Rose-Hulman Undergraduate Mathematics Journal
Volume 19 Issue 1 Article 8
Sums Involving the Number of Distinct Prime Factors Function
Tanay Wakhare University of Maryland, College Park, [email protected]
Follow this and additional works at: https://scholar.rose-hulman.edu/rhumj
Part of the Number Theory Commons
Recommended Citation Wakhare, Tanay (2018) "Sums Involving the Number of Distinct Prime Factors Function," Rose-Hulman Undergraduate Mathematics Journal: Vol. 19 : Iss. 1 , Article 8. Available at: https://scholar.rose-hulman.edu/rhumj/vol19/iss1/8 Sums Involving the Number of Distinct Prime Factors Function
Cover Page Footnote I wrote this when I was 17, and looking back now I'm still amazed. This was my first ve er paper, and helped launch my research career -- and I couldn't have done it were it not for literally dozens of people around me. Many thanks go to my research supervisors at the National Institute of Standards and Technology: Guru Khalsa, Mark Stiles, Kyoung-Whan Kim, and Vivek Amin. I am also indebted to Howard Cohl, who took a frankly horrifying manuscript and turned it into a publishable paper. Without him this project wouldn't have gotten off the ground. Bruce Berndt, Krishna Alladi, and the anonymous referee also provided valuable feedback. Finally, many thanks go out to my old high school teachers -- William Wuu, Joseph Boettcher, Colleen Adams, Joshua Schuman, Jamie Andrews, and others -- who miraculously put up with me through the years.
This article is available in Rose-Hulman Undergraduate Mathematics Journal: https://scholar.rose-hulman.edu/rhumj/ vol19/iss1/8 Rose- Hulman Undergraduate Mathematics Journal
Sums Involving the Number of Distinct Prime Factors Function
Tanay Wakhare a b
Volume 19, No. 1, Fall, 2018
Sponsored by
Rose-Hulman Institute of Technology Department of Mathematics Terre Haute, IN 47803 aQuince Orchard High School, Gaithersburg MD 20878 b [email protected] Center for Nanoscale Science and Technology, National Institute of scholar.rose-hulman.edu/rhumj Standards and Technology, Gaithersburg MD 20899 Rose-Hulman Undergraduate Mathematics Journal Volume 19, No. 1, Fall, 2018
Sums Involving the Number of Distinct Prime Factors Function
Abstract. We nd closed form expressions for nite and in nite sums that are weighted by ω(n), where ω(n) is the number of distinct prime factors of n. 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 symmetric polynomials as divisor sums and sums over the natural numbers. Page 2 RHIT Undergrad. Math. J., Vol. 19, No. 1
1 Introduction
The number of distinct prime factors of a natural number, ω(n), is a fundamental arithmetic quantity, and has been the subject of extensive study from Hardy and Ramanujan [5] to Erd®s and Kac [2]. We seek to expand the list of known series identities involving ω(n), with potential applications to the asymptotics of ω(n) and its higher moments ωk(n). The prototype for our results is the Dirichlet series
X ω(n) = ζ(s)P (s), ns n∈N where is the Riemann zeta function and is the prime zeta function, P 1 . ζ(s) P (s) P (s) := p ps Results such as this allow us to study the asymptotic behavior of the partial sums P . n≤x ω(n) However, sometimes it is simpler to study smoothed asymptotics P , for some n≤x ω(n)f(n) suitable arithmetic function f. Therefore, in this work we compute closed forms for Dirichlet series involving ω(n)f(n), for any arbitary multiplicative weights f, and explicitly calculate many special cases. In the last section, we also lay out an extension for higher powers ω(n)kf(n), and explicitly calculate some examples for k = 2. We also describe nite versions of the above identities, which involve divisor sums over ω(n)f(n) for arbitary multiplicative f. This again helps expand our knowledge of the ω function, and the many of the special cases available here do not seem to be present anywhere else in the literature, perhaps because the easiest arithmetic functions to study are multiplicative, while ω(n) is additive. Identities mixing both additive and multiplicative functions are not readily available using standard methods, so we use nonstandard symmetric function expansions to study sums involving both. The methods readily generalize, and can be applied to other additive functions. The paper is organized as follows. After standardizing some notation, we derive identities for the elemtary symmetric functions in Section 3. In Section 4, we explore divisor sums of ω(n) weighted by other functions as corollaries of our results on symmetric polynomials. In Section 5, Dirichlet series of ω(n) weighted by other functions are explored. We also compile an extensive list of corollaries, which may prove handy when considering exotic divisor sums. Finally, in Section 6 we present an extension to higher orders, and present a short summary in Section 7.
2 Background and notation
We recall some de nitions and background material which will be essential in later parts of this paper. Letting Q αi denote the prime factorization of , we have . n = 1≤i≤k pi n ω(n) = k For example, ω(5) = 1, ω(10) = 2, and ω(100) = ω(2252) = 2. Recall that a multiplicative function satis es f(nm) = f(n)f(m) for (n, m) = 1, where (n, m) gives the greatest common divisor of n and m. A completely multiplicative function satis es f(nm) = f(n)f(m) for any n and m. We assume that f(1) = 1. An additive function satis es f(nm) = f(n)+f(m) RHIT Undergrad. Math. J., Vol. 19, No. 1 Page 3
for (n, m) = 1, while a completely additive function satis es f(nm) = f(n) + f(m) for any n and m. Now, we address some notation. We let P and Q denote sums and products over p ap p ap all primes , beginning with . We let P and Q denote sums and products p p = 2 p|n ap p|n ap over the distinct primes that divide a positive integer . Finally, we let P and Q n d|n ad d|n ad denote sums and products over the positive divisors of a positive integer n, including 1 and . For example, if , Q and Q . For the duration n n = 12 d|12 ad = a1a2a3a4a6a12 p|12 ap = a2a3 of this paper, the symbol N := {1, 2, 3 ...} will denote the set of positive integers and will be referred to as the natural numbers. We obey the convention that for s ∈ C, s = σ + it with . We nally de ne . σ, t ∈ R Cα := {z ∈ C : z 6= α} We also utilize the theory of Euler products. An Euler product, described in [8, (27.4.1- 2)], is the product form of a Dirichlet series. For any multiplicative function f we have [1, (11.8)] ∞ X f(n) Y X f(pm) = (1 + a (p)) , where a (p) := . (1) ns f,s f,s pms n∈N p m=1 The left-hand side of the previous equation is a Dirichlet series, and the right-hand is an Euler product. Furthermore, letting s = σ + it, the abscissa of absolute convergence is the unique real number such that P f(n) absolutely converges if and only if . n∈ ns σ > σa Some multiplicative functions whichN will be used in this paper include denotes the Euler totient function φ(n), which counts the number of natural numbers less than or equal to n which are coprime to n [8, (27.2.7)]. We will frequently use the multiplicative representation [8, (27.3.3)] Y 1 φ(n) = n 1 − . (2) p p|n A standard generalization of φ is Jordan's totient function [8, (27.3.4)], given by
Y 1 J (n) := nk 1 − . (3) k pk p|n
Using ω(n), we can also de ne the Möbius function, which gives the parity of the number of prime factors in a squarefree number, and is formally given by [8, (27.2.12)]
1, if n=1, µ(n) := 0, if n is non-squarefree, (4) (−1)ω(n), if n is squarefree. 3 Factorization identities
The main results of this paper are based on the following standard symmetric function identity. Page 4 RHIT Undergrad. Math. J., Vol. 19, No. 1
Proposition 3.1. Let and . Then n ∈ N x1, . . . , xn ∈ C1
n ! n ! n Y X xi X k (1 − xi) = (−1) kek, (5) xi − 1 i=1 i=1 k=1 where ek = ek(x1, . . . , xn) denotes the elementary symmetric polynomial de ned as
X (6) ek := xi1 xi2 ··· xik .
1≤i1 n n Y X k k E(t) = (1 − txi) = (−1) t ek, (7) i=1 k=0 with . Taking a logarithmic derivative of for indeterminate yields xi ∈ C E(t) t n X xi E0(t) = E(t) . (8) txi − 1 k=1 Taking t = 1 completes the proof. Corollary 3.2. Let and . Then n ∈ N x1, . . . , xn ∈ C−1 n ! n ! n Y X xi X (1 + xi) = kek. (9) xi + 1 i=1 i=1 k=1 Proof. We repeat the previous argument in the proof of Proposition 3.1 but map t to −t before taking the logarithmic derivative. Note that by comparing coe cients of tn we recover the Newton-Girard identities [6, (Chapter 1)], but these are in fact identities for the generating functions for various symmetric functions. 4 Divisor sums The following theorems are obtained by reinterpreting (5). We consider divisor sums of a multiplicative function f weighted by ω(n) and µ(n) or |µ(n)| = µ2(n). Throughout, will refer to the value of evaluated at any prime . We also de ne P as f(p) f p p|n g(p) 0 if n = 1, where g is any function, not necessarily multiplicative, because 1 has no distinct prime factors. Under this convention, the theorems in this section also hold for n = 1. RHIT Undergrad. Math. J., Vol. 19, No. 1 Page 5 Theorem 4.1. Let f(n) be a multiplicative function de ned such that f(p) 6= 1 for any prime p that divides n. Then X Y X f(p) µ(d)ω(d)f(d) = (1 − f(p)) . (10) f(p) − 1 d|n p|n p|n Proof. First take any squarefree natural number , so that Qω(n) by the fundamental n n = i=1 pi theorem of arithmetic. We then let xi = f(pi) in (5), such that each xi is an arithmetic function evaluated at each distinct prime that divides n. We can evaluate ek, yielding X ek = f(pi1 )f(pi2 ) ··· f(pik ). 1≤i1 Since f is multiplicative and each pi is coprime to the others by de nition, we have X ek = f(pi1 pi2 ··· pik ). 1≤i1 Now we can regard ek as the sum of f(n) evaluated at the divisors of n with k prime factors, since each term in ek trivially has k prime factors, and every possible product of k primes that divide n is included in ek. This is equivalent to partitioning the divisors of n based on their number of distinct prime factors. Then (5) transforms into ω(n) ω(n) Y X f(pi) X ω(d) (1 − f(pi)) = ω(d)(−1) f(d). f(pi) − 1 i=1 i=1 d|n Each divisor d is squarefree since n is squarefree, so we can replace (−1)ω(d) by µ(d), where µ(d) is the Möbius function which is de ned by (4). Rewriting the product and sum over pi, 1 ≤ i ≤ ω(n), as a product and sum over p gives (10) for squarefree numbers. However, we can immediately see that if n is non-squarefree, µ(d) eliminates any non-squarefree divisors on the left-hand side. Meanwhile, the right-hand side is evaluated over the distinct primes that divide n so changing the multiplicities of these primes will not aect the sum in any way. Therefore (10) is valid for all n ∈ N, which completes the proof. Corollary 4.2. Let f(n) be a multiplicative function, n ∈ N, and f(p) 6= 1 for any prime p that divides n. Then X X X f(p) µ(d)ω(d)f(d) = µ(d)f(d) . (11) f(p) − 1 d|n d|n p|n Proof. A general theorem for any multiplicative function f(n), found in [1, (2.18)], is that X Y µ(d)f(d) = (1 − f(p)) . (12) d|n p|n Substituting this relation into (10) completes the proof. Page 6 RHIT Undergrad. Math. J., Vol. 19, No. 1 Theorem 4.3. Let f(n) be a multiplicative function, n ∈ N, and f(p) 6= −1 for any prime p that divides n. Then X Y X f(p) |µ(d)|ω(d)f(d) = (1 + f(p)) . (13) 1 + f(p) d|n p|n p|n Proof. We follow the proof of Theorem 4.1 but substitute xi = f(pi) into (9) instead of (5). Hence, we only have to take the divisor sum over squarefree divisors without multiplying by (−1)ω(d). We do this by multiplying the divisor sums by |µ(d)|, the characteristic function of the squarefree numbers. This completes the proof. Corollary 4.4. Let f(n) be a multiplicative function with f(1) = 1, f(p) 6= −1, and n ∈ N. Then X X X f(p) |µ(d)|ω(d)f(d) = |µ(d)|f(d) . (14) 1 + f(p) d|n d|n p|n Proof. We substitute f(n) = µ(n)g(n) into (12), where g(n) is multiplicative, ensuring that f(n) is also multiplicative. Noting that µ2(n) = |µ(n)| since the Möbius function only takes values of ±1 and 0 yields X Y |µ(d)|g(d) = (1 + g(p)) . d|n p|n Mapping g to f to maintain consistent notation and substituting into (13) completes the proof. Specializing f(n) yields an assortment of new formulae involving convolutions with ω(n). Below we use a variety of functions f(n) along with (10), (11), and (13) to nd new expressions for divisor sums involving ω(n). Corollary 4.5. Let n ∈ N. Then X |µ(d)|ω(d) = ω(n)2ω(n)−1. (15) d|n Proof. Substituting into (13) gives P Q P 1 Since f(n) = 1 d|n |µ(d)|ω(d) = p|n 2 p|n 2 . the product and sum on the right-hand side are over the distinct primes that divide n, each is evaluated times. This simpli es to P ω(n) ω(n) . ω(n) d|n |µ(d)|ω(d) = 2 2 Corollary 4.6. Let n ∈ N. Then X nk X 1 µ(d)ω(d) = J (n) . (16) d k 1 − pk d|n p|n RHIT Undergrad. Math. J., Vol. 19, No. 1 Page 7 Proof. Substituting 1 into (10) gives f(n) = nk X µ(d)ω(d) Y 1 X 1 = 1 − . (17) dk pk 1 − pk d|n p|n p|n Using (3) to see that Q 1 Jk(n) and substituting this relation into (17) completes p|n(1 − pk ) = nk the proof. Corollary 4.7. Let n ∈ N. Then k X n J2k(n) X 1 (18) |µ(d)|ω(d) = k . d Jk(n) 1 + p d|n p|n Proof. Substituting 1 into (13) gives f(n) = nk X |µ(d)|ω(d) Y 1 X 1 = 1 + . dk pk 1 + pk d|n p|n p|n Now we complete the proof by using (3) to see that 2k Q 1 Y 1 n p|n(1 − p2k ) 1 J2k(n) 1 + = = . k 2k Q 1 k p n (1 − ) n Jk(n) p|n p|n pk Corollary 4.8. Let n ∈ N with n squarefree. Then J (n) pk X k 2k X (19) ω(d)d = k . Jk(n) 1 + p d|n p|n Proof. Substituting f(n) = nk into (13) gives X Y X pk |µ(d)|ω(d)dk = (1 + pk) . (20) 1 + pk d|n p|n p|n Now if we let be squarefree, then Q . We also know that (3) transforms into n n = p|n p Q k Y 1 p|n(p − 1) J (n) = nk 1 − = nk . k pk Q pk p|n p|n However we now have k Q k, so Q k . Then n = p|n p Jk(n) = p|n(p − 1) Q 2k 2k J2k(n) p|n(p − 1) Y (p − 1) Y = = = (pk + 1). (21) Q k k Jk(n) (p − 1) (p − 1) p|n p|n p|n Letting n be a squarefree natural number in (13), we can also eliminate |µ(d)| from the sum on the left-hand side since every divisor of n will already be squarefree. Substituting (21) into (20) completes the proof. Page 8 RHIT Undergrad. Math. J., Vol. 19, No. 1 5 In nite sums The propositions (5) and (9) hold for a nite number of elements xi. However, we can take the limit i → ∞ to extend this sum. This enables us to nd closed form product expressions for Dirichlet series, which are described in [8, (27.4.4)], of the form P ω(n)f(n) for many n∈N ns commonly encountered multiplicative functions f(n). We introduce the prime zeta function [3, (0.1)], denoted by P (s). We de ne it by X 1 P (s) := , ps p and note that it converges for <(s) > 1. It is an analog of the Riemann zeta function, described in [8, (25.2.1)], with the sum taken over prime numbers instead of all natural numbers. We also de ne the shifted prime zeta function P (s, a) as X 1 P (s, a) := , ps + a p such that P (s, 0) = P (s). Lemma 5.1. Let a ∈ C and |a| < 2. Then P(s,a) converges absolutely if and only if s ∈ C, <(s) > 1. Proof. Consider a ∈ C. Now if <(s) = 1, we prove that P (s, a) diverges if |a| < 2. This implies that P (s, a) diverges for all s such that <(s) < 1 for these values of a. To prove divergence, it su ces to let a ∈ R since R ⊂ C. We let s = 1, a ∈ R, and 2 > a ≥ 0. Peeling o the i = 1 term, where p1 = 2, we have pi+1 − pi ≥ 2 > a ≥ 0 since the minimum prime gap for primes not equal to 2 and 3 is 2. This means ∞ ∞ ∞ X 1 1 X 1 1 X 1 1 5 P (1, a) = = + ≥ + = + P (1) − . p + a 2 + a p + a 2 + a p 2 + a 6 i=1 i i=2 i i=2 i+1 However, [3] gives us that P (1) diverges. Now we let s = 1 and −2 < a ≤ 0. Then by the direct comparison test, 1 1 . This means is bounded below by and the sum p+a > p P (s, a) P (1) diverges. The restriction a > −2 is because there is a simple pole if s = 1 and a = −2. Now we let <(s) > 1, a ∈ C, and 0 ≤ |a| < 2. By the triangle inequality, |ps| = |ps + a − a| ≤ |ps + a| + | − a| = |ps + a| + |a| so then |ps + a| ≥ |ps| − |a| = p<(s) − |a| > 0, since p ∈ N and p ≥ 2. Then we can bound P (s, a) so that X 1 X 1 |P (s, a)| ≤ ≤ . ps + a p<(s) − |a| p p We have now reduced the consideration of a complex s to considering a real s. Letting for convenience, we have that σ σ . This allows us to form the <(s) = σ pi − pi−1 ≥ 2 > |a| ≥ 0 RHIT Undergrad. Math. J., Vol. 19, No. 1 Page 9 bound ∞ ∞ ∞ X 1 1 X 1 1 X 1 1 P (s, a) ≤ = + ≤ + = + P (σ). pσ − |a| 2σ − |a| pσ − |a| 2σ − |a| pσ 2σ − |a| i=1 i i=2 i i=2 i−1 We know that P (s) converges for <(s) = σ > 1, so P (s, a) will also converge for 0 ≤ |a| < 2, completing the proof. Lemma 5.2. Let with . Then P pk converges absolutely if and only if a, s, k ∈ C |a| < 2 p ps+a <(s) > max (1, 1 + <(k)). Proof. First we assume <(k) ≤ 0. Then k k X p X p X 1 ≤ ≤ , ps + a ps + a ps + a p p p which by (5.1) converges absolutely if and only if s > 1. Now we assume <(k) > 0. Then k X p X 1 ≤ . ps + a ps−k + ap−k p p Since a < 2 < 2, the conditions for (5.1) are satis ed and the series converges absolutely if pk 2 and only if <(s − k) > 1. Combining the two requirements for <(s) completes the proof. Our results on in nite series are based on a certain symmetric polynomial evaluation, which is a re nement of the Euler product/Dirichlet series duality ∞ X f(n) Y X f(pm) = (1 + a (p)) , where a (p) := . ns f,s f,s pms n∈N p m=1 In particular, given the evaluation of ek from Lemma 5.3, we can formally consider the generating product ∞ ∞ X k Y E(t) = ekt = (1 + xit), k≥0 i=1 set t = 1 and xi = af,s(pi), and rearrange the resulting series as ∞ ∞ Y X X f(n) X f(n) (1 + a (p )) = = . f,s i ns ns i=1 k=0 n≥1 n≥1 ω(n)=k This is precisely the de nition of an Euler product, though we have ignored questions of convergence in this comparison. The following lemma precisely encodes the correspondence above, and is the key element in the proof of Theorem 5.4. Page 10 RHIT Undergrad. Math. J., Vol. 19, No. 1 th Lemma 5.3. Let 1 ≤ i < ∞, i ∈ , pi denote the i prime, f denote any multiplicative m N function, and P∞ f(pi ) . Furthermore, let and xi = ms k ∈ m=1 pi N (22) Sk := {n ∈ N : ω(n) = k}, so that S is the set of natural numbers with k distinct prime factors. If P f(n) has an k n∈N ns abscissa of absolute convergence σa, then X X f(n) e := x x ··· x = , (23) k i1 i2 ik ns 1≤i1 Here i varies because it goes over every single power of p which is present in xi. Since xi is the sum of a subsequence of P f(n) , it will also converge absolutely for σ > σ and any n∈N ns a rearrangement of its terms will not change the value of the sum. Since f is multiplicative and each pi is coprime to the others by de nition, we have f p i1 p i2 ··· p ik X i1 i2 ik ek = s . p i1 p i2 ··· p ik 1≤i1 that every product of distinct primes in the expression for ek corresponds to a natural number n. Taking it all together, it follows that ek runs over every natural number n with k distinct prime factors. Rewriting each product of primes as n then gives equation (23). Since P f(n) is the sum over a subsequence of P f(n) , it will also converge absolutely s s n∈Sk n n∈N n for σ > σa. On a high level, the proof of the following theorem follows by substituting the symmetric polynomial evaluation of Lemma 5.3 into the symmetric polynomial identity (9). We can then rewrite the sum over k in (9) as a sum over all natural numbers, while recognizing the linear k weight as an occurrence of ω(n). However, there are several technical restrictions we need in order to ensure convergence of the resulting Dirichlet series. RHIT Undergrad. Math. J., Vol. 19, No. 1 Page 11 Theorem 5.4. Let be a multiplicative function, , P∞ f(pm) , and f(n) s ∈ C af,s(p) := m=1 pms for any prime . If P f(n) and P af,s(p) both converge absolutely for af,s(p) 6= −1 p s n∈N n p 1+af,s(p) σ > σa, then ! ! X ω(n)f(n) X f(n) X af,s(p) (24) s = s , σ > σa. n n 1 + af,s(p) n∈N n∈N p th Proof. We let pi denote the i prime number. We must choose a suitable xi to substitute into (9), so we let xi = af,s(pi), 1 ≤ i < ∞. We still retain the condition xi = af,s(pi) 6= −1. Substituting this specialization of xi into (9) gives ∞ X Y X af,s(p) kek = (1 + af,s(p)) . 1 + af,s(p) k=1 p p The product and sum on the right now run through every prime p since each xi is in a one-to-one correspondence with a sum over the ith prime. We also note that the product over primes is the Euler product for the Dirichlet series P f(n) . We can then appeal to n∈N ns the result of Lemma 5.3, but we now interpret k as the weight ω(n). We have the series rearrangement ∞ ∞ ∞ X X X f(n) X X f(n) X ω(n)f(n) ke = k = ω(n) = . k ns ns ns k=1 k=1 n∈Sk k=0 n∈Sk n∈N The most important fact about this proof is that k is constant on each set Sk by de nition; in fact, k is equal to the number of distinct prime factors of every integer in Sk. Therefore, we can move it inside the second sum and rewrite it as ω(n). The inner sum converges absolutely for σ > σa, but the sum over n does not converge on this half plane in general. This rearrangement is valid if P ω(n)f(n) converges absolutely. However, as a Dirichlet series it n∈ ns is guaranteed to have an abscissaN of absolute convergence and therefore the rearrangement is valid for some σ. Simplifying (9) nally shows that ! ! X ω(n)f(n) X f(n) X af,s(p) s = s . n n 1 + af,s(p) n∈N n∈N p The left-hand side has the same convergence criteria as the right-hand side. Therefore if P f(n) has an abscissa of absolutely convergence and P af,s(p) converges absolutely s σa n∈N n p 1+af,s(p) for some σ > σb, the left-hand side will converge absolutely for σ > max (σa, σb). This shows that weighting the terms of the Dirichlet series of any multiplicative function f(n) by ω(n) multiplies the original series by a sum of f over primes. Page 12 RHIT Undergrad. Math. J., Vol. 19, No. 1 Theorem 5.5. Let f(n) be a completely multiplicative function, s ∈ C, and n ∈ N. If P f(n) has an abscissa of absolute convergence σ , then n∈N ns a ! ! X ω(n)f(n) X f(n) X f(p) = , σ > σ . (25) ns ns ps a n∈N n∈N p f(pm) Proof. If , then P∞ i converges absolutely and therefore f(p) . Then σ > σa ms s < 1 m=1 pi p ∞ m ∞ m X f(pi ) X f(pi) 1 xi = ap = = = − 1. i ms ms f(pi) pi pi 1 − s m=1 m=1 pi −1 Replacing by f(p) in (24) and simplifying completes the proof. We also af,s(p) 1 − ps − 1 note that P f(p) is the sum over a subsequence of P f(n) , so it will also converge n∈N ps n∈N ns absolutely for σ > σa and we can simplify our convergence criterion. Theorem 5.6. Let be a multiplicative function. Let P |µ(n)|f(n) and P f(p) both f(n) n∈ ns p ps+f(p) converge absolutely for . Assume that for all prime andN such that we σ > σa p s ∈ C σ > σa have f(p) 6= −ps. Then ! ! X |µ(n)|ω(n)f(n) X |µ(n)|f(n) X f(p) = , (26) ns ns ps + f(p) n∈N n∈N p which converges absolutely for σ > σa. Proof. To begin, we note that a product of multiplicative functions is also multiplicative. Letting f(n) = |µ(n)|g(n) in (24), where g is any multiplicative function which guarantees m that f is multiplicative, we can simplify af,s(p). We have that |µ(p )| is 0 for m ≥ 2 and 1 for m = 1, since |µ(n)| is the characteristic function of the squarefree integers. If m = 1, we also have that , which means that P∞ f(pm) g(p) . We f(p) = |µ(p)|g(p) = g(p) af,s(p) = m=1 pms = ps still retain the g(p) condition. Assuming that the sum over primes converges, af,s(p) = ps 6= −1 substituting into (24) gives ! g(p) ! X |µ(n)|ω(n)g(n) X |µ(n)|g(n) X ps = . ns ns 1 + g(p) n∈N n∈N p ps We utilize the same convergence criterion as Theorem 5.4. Simplifying the fraction and letting g be represented by f in order to maintain consistent notation completes the proof. Theorem 5.7. Let be a multiplicative function. Let P µ(n)f(n) and P f(p) both f(n) n∈ ns p ps−f(p) converge absolutely for . Assume that for all prime andN such that we σ > σa p s ∈ C σ > σa have f(p) 6= ps. Then ! ! X µ(n)ω(n)f(n) X µ(n)f(n) X f(p) = , (27) ns ns f(p) − ps n∈N n∈N p which converges absolutely for σ > σa. RHIT Undergrad. Math. J., Vol. 19, No. 1 Page 13 Proof. We let f(n) = µ(n)g(n) in (24), where g is any multiplicative function which guarantees that is multiplicative. We can then simplify , since P∞ f(pm) g(p) . f af,s(p) af,s(p) := m=1 pms = − ps We still retain the g(p) condition. Assuming that the sum over primes af,s(p) = − ps 6= −1 converges, substituting into (24) gives ! g(p) ! X µ(n)ω(n)g(n) X µ(n)g(n) X − ps = . ns ns 1 − g(p) n∈N n∈N p ps We utilize the same convergence criterion as Theorem 5.4. Simplifying the fraction and letting g be represented by f in order to maintain consistent notation completes the proof. We note that the following proposition, the simplest application of (24), can be found in [4, (D-17)]. Corollary 5.8. Let ζ(s) be the Riemann zeta function. For s ∈ C such that σ > 1 , X ω(n) = ζ(s)P (s). (28) ns n∈N Proof. We let f(n) = 1 in (25), since this is a completely multiplicative function. We then note that P 1 is the prime zeta function. The zeta and prime zeta functions both converge p ps absolutely for σ > 1, so the left-hand side will too. Corollary 5.9. Let λ(s) be Liouville's function. For s ∈ C such that σ > 1, X ω(n)λ(n) ζ(2s) = − P (s). (29) ns ζ(s) n∈N Proof. We let f(n) = λ(n) in (25). Liouville's function, found in [8, (27.2.13)], is completely multiplicative. We note that λ(p) = −1 for every prime p, since they trivially only have a single prime divisor with multiplicity 1. This gives ! ! X ω(n)λ(n) X λ(n) X (−1) = . ns ns ps n∈N n∈N p We then note that [8, (27.4.7)] gives P λ(n) = ζ(2s) and states that it converges for n∈N ns ζ(s) σ > 1, which completes the proof since both the zeta and prime zeta functions converge for σ > 1. Corollary 5.10. Let χ(n) denote a Dirichlet character. For s ∈ C such that σ > 1 , X ω(n)χ(n) X χ(p) = L(s, χ) . (30) ns ps n∈N p Page 14 RHIT Undergrad. Math. J., Vol. 19, No. 1 Proof. We let f(n) = χ(n) in (25). Here χ(n) is a Dirichlet character, found in [8, (27.8.1)], which is a completely multiplicative function that is periodic with period k and vanishes for (n, k) > 1. Substituting it in (25) gives ! ! X ω(n)χ(n) X χ(n) X χ(p) = . ns ns ps n∈N n∈N p We then note that a Dirichlet L-series, an important number theoretic series, is de ned in [8, (25.15.1)] as L(s, χ) = P χ(n) . Simplifying to write the sum over n as an L-series while n∈N ns noting that [8, (25.15.1)] states that an L-series converges absolutely for σ > 1 completes the proof, since both the L series and prime L series converge absolutely for σ > 1. Corollary 5.11. For s ∈ C such that σ > 1 , X |µ(n)|ω(n) ζ(s) = P (s, 1). (31) ns ζ(2s) n∈N Proof. We let f(n) = 1 in (26) and note that [8, (27.4.8)] states that P |µ(n)| = ζ(s) n∈N ns ζ(2s) and that this converges for σ > 1. We note that (5.1) states that P (s, 1) will also converge absolutely for σ > 1, which completes the proof. Corollary 5.12. For s ∈ C such that σ > 1 , X µ(n)ω(n) 1 = − P (s, −1). (32) ns ζ(s) n∈N Proof. We let f(n) = 1 in (27) and note that [8, (27.4.5)] states P µ(n) = 1 and that n∈N ns ζ(s) it converges for σ > 1. We note that (5.1) states that P (s, −1) will also converge absolutely for σ > 1, which completes the proof. While the previous sums have involved completely multiplicative functions or convolutions with the Möbius function, we can sometimes directly evaluate P∞ f(pm) . Taking (24) but m=1 pms converting P f(n) back to its Euler product means that n∈N ns X ω(n)f(n) Y X af,s(p) s = (1 + af,s(p)) , n 1 + af,s(p) n∈N p p where P∞ f(pm) . We can extract the coe cient through a variety of af,s(p) := m=1 pms af,s(p) methods. Corollary 5.13. For s ∈ C such that σ > 1, X ω(n)2ω(n) ζ2(s) = 2 P (s, 1). (33) ns ζ(2s) n∈N RHIT Undergrad. Math. J., Vol. 19, No. 1 Page 15 2 Proof. We know from [8, (27.4.9)] that P 2ω(n) = ζ (s) and that it converges for <(s) > 1, n∈N ns ζ(2s) ω(n) where 2 is the number of squarefree divisors of n. We now directly evaluate af,s(p), summing it as a geometric series. We note that 2ω(pm) = 2, since pm trivially has a single distinct prime factor. Substituting into the formula for af,s(p) shows that ∞ m ∞ m X 2ω(p ) X 1 2 a (p) = = 2 = . f,s pms ps ps − 1 m=1 m=1 Then af,s(p) 2 . Substituting into (24) shows that = s 1+af,s(p) p +1 ! ! X ω(n)2ω(n) X 2ω(n) X 2 = . ns ns ps + 1 n∈N n∈N p Rewriting the right-hand side in terms of zeta and prime zeta functions while noting that they will both converge if σ > 1 completes the proof. Corollary 5.14. Let denote Jordan's totient function. For such that Jk(n) s, k ∈ C σ > max (1, 1 + <(k)), k X ω(n)Jk(n) ζ(s − k) X p − 1 = . (34) ns ζ(s) ps − 1 n∈N p k −1 Proof. Taking p −1 , we have 1 1 and af,s(p) af,s(p) = s k (1 + af,s(p)) = 1 − s 1 − s−k = p −p p p 1+af,s(p) pk−1 . Substituting into (24) shows that ps−1 −1 k ! X ω(n)Jk(n) Y 1 1 X p − 1 = 1 − 1 − . ns ps ps−k ps − 1 n∈N p p The series P pk−1 does not have a representation as a sum of prime zeta and shifted prime p ps−1 zeta functions in general, but in special cases such as s = 2k, k > 1, it does. Rewriting in terms of zeta functions and taking convergence criteria based on (5.2) completes the proof. We conclude with some additional nontrivial examples, which will perhaps be of use when considering exotic divisor sums. They are all special cases of Theorem 5.4 and can be proved similarly to Corollary 5.14. In each case we begin with a known Euler product for a function f(n), then use that to extract af,s(p). Corollary 5.15. For s, k ∈ C such that σ > max (1, 1 + <(k)), X ω(n)σk(n) = ζ(s)ζ(s − k)(P (s) + P (s − k) − P (2s − k)) . (35) ns n∈N Page 16 RHIT Undergrad. Math. J., Vol. 19, No. 1 Proof. We know from [8, (27.4.11)] that P σk(n) = ζ(s)ζ(s − k), where σ (n) is the n∈N ns k sum of the kth powers of the divisors of n. Expanding the Euler product gives us that −1 −1 Q 1 1 Q . We obtain ζ(s)ζ(s − k) = p 1 − ps 1 − ps−k = p (1 + af,s(p)) 1 + af,s(p) = −1 −1 1 1 . Simplifying gives that af,s(p) 1 1 1 . Substituting 1 − s 1 − s−k = s + s−k − 2s−k p p 1+af,s(p) p p p into (24) shows that ! X ω(n)σk(n) X σk(n) X 1 1 1 = + − . ns ns ps ps−k p2s−k n∈N n∈N p Rewriting the right hand side in terms of zeta and prime zeta functions while noting that they will both converge if <(s) > max (1, 1 + <(k)) completes the proof. Corollary 5.16. Let s ∈ C. Then X ω(n)d(n2) ζ3(s) = (4P (s, 1) − P (s)) , (36) ns ζ(2s) n∈N which converges if <(s) > 1. Proof. We know from [7, (3.41)] that 2 X σa(n ) ζ(s)ζ(s − a)ζ(s − 2a) = . ns ζ(2s − 2a) n∈N As in the proof of (35), expanding the Euler product gives us that −1 −1 −1 ζ(s)ζ(s − a)ζ(s − 2a) Y 1 1 1 1 = 1 − 1 − 1 − 1 − . ζ(2s − 2a) ps ps−a ps−2a p2s−2a p The term inside the product is equal to (1 + af,s(p)) for every prime p, so on simplifying we get that −1 af,s(p) 1 1 1 1 2 1 1 1 = s + s−a + s−2a − 2s−a − 2s−2a − 2s−3a + 3s−3a 1 − 2s−2a . 1 + af,s(p) p p p p p p p p However, there is no convenient decomposition into a sum of terms of the form 1 and 1 pα pα±1 2 2 2 so we restrict our attention to the case a = 0. Then σ0(n ) = d(n ), where d(n ) counts the number of divisors of n2. With a = 0, we have −1 af,s(p) 3 4 1 1 = s − 2s + 3s 1 − 2s . 1 + af,s(p) p p p p RHIT Undergrad. Math. J., Vol. 19, No. 1 Page 17 Using the identity p2s − 1 = (ps − 1)(ps + 1) allows us to simplify this to af,s(p) 4 1 = s − s . 1 + af,s(p) p + 1 p Substituting this into (24) gives us that ! X ω(n)d(n2) X d(n2) X 4 1 = − . ns ns ps + 1 ps n∈N n∈N p Rewriting in terms of zeta and prime zeta functions while nding convergence criteria to match both completes the proof. Corollary 5.17. Let s ∈ C. Then X ω(n)d2(n) ζ4(s) = (8P (s, 1) + P (2s) − 4P (s)) , (37) ns ζ(2s) n∈N which converges if <(s) > 1. Proof. Mirroring the proof of the last result, [7, (3.64)] shows that X σa(n)σb(n) ζ(s)ζ(s − a)ζ(s − b)ζ(s − a − b) = . ns ζ(2s − a − b) n∈N As in the proof of (35), expanding the Euler product gives us that −1 −1 −1 ζ(s)ζ(s − a)ζ(s − b)ζ(s − a − b) Y 1 1 1 = 1 − 1 − 1 − ζ(2s − a − b) ps ps−a ps−b p 1 −1 1 × 1 − 1 − . ps−a−b p2s−a−b The term inside the product is equal to (1 + af,s(p)) for every prime p, so on simplifying we get that af,s(p) 1 1 1 1 1 1 3 1 1 = s + s−a + s−b + s−a−b − 2s−a − 2s−b − 2s−a−b − 2s−a−2b − 2s−2a−b 1 + af,s(p) p p p p p p p p p 1 1 1 1 1 1 −1 + + + + − 1 − . p3s−a−b p3s−2a−b p3s−a−2b p3s−2a−2b p4s−2a−2b p2s−a−b However, there is no good decomposition into a sum of terms of the form 1 and 1 . Even pα pα±1 the case a = b proves intractable so we restrict our attention to the case a = b = 0. Then Page 18 RHIT Undergrad. Math. J., Vol. 19, No. 1 2 2 σ0(n)σ0(n) = d (n), where d (n) is the square of the number of divisors of n. With a = b = 0, we have −1 af,s(p) 4 7 4 1 1 = s − 2s + 3s − 4s 1 − 2s . 1 + af,s(p) p p p p p Using the identity p2s − 1 = (ps − 1)(ps + 1) allows us to simplify this to af,s(p) 8 1 4 = s + 2s − s . 1 + af,s(p) p + 1 p p Substituting this into (24) gives us that ! X ω(n)d2(n) X d2(n) X 8 1 4 = + − . ns ns ps + 1 p2s ps n∈N n∈N p Rewriting in terms of zeta and prime zeta functions while nding convergence criteria to match both completes the proof. 6 Extensions Lastly, we show how to generalize the methods of this paper to second and higher order derivatives. We only sketch the following proofs, which can be made rigorous by paying su cient attention to convergence criteria. Proposition 6.1. We have the symmetric function identity n n ! n !2 n x x X 2 k Y X i X i (38) k ekt = (1 + xi) + 2 . 1 + xi k=1 i=1 i=1 i=1 (1 + xi) Proof. Starting with the generating product for , Qn Pn k, we study the ek k=1(1+txi) = k=0 ekt action of the dierential operator d . Applying it once yields D := t dt n ! n Y X k D (1 + txi) = kekt , k=1 k=1 from which we recover the familiar (5) after evaluating at t = 1. Applying D a second time results in n n n ! X 2 k Y X xi k ekt = D t (1 + txi) (39) 1 + txi k=1 k=1 k=1 2 n ! n ! n n 2 Y X xi X xi X x = (1 + x ) + − i . (40) i 1 + x 1 + x 2 i=1 i=1 i i=1 i i=1 (1 + xi) Some algebra completes the proof. RHIT Undergrad. Math. J., Vol. 19, No. 1 Page 19 In general, iterating the application of D will yield higher order identities for the elementary symmetric polynomials. Proposition 6.2. Let P∞ f(pm) as usual. If is multiplicative and has af,s(p) := m=1 pms f s su ciently large real part to ensure convergence of all the following series, then 2 ! !2 X ω(n) f(n) X f(n) X af,s(p) X af,s(p) (41) s = s + 2 . n n 1 + af,s(p) (1 + af,s(p)) n∈N n∈N p p If f is instead completely multiplicative, we have 2 ! !2 2 X ω(n) f(n) X f(n) X f(p) X f(p) X f(p) = + − . (42) ns ns ps ps ps n∈N n∈N p p p Proof. Letting xi = af,s(pi) in (38) while letting t → 1 and n → ∞ gives the rst result. Summing af,s(p) as a geometric series gives the second. As an example, letting f(n) = 1 gives 2 X ω(n) = ζ(s) P 2(s) + P (s) − P (2s) , (43) ns n∈N which converges for <(s) > 1. This result, though the simplest application of Proposition 6.2, appears to be new. In general, applying D to E(t) k times will result in a Dirichlet series of the form P ω(n)kf(n) . This may prove useful when considering asymptotics for n∈N ns the weighted k-th moments X ωk(n)f(n), n≤x for some multiplicative f. 7 Conclusion These theorems apply to any multiplicative functions. Together, this allows for a large class of in nite and divisor sums weighted by ω(n) to be addressed for the rst time. This leads to some surprising results such as that the Dirichlet series for products of ω(n) and other multiplicative functions often have a convenient closed form expression in terms of zeta and prime zeta functions. The methods of this paper also suggest an obvious generalization; taking the kth derivative of E(t) will lead to nite and in nite sums involving ω(n)k. This also suggests deep connections between the theory of symmetric functions and Dirichlet series, since with the right choice of xi we can interpret a Dirichlet series as a sum over symmetric polynomials. Dierent identities for symmetric polynomials with correspond to general expressions for Dirichlet series weighted by dierent functions. Due to this, a systematic study of identities for symmetric functions should correspond to surprising identities for Dirichlet series. Page 20 RHIT Undergrad. Math. J., Vol. 19, No. 1 References [1] Apostol, T. M. Introduction to Analytic Number Theory. Springer-Verlag, New York, NY, 1976. [2] Erdös, P. and Kac, M.. The Gaussian law of errors in the theory of additive number theoretic functions. American Journal of Mathematics, 62:738742, 1940. [3] Fröberg, C.-E. On the prime zeta function. Nordisk Tidskr. Informationsbehandling (BIT), 8:187202, 1968. [4] Gould, H. W. and Shonhiwa, T.. A catalog of interesting Dirichlet series. Missouri J. Math. Sci., 1:218, 2008. [5] Hardy, G. H. and Ramanujan, S. The normal number of prime factors of a number n [Quart. J. Math. 48 (1917), 7692]. Collected papers of Srinivasa Ramanujan. AMS Chelsea Publ., Providence, RI, 2000. [6] I. G. Macdonald, Symmetric functions and Hall polynomials, 2nd ed., Oxford University Press, New York, 1995. [7] Mathar, R. J. Survey of Dirichlet series of multiplicative arithmetic functions. Pre-print, 2012. http://arxiv.org/abs/1106.4038. [8] NIST Digital Library of Mathematical Functions. http://dlmf.nist.gov, Release 1.0.9 of 2014-08-29. Online companion to [9]. [9] Olver, F. W. J. and Lozier, D. W. and Boisvert, R. F. and Clark, C. W., editors. NIST Handbook of Mathematical Functions. Cambridge University Press, New York, NY, 2010. Print companion to [8].