Hardy-Ramanujan Journal 40 (2017), 31-42 submitted 29/10/2017, accepted 29/11/2017, revised 29/11/2017
Non-vanishing of Dirichlet series without Euler products
William D. Banks
Abstract. We give a new proof that the Riemann zeta function is nonzero in the half-plane s C : > 1 .Anovelfeatureof { 2 } this proof is that it makes no use of the Euler product for ⇣(s). Keywords. Riemann zeta function, Euler product, zeros. 2010 Mathematics Subject Classification. 11M06
1. Introduction
Let s = + it be a complex variable. In the half-plane
H .= s C : > 1 { 2 } the Riemann zeta function can be defined either as a Dirichlet series
. s ⇣(s) .= n n N X2 or (equivalently) as an Euler product
. s 1 ⇣(s) .= (1 p ) . p primeY Since a convergent infinite product of nonzero factors is not zero, the zeta function does not vanish in H . This can also be seen by applying the logarithm to the Euler product:
s 1 log ⇣(s)= (mp ) . p m N X X2 Indeed, since the double sum on the right converges absolutely in H , it follows that ⇣(s) = 0 for all 6 s H . Alternatively, since the M¨obius function µ is bounded, it follows that the series 2
1 1 s ⇣(s) = µ(n)n n=1 X converges absolutely when > 1, so ⇣(s) cannot vanish. Of course, to prove that the M¨obius function is bounded, one exploits the multiplicativity of µ, so this argument also relies (albeit implicitly) on the Euler product for ⇣(s). It is crucial to our understanding of the primes to extend the zero-free region for ⇣(s) as far to the left of = 1 as possible.1 According to Titchmarsh [Ti86, §3.1] this means extending the “sphere of influence” of the Euler product: We thank episciences.org for providing open access hosting of the electronic journal Hardy-Ramanujan Journal 1At present, the strongest result in this direction is due to Mossingho↵ and Trudjian [MosTru15]; see also the earlier papers [Fo00, JaKw14, Kad05] and references therein. 32 2. Statement of results
The problem of the zero-free region appears to be a question of extending the sphere of influence of the Euler product beyond its actual region of convergence; for examples are known of functions which are extremely like the zeta-function in their representation by Dirichlet series, functional equation, and so on, but which have no Euler product, and for which the analogue of the Riemann hypothesis is false. In fact the deepest theorems on the distribution of the zeros of ⇣(s) are obtained in the way suggested. But the problem of extending the sphere of influence of [the Euler product] to the left of =1in any e↵ective way appears to be of extreme di culty.
But let’s play the devil’s advocate for a moment! Is it really the case that the non-vanishing of the Riemann zeta function in H (and in wider regions) fundamentally relies on the existence of an Euler product? Our aim in this paper is to provide some evidence to the contrary.
2. Statement of results
For a given arithmetical function F with F (1) = 0, let F denote the Dirichlet inverse of F ; this can 6 be defined via the M¨obius relation e 1ifn = 1; F (a)F (b)=I(n) .= (0 otherwise. abX=n e . s To prove that a Dirichlet series D(s) .= F (n)n is nonzero in H , it is enough to show that D(s) 1 = F (n)n s converges in H . Using partial summation, this is a consequence of any bound P of the form P e F (n) x1+o(1) (x ). (2.1) ⌧ !1 n x X e Our proof of the next theorem establishes (2.1)wheneverF is supported on a set of -free numbers.2
. s Theorem 2.1. Let D(s) = n N F (n)n be a Dirichlete series such that F is bounded, F (1) =0, 2 6 and the Dirichlet inverse F is supported on the set of -free numbers for some 2. Then D(s) =0 P 6 in H . e This theorem is proved in §5. It establishes the property of non-vanishing in H for a large class of Dirichlet series, almost all of which do not have an Euler product (but some do). For a Dirichlet series D(s) attached to a bounded completely multiplicative function F (for ex- ample, the Riemann zeta function), Theorem 2.1 provides a novel route to showing that D(s)is nonzero in H . For such F , one can easily show that F is supported on the set of squarefree numbers provided that one has the luxury of using the Euler product for D(s). For this reason, it is important to note that our proof of the next theorem makes noe use of the Euler product for D(s). Instead, a combinatorial identity is employed to show that F has the required support.
Theorem 2.2. Let F be an arithmetical functione that is bounded and completely multiplicative. Then the Dirichlet inverse F is supported on the set of squarefree numbers.
This theorem is provede in §6. In particular, Theorems 2.1 and 2.2 together yield the following result without any use of the Euler product for ⇣(s).
Corollary 2.3. The Riemann zeta function does not vanish in H .
2 k For a given integer 2, a natural number n is said to be -free if p - n for every prime p. W. D. Banks, Non-vanishing of Dirichlet series without Euler products 33
To further illustrate how our results can be applied, we introduce and study a special family of . Dirichlet series D .= Dz(s):z C with the following properties: { 2 } (i) The Riemann zeta function belongs to D;
(ii) Every series Dz(s) is meromorphic and nonzero in the region H ;
(iii) Only two series in D have an Euler product, namely the Riemann zeta function and the constant function 1 (s) = 1 for all s C. C 2 Viewing ⇣(s) in relation to the other members of D, the existence of an Euler product seems quite unusual, whereas non-vanishing in the half-plane H is a property enjoyed by every member of D.
3. Preliminaries
Throughout the paper, we fix an integer parameter 2 and denote by N the set of -free numbers. We denote by 1N the indicator function of N:
1ifn ; . N 1N (n) = 2 (0 otherwise.
We denote by !(n) the number of distinct prime factors of n and by ⌦(n) the number of prime factors of n, counted with multiplicity. For any integer k 2 we denote by log x the k-th iterate of the function x max log x, 1 .In k 7! { } particular, log2 x = log log x and log3 x = log log log x when x is su ciently large. We use the equivalent notations f(x)=O(g(x)) and f(x) g(x) to mean that the inequality ⌧ f(x) cg(x) holds with some constant c. Throughout the paper, any implied constants in the | | symbols O and may depend (where obvious) on the parameters , " but are absolute otherwise. ⌧ Two classical results of Hardy and Ramanujan [HR17, Lemmas B and C] assert the existence of constants c1,c2 > 0 such that the inequalities
c x (log x + c )` 1 n x : !(n)=` 1 2 2 (3.2) { } log x (` 1)! and ` 1 j c1x ` 1 j (log2 x + c2) n x : ⌦(n)=` 9 (3.3) { } log x 10 j! j=0 X hold for all real x 2. In the next lemma, we study the counting function . N,`(x) .= n x : n N and ⌦(n)=` . (3.4) { 2 } Although this function might seem closely related to that on the left side of (3.3), we prove that it satisfies a bound nearly as strong as (3.2).
Lemma 3.1. There are absolute constants C1,C2 > 0 with the following property. For any integers 2 and ` 1, the counting function defined by (3.4) satisfies the upper bound C x (( 1) log x +( 1)C )` 1 N (x) 1 2 2 (x 2). (3.5) ,` log x (` 1)! 34 3. Preliminaries
Proof. Our proof is an adaptation of arguments from [HR17]. When ` , the condition ⌦(n)=` implies that n N. Using (3.3) it follows that 2 ec x(log x + c )` 1 N (x)= n x : ⌦(n)=` 1 2 2 (x 2), ,` { } log x hence (3.5) holds for any choice of C ec and C c . 1 1 2 2 From now on, we assume that ` > . To simplify the notation slightly, we put .= 1. 1 Let p(1) .=2
1 copies 1 copies 1 copies in which the primes appear in increasing| {z order,} | {z each} being| {z repeated} 1 times. Let n N, n 2, and suppose that ⌦(n)=`. Among all of the ordered `-tuples (j1,...,j`) 2 having j < Let be the set of ordered pairs (j, m) such that j J(m), m N, andp ˜(j)m x.The S 2 condition j J(m) implies that the primep ˜(j) does not exceed the largest prime factor of m, and thusp ˜(j) m x/p˜(j); consequently, N,` 1(x/p˜(j)). (3.7) |S| j :˜p(j)2 x X On the other hand, suppose that n x, n N and ⌦(n)=`. Factoring n as in (3.6)with 2 (j ,...,j )= (n), one verifies that the pair (j ,n/p˜(j )) lies in for each i =1,...,` 1 Hence, n 1 ` i i S can be expressed in ` 1di↵erent ways as the product of the entries of an ordered pair in ,which S implies that (` 1)N (x) . (3.8) ,` |S| Combining (3.7) and (3.8), and using induction, we have 1 N,`(x) N,` 1(x/p˜(j)) ` 1 j :˜p(j)2 x X ` 2 1 C1(x/p˜(j)) (1 log2(x/p˜(j)) + 1C2) ` 1 log(x/p˜(j)) (` 2)! j :˜p(j)2 x X C x( log x + C )` 2 1 1 1 2 1 2 . (` 1)! p˜(j) log(x/p˜(j)) j :˜p(j)2 x X W. D. Banks, Non-vanishing of Dirichlet series without Euler products 35 As each prime is repeated precisely 1 times in p˜(j) we have j N 2 1 1 = . p˜(j) log(x/p˜(j)) 1 p log(x/p) j :˜p(j)2 x p : p2 x X X The proof is completed using the fact that 1 log x + C < 2 2 (x 2) 2 p log(x/p) log x p :Xp 4. Factorisatio numerorum For any n 2, let f(n) denote the number of representations of n as a product of integers exceeding one, where two representations are considered equal only if they contain the same factors in the same order. For technical reasons, we also set f(1) .= 1. One can define f(n) as follows. For every positive integer k,letfk(n) denote the number of ordered k-tuples (n ,...,n ) such that each n 2 and the product n n equals n.Then 1 k j 1 ··· k . f(n) .= I(n)+ fk(n)(n N), (4.9) 2 k 1 X where 1ifn = 1; I(n) .= (0 otherwise. Note that the sum in (4.9)isfinitesincefk(n)=0whenk>⌦(n). In one of the earliest papers about the function f(n), Kalm´ar[Kal30] establishes the asymptotic formula x f(n) (x ), (4.10) ⇠ ⇣ ( ) !1 n x 0 X where =1.728647 is the unique positive root of ⇣( ) = 2. In particular, this implies that ··· f(n) n (n N). (4.11) ⌧ 2 The bound (4.11) is essentially optimal since Hille [Hi36] has shown that for every " > 0 the lower bound f(n) n " holds for infinitely many n; see also Erd˝os[Er41]. The next proposition is fundamental in the sequel as it leads to a significant strengthening of (4.11) for -free numbers n. Proposition 4.1. For any integer n 2 we have f(n) exp(` log ` + O(` log ` log `)) with ` .= ⌦(n). 2 3 Proof. Let (n) be the set of ordered tuples (n ,...,n ) of any length r such that every n 2 and T 1 r j n n = n.Thus, (n) = f(n). 1 ··· r |T | Let (`) be the set of ordered partitions of `; these are ordered tuples =( ,..., ) of any P 1 r length r such that 1 and + + = `. 1 ··· r 1 ··· r We begin by constructing a map : (n) (`) as follows. For any given ⌘ =(n ,...,n )in T ! P 1 r (n), let (⌘) denote the tuple (⌦(n ),...,⌦(n )), and set T ⌦ 1 r (n) .= (⌘):⌘ (n) ; U { ⌦ 2 T } 36 4. Factorisatio numerorum thus, : (n) (n). Next, for any w =(w ,...,w )in (n), let (w)bethetuple( ,..., ) ⌦ T ! U 1 r U 1 r that is obtained by rearranging the entries of w into nondecreasing order; then : (n) (`). U ! P The map : (n) (`) is defined to be the composition . T ! P ⌦ Next, for any (`)letd (n) be the cardinality of the set 1( ) of preimages of in (n).3 2 P { } T Since any product counted by f(n) gives rise to a unique partition via the map ,wehave . f(n) .= I(n)+ d (n). (`) 2XP In view of the celebrated estimate of Hardy and Ramanujan [HR18] 1 (`) (4`p3) exp ⇡ 2`/3 (` ), |P | ⇠ !1 to prove the proposition it su ces to show that the p individual bound d (n) exp(` log ` + O(` log ` log `)) (4.12) 2 3 holds for every (`). 2 P To this end, let =( ,..., ) be a fixed element of (`). For any natural number k,letm be 1 r P k the multiplicity with which k occurs in the partition ,i.e., . mk .= j : j = k (k N). { } 2 . Note that ` = k kmk.Settingm = k mk, a simple combinatorial argument shows that P P `! m! d (n) (4.13) mk k(k!) · k mk! Q Q 1 (roughly speaking, the second factor is the cardinality of the set ( ) of preimages of in { } 1 (n), whereas for any such preimage w the first factor bounds the cardinality of the set ( w ) of U ⌦ { } preimages of w in (n)). We remark that (4.13) holds with equality whenever n is squarefree. Since T `! ``, to establish (4.12) it is enough to show that m! log ` log2 ` log3 `. (4.14) (k!)mk m ! ⌧ ✓ k k k ◆ Since m ` and log j!=j log j + OQ(j) for allQ positive integers j, the left side of (4.14)is m log ` m (k log k + O(k)) (m log m + O(m )) k k k k Xk Xk ` = mk log k + O(`)=S1 + S2 + O(`), (say) k mk k : m =0 Xk6 ⇣ ⌘ where ` ` . . S1 = mk log k and S2 = mk log k k mk k mk k : mk=0 k : mk=0 m >X`/g6 (`) ⇣ ⌘ m X`/g6 (`) ⇣ ⌘ k k and (log `)2 . g(`) = 2 . (log2 `) log3 ` 3 A more descriptive but less precise definition is the following. If =( 1,..., r) (`), then d (n)isthenumberof 2 P r-tuples (n1,...,nr) for which the product n1 nr equals n and such that, after a suitable permutation of the indices, ··· one has ⌦(nj )= j for each j (that is, the multisets ⌦(nj ) and j are the same). { } { } W. D. Banks, Non-vanishing of Dirichlet series without Euler products 37 For each k in the sum S1,wehavemk > `/g(`) and kmk j ` jmj = `; therefore, ` g(`) P 1 S log ` log g(`) 1 k kk k k k : kk g(`) X ⇣ ⌘ X ` log g(`) log g(`) ` log ` log `. ⌧ 2 ⌧ 2 3 For each k in the sum S ,wehave1 m `/g(`); thus, 2 k ` ` ` log ` `(log `)2 S2 log k 1 ` log2 ` log3 `. g(`) k g(`) ⌧ g(`) log2 ` ⌧ k k : kk ` X ⇣ ⌘ X Combining the above bounds on S1 and S2,wederive(4.14), and in turn (4.12), finishing the proof. The following corollary is crucial in the next section. Corollary 4.2. For any constant C>0 we have C⌦(n)f(n)1 (n) x1+o(1) (x ), (4.15) N !1 n x X where the function implied by o(1) depends only on C and . Proof. Let Q denote the quantity on the left side of (4.15). For any n N we have ⌦(n) !(n). Also, !(n) 2(log x)/ log2 x for all n x once x is 2 . su ciently large. Hence, defining B(x) .=2(log x)/ log2 x it follows from Proposition 4.1 that Q C` exp(` log ` + O(` log ` log `)) N (x), 2 3 · ,` ` B(x) X where N,`(x) is the counting function given by (3.4). By Lemma 3.1 we have C x (( 1) log x +( 1)C )` 1 Q C` exp(` log ` + O(` log ` log `)) 1 2 2 2 3 · log x (` 1)! ` B(x) X (( 1) log x +( 1)C )` 1 x1+o(1) exp(` log `) 2 2 (x ). · (` 1)! !1 ` B(x) X Using the estimates (` 1)! = exp(` log ` + O(`)) = xo(1) exp(` log `) and ` 1 o(1) (( 1) log x +( 1)C ) =exp(O(` log x)) = x , 2 2 3 which hold uniformly for all ` B (x), the result follows. 5. Reciprocal of a Dirichlet series s Theorem 5.1. Suppose that F is bounded on N,andF (1) =0. Then F (n)1 (n)n con- 6 n N N verges absolutely in H . 2 P e 38 5. Reciprocal of a Dirichlet series Proof. Without loss of generality, we can assume that F (1) = F (1) = 1. Let C 1 be a number such that F (n) C (n N).e (5.16) | | 2 For every positive integer k,let (n) be the set of ordered k-tuples (n ,...,n ) such that every Tk 1 k n 2 and n n = n.Then (n) = f (n) in the notation of §4. We denote j 1 ··· k |Tk | k f (F ; n) .= F (n ) F (n ). (5.17) k 1 ··· k (n1,...,n ) (n) Xk 2Tk Using (5.16) we derive that f (F ; n) Ckf (n). (5.18) | k | k Since f (F ; n) = 0 for all k>⌦(n), and the inequality ⌦(n) (log n)/ log 2 holds for all n, it follows k that f (F ; n) nBf (n)withB .= max 0, (log C)/ log 2 . | k | k { } Summing over k and using (4.11), we see that B+ fk(F ; n) n (n N). (5.19) | | ⌧ 2 k 1 X Next, put 1 . s . s D(s) .= F (n)n =1+Z(F ; s)withZ(F ; s) .= F (n)n . n=1 n 2 X X For every positive integer k we have k s Z(F ; s) = fk(F ; n)n . n 2 X In view of (5.19)theidentity 1 k k k s =1+ ( 1) Z(F ; s) =1+ ( 1) f (F ; n)n (5.20) 1+Z(F ; s) k k 1 n 2 k 1 X X X holds throughout the half-plane >B+ since all sums converge absolutely in that region. Noting { 1 } s that the left side of (5.20)isD(s) = n N F (n)n , we conclude that 2 P k F (n)=I(n)+ (e 1) fk(F ; n)(n N). (5.21) 2 k 1 X e To prove the theorem, we need to show that the Dirichlet series s k s F (n)1 (n)n =1+ ( 1) f (F ; n)1 (n)n N k N n N n 2 k 1 X2 X X e converges absolutely in H . For any natural number n it is clear that fk(F ; n)=0wheneverk>⌦(n), hence using (5.18) we see that ( 1)kf (F ; n)1 (n) I(n)+ Ckf (n)1 (n) C⌦(n)f(n)1 (n). k N k N N k 1 k ⌦(n) X X W. D. Banks, Non-vanishing of Dirichlet series without Euler products 39 Consequently, it su ces to show that the sum ⌦(n) C f(n)1N (n)n (5.22) n N X2 converges for > 1. However, since the summatory function . ⌦(n) S(x) = C f(n)1N (n) n x X satisfies the bound S(x) x1+" for every " > 0 by Corollary 4.2, the convergence of (5.22) for > 1 ⌧ follows by partial summation. Proof of Theorem 2.1. Since the Dirichlet inverse F has its support in N for some 2, we have F (n)=F (n)1N (n) for all n. By Theorem 5.1, e s s 1 e e F (n)1N (n)n = F (n)n = D(s) n N n N X2 X2 e e converges absolutely in H , and the result follows. 6. The family D For any fixed z C,letFz be the arithmetical function defined by 2 . 1ifn = 1; Fz(n) .= ( z otherwise. Then s F (n)n =1 z(⇣(s) 1) (s H ). z 2 n N X2 Taking F .= F in (5.19) and (5.21)wehaveF (n) nB+ ,whereB .= max 0, (log z )/ log 2 and z z ⌧ { | | } =1.728647 as before. This implies that the formal identity ··· e s 1 F (n)n = (6.23) z 1 z(⇣(s) 1) n N X2 e holds rigorously when >B+ +1. Moreover, it is clear that the Dirichlet series can be analytically 1 continued to the region > z ,where z is the unique positive root of ⇣( z)=1+ z if z = 0, . { } | | 6 and 0 .= . It is worth mentioning that for any fixed z = 0 or 1, the function on the right 1 6 1 side of (6.23) has infinitely many poles in H since the equation ⇣(s)=1+z has infinitely many solutions in any strip 1 < < 1+" ; see, e.g., Titchmarsh [Ti86, Theorem 11.6 (C)]. { } Next, we introduce two Dirichlet series given by . s Dz†(s) = Fz(n)1N2 (n)n n N X2 e and . 1 s Dz(s) .= Dz†(s) = Gz(n)n , n N X2 where G is the Dirichlet inverse of F 1 . According to Theorem 5.1, Dz†(s) converges absolutely z z · N2 in H , hence it is analytic in that region. This implies that Dz(s) has a meromorphic extension to H , and D (s) =0inH ; thus, we havee verified property (ii) of §1. z 6 40 6. The family D From the above definitions, one sees that F 1(n)=1N(n), and thus D 1(s)=⇣(s). This estab- lishes property (i) of §1. We also have F0(n)=I(n), so that D0(s)=1C(s). Finally, we establish property (iii) of §1. Observe that the M¨obius relations 2 Fz(a)Fz(b)=I(n) and Fz(a)1N2 (a) Gz(b)=I(n) abX=n abX=n e e immediately imply that G (p)=G (q)=G (pq)= z for any two di↵erent primes p and q.4 If z z z Dz(s) has an Euler product, then Gz is multiplicative, and therefore ( z)2 = G (p)G (q)=G (pq)= z, z z z which is only possible for z = 0 or 1. Lemma 6.1. Let z be a complex number, n a natural number, and p a prime number not dividing n. For any integer ↵ 1 we have ↵ ↵ 1 ` 1 ↵ + ` 1 F (p n)=(z + 1) z z + `↵ (z + 1) f (n). z ` ` ` 1 ✓ ◆ X e Proof. Using (5.17), (5.21) and the definition of Fz, it follows that ↵ k ↵ Fz(p n)= z fk(p n). (6.24) k 1 X e ↵ ↵j The quantity fk(p n) is the number of ordered k-tuples (m1,...,mk) such that mj = p nj for each j,with↵ 1 or n 2, ↵ + + ↵ = ↵, and n n = n. To construct such a k-tuple, first j j 1 ··· k 1 ··· k choose an integer ` in the range 1 ` k and an ordered `-tuple (ˆn ,...,nˆ ) with eachn ˆ 2 1 ` j andn ˆ nˆ = n; for any choice of ` there are precisely f (n)such`-tuples. Next, maintaining the 1 ··· ` ` ordering of the integersn ˆ in (ˆn ,...,nˆ ), we construct (n ,...,n )byinsertingk ` extra entries, j 1 ` 1 k each equal to one (thus, every ni in the resulting k-tuple is one of the numbersn ˆj, or else ni = 1); k k there are k ` = ` ways to insert these extra ones to form (n1,...,nk). To guarantee that every m = p↵j n 2 in the final k-tuple (m ,...,m ), so it is counted in the computation of f (p↵n), j j 1 k k we have to replace each entry ni =1in(n1,...,nk) with a copy of the prime p. As there are only ↵ copies of p available, it must be the case that ↵ k ` else this choice of ` is unacceptable. The remaining ↵ k + ` copies of the prime p can be distributed arbitrarily. As the number of ways that ↵+` 1 one can distribute ↵ k + ` objects into k boxes is k 1 , putting everything together we have k k ↵ + ` 1 f (p↵n)= f (n). k ` k 1 ` `=max 1,k ↵ ✓ ◆✓ ◆ X{ } Combining this result with (6.24), we derive that ↵+` k ↵ + ` 1 F (p↵n)= f (n) zk z ` ` k 1 ` 1 k=` ✓ ◆✓ ◆ X X e Making the change of variables k k + ` in the inner summation, it follows that 7! ↵ ` Fz(p n)= z f`(n)B(z,↵, `), ` 1 X 4 e A more elaborate argument shows that Fz(n)=Gz(n)= z for all squarefree numbers n. W. D. Banks, Non-vanishing of Dirichlet series without Euler products 41 where ↵ k + ` ↵ + ` 1 B(z,↵, `) .= zk . ` k + ` 1 Xk=0 ✓ ◆✓ ◆ In view of the combinatorial identity k + ` ↵ + ` 1 ↵ + ` 1 ↵ 1 ` ↵ = + ` k + ` 1 ` k 1 ↵ k ✓ ◆✓ ◆ ✓ ◆✓✓ ◆ ✓ ◆◆ ↵ 1 (where k 1 =0whenk = 0), it follows that ↵ ↵ + ` 1 ↵ 1 ` ↵ B(z,↵, `)= zk + ` k 1 ↵ k ✓ ◆ Xk=0 ✓✓ ◆ ✓ ◆◆ ↵ + ` 1 ↵ 1 1 ↵ = z(z + 1) + `↵ (z + 1) , ` ✓ ◆ and we obtain the stated result. Proof of Theorem 2.2. Since F is completely multiplicative, from (5.17) it follows that fk(F ; n)=F (n)fk(n)(k, n N). 2 With two applications of (5.21) we deduce that k F (n)=I(n)+F (n) ( 1) fk(n)=F (n)F 1(n)(n N). (6.25) 2 k 1 X e e ↵ Applying Lemma 6.1 with z = 1, we see that F 1(p n) = 0 for any n N, any prime p not dividing 2 5 n, and all ↵ 2. This implies that F 1 is supported on the set of squarefree numbers, and (6.25) shows that the same is true of F . e e 7. Remarks e Theorems 2.1 and 5.1 can be extended to cover all functions satisfying the polynomial growth condition F (n) nA provided that one is willing to replace H with the half-plane s C : >A+1 in ⌧ { 2 } those theorems. It would be interesting to see whether the ideas of this paper can be developed to produce zero-free regions for ⇣(s) and other Dirichlet series inside the critical strip. Sarnak [Sa11] has recently considered a general pseudo-randomness principle related to a famous conjecture of Chowla [Ch65]. Roughly speaking, the principle asserts that the M¨obius function µ(n) does not correlate with any function ⇠(n) of low complexity. In other words, µ(n)⇠(n)=o ⇠(n) (x ). (7.26) | | !1 n x ✓ n x ◆ X X Combining Kalm´ar’s result (4.10) with Corollary 4.2, we see that (7.26) is verified for the function ⇠(n) .= f(n). However, this is not due to the randomness of µ(n) but instead to the fact f(n) takes smaller values on squarefree numbers than it does on natural numbers in general. It would be . 2 interesting see whether (7.26) holds for ⇠(n) = f(n)1N2 (n) . 5 As we have already seen, F 1 is the M¨obius function µ. e 42 7. Remarks Let feven (n)[resp.fodd (n)] denote the number of representations of n as a product of an even [resp. odd ] number of integers exceeding one, where two representations are considered equal only if they contain the same factors in the same order. In other words, . . feven (n) .= I(n)+ fk(n) and fodd (n) .= fk(n). k 1 k 1 kXeven kXodd Clearly, f(n)=feven (n)+fodd (n), but it is somewhat less obvious that µ(n)=feven (n) fodd (n)(n N). (7.27) 2 . Indeed, taking F = F 1 = 1N we have fk(F ; n)=fk(n) for all n by (5.17), and then (7.27) follows immediately from (5.21). The identity (7.27) is originally due to Linnik [Li63]; see also Friedlander and Iwaniec [FrIw10, Chapter 17]. Acknowledgment. The author thanks Andrew Granville and Igor Shparlinski for their insightful comments on the original draft, which led to improvements in the results and the exposition. References [Ch65] S. Chowla, The Riemann hypothesis and Hilbert’s tenth problem. Mathematics and its applications,4,Gordonand Breach, New York, London, Paris, 1965. [Er41] P. Erd˝os, On some asymptotic formulas in the theory of the “factorisatio numerorum.” Ann. of Math. (2) 42, (1941), 989–993. [Fo00] K. Ford, Zero-free regions for the Riemann zeta function. 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Banks University of Missouri, Department of Mathematics, 102 Mathematical Sciences Bldg. Columbia, MO 65203 USA e-mail: [email protected]