

###### Theorem \thelemma.

Let be a multiplicative function. Let and both converge absolutely for . Assume that for all prime and such that we have . Then

 ∑n∈N|μ(n)|ω(n)f(n)ns=(∑n∈N|μ(n)|f(n)ns)(∑pf(p)ps+f(p)), (29)

which converges absolutely for .

###### Proof.

To begin, we note that a product of multiplicative functions is also multiplicative. Letting in (25), where is any multiplicative function which guarantees that is multiplicative, we can simplify . We have that is for and for , since is the characteristic function of the squarefree integers. If , we also have that , which means that . We still retain the condition. Assuming that the sum over primes converges, substituting into (25) gives

 ∑n∈N|μ(n)|ω(n)g(n)ns=(∑n∈N|μ(n)|g(n)ns)⎛⎜ ⎜⎝∑pg(p)ps1+g(p)ps⎞⎟ ⎟⎠.

We utilize the same convergence criterion as Theorem 4. Simplifying the fraction and letting be represented by in order to maintain consistent notation completes the proof. ∎

###### Theorem \thelemma.

Let be a multiplicative function. Let and both converge absolutely for . Assume that for all prime and such that we have . Then

 ∑n∈Nμ(n)ω(n)f(n)ns=(∑n∈Nμ(n)f(n)ns)(∑pf(p)f(p)−ps), (30)

which converges absolutely for .

###### Proof.

We let in (25), where is any multiplicative function which guarantees that is multiplicative. We can then simplify , since . We still retain the condition. Assuming that the sum over primes converges, substituting into (25) gives

 ∑n∈Nμ(n)ω(n)g(n)ns=(∑n∈Nμ(n)g(n)ns)⎛⎜ ⎜⎝∑p−g(p)ps1−g(p)ps⎞⎟ ⎟⎠.

We utilize the same convergence criterion as Theorem 4. Simplifying the fraction and letting be represented by in order to maintain consistent notation completes the proof. ∎

We note that the following proposition, the simplest application of (25), can be found in [8, (D-17)].

###### Theorem \thelemma.

Let be the Riemann zeta function. For such that ,

 ∑n∈Nω(n)ns=ζ(s)P(s). (31)
###### Proof.

We let in (28), since this is a completely multiplicative function. We then note that is the prime zeta function. The zeta and prime zeta functions both converge absolutely for , so the left-hand side will too. ∎

###### Theorem \thelemma.

Let be Liouville’s function. For such that ,

 ∑n∈Nω(n)λ(n)ns=−ζ(2s)ζ(s)P(s). (32)
###### Proof.

We let in (28). Liouville’s function, found in [3, (27.2.13)], is completely multiplicative. We note that for every prime , since they trivially only have a single prime divisor with multiplicity . This gives

 ∑n∈Nω(n)λ(n)ns=(∑n∈Nλ(n)ns)(∑p(−1)ps).

We then note that [3, (27.4.7)] gives and states that it converges for , which completes the proof since both the zeta and prime zeta functions converge for . ∎

###### Theorem \thelemma.

Let denote a Dirichlet character. For such that ,

 ∑n∈Nω(n)χ(n)ns=L(s,χ)∑pχ(p)ps. (33)
###### Proof.

We let in (28). Here is a Dirichlet character, found in [3, (27.8.1)], which is a completely multiplicative function that is periodic with period and vanishes for . Substituting it in (28) gives

We then note that a Dirichlet -series, an important number theoretic series, is defined in [3, (25.15.1)] as Simplifying to write the sum over as an -series while noting that [3, (25.15.1)] states that an -series converges absolutely for completes the proof, since both the series and prime series converge absolutely for . ∎

###### Theorem \thelemma.

For such that ,

 ∑n∈N|μ(n)|ω(n)ns=ζ(s)ζ(2s)P(s,1). (34)
###### Proof.

We let in (29) and note that [3, (27.4.8)] states that and that this converges for . We note that (4) states that will also converge absolutely for , which completes the proof. ∎

###### Theorem \thelemma.

For such that ,

 ∑n∈Nμ(n)ω(n)ns=−1ζ(s)P(s,−1). (35)
###### Proof.

We let in (30) and note that [3, (27.4.5)] states and that it converges for . We note that (4) states that will also converge absolutely for , which completes the proof. ∎

While the previous sums have involved completely multiplicative functions or convolutions with the Möbius function, we can sometimes directly evaluate . Taking (25) but converting back to its Euler product means that

 ∑n∈Nω(n)f(n)ns=∏p(1+ap)∑p(ap1+ap),

where