PiatetskiShapiro primes from almost primes


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1 PiatetskiShapiro primes from almost primes Roger C. Baker Department of Mathematics, Brigham Young University Provo, UT USA William D. Banks Department of Mathematics, University of Missouri Columbia, MO USA Zhenyu V. Guo Department of Mathematics, University of Missouri Columbia, MO USA Aaron M. Yeager Department of Mathematics, University of Missouri Columbia, MO USA Abstract Let be the floor function. In this paper, we show that for any fixed c (1, ) there are infinitely many primes of the form p = nc, where n is a natural number with at most eight prime factors (counted with multiplicity). 1
2 1 Introduction The PiatetskiShapiro sequences are those sequences of the form ( n c ) n N (c > 1, c N), where t denotes the integer part of any t R. Such sequences are named in honor of PiatetskiShapiro, who showed in [6] that for any number c (1, 12) 11 the set P (c) := { p prime : p = n c for some n N } is infinite. The admissible range for c in this result has been extended many times over the years and is currently known for all c (1, 243 ) thanks to 205 Rivat and Wu [8]. For any natural number r, let N r denote the set of ralmost primes, i.e., the set of natural numbers having at most r prime factors, counted with multiplicity. In this paper, we introduce and study sets of PiatetskiShapiro primes of the form P (c) r := { p prime : p = n c for some n N r }. Our main result is the following: Theorem 1. For any fixed c (1, 77 ) the set P(c) 76 8 is infinite. More precisely, { n x : n N 8 and n c is prime } x (log x) 2, where the implied constant in the symbol depends only on c. 2 Notation Throughout the paper, we set γ := 1/c for a given real number c > 1. The parameters ε, δ are positive real numbers that are sufficiently small for all of our purposes. Any implied constants in the symbols O,,, may depend on c, ε, δ but are absolute otherwise. The letter p always denotes a prime number, and Λ is used to denote the von Mangoldt function. We use notation of the form m M as an abbreviation for M < m 2M. 2
3 As is customary, we put e(t) := e 2πit and {t} := t t (t R). Throughout the paper, we make considerable use of the sawtooth function defined by ψ(t) := t t 1 2 = {t} 1 2 (t R) (1) as well as the well known approximation of Vaaler [10]: for any H 1 there exist numbers a h and b h such that ψ(t) a h e(th) b h e(th), a h 1 h, b h 1 H. (2) 0< h H h H 3 Proof of Theorem Initial approach We analyze exponential sums that are relevant for finding primes in P (c) r the number r as small as possible. The set that we sieve is A := { n x : n c is prime }. with For any d D, where D is a fixed power of x to be specified later, we must estimate accurately the cardinality of Since md A if and only if A d := { n A : d n }. p (md) c < p + 1 and md x, to within O(1) the cardinality of A d is equal to the number of primes p x c for which the interval [ p γ d 1, (p + 1) γ d 1) contains a natural number; thus, A d = { p x c : (p + 1) γ d 1 < m p γ d 1 for some m N } + O(1) = p x c ( p γ d 1 (p + 1) γ d 1 ) + O(1) = X d 1 + p x c ( ψ( (p + 1) γ d 1 ) ψ( p γ d 1 ) ) + O(1), 3
4 where ψ is given by (1), and X := p x c ((p + 1) γ p γ ) = p x c γp γ 1 + O(1) x c log x (x ). It is unnecessary to evaluate X more precisely than this; however, for any sufficiently small ε > 0 we need to show that Ad X d 1 X x ε/3 x1 ε/3 log x d D (x ). (3) Splitting the range of d into dyadic subintervals and using partial summation in a standard way, it suffices to prove that the bound Λ(n) ( ψ( (n + 1) γ d 1 ) ψ( n γ d 1 ) ) x1 ε/2 (4) d D 1 holds uniformly for D 1 D, N x c, N 1 N. In turn, (4) is an immediate consequence of the uniform bound Λ(n) ( ψ( (n + 1) γ d 1 ) ψ( n γ d 1 ) ) x 1 ε/2 d 1 (5) for d D, N x c, N 1 N. Our aim is to establish (5) with D as large as possible, and in this subsection we show that (5) holds when D x 1 136c/157 (6) and ε > 0 is sufficiently small. Suppose this has been done, and observe that 1 136c 157 > 8 63 whenever c < Then, for any fixed c (1, 8635) and α ( 8, 1 136c ), the inequality (3) with D := x α implies the bound A d X d 1 d D X (log X ) 2, thus we can apply the weighted sieve in the form [4, Ch. 5, Prop. 1] with the choices R := 8, δ R := and g :=
5 Note that g < R δ R, and (if x is large enough) the inequality a < D g holds for all a A since αg > 1; thus, the conditions of [4, Ch. 5, Prop. 1] are met, and we conclude that A contains at least X / log X numbers with at most eight prime factors. This yields the statement of the theorem for all c in the interval (1, 8635) We now turn to the proof of (5) for all D satisfying (6). Let S denote the sum on the left side of (5). From Vaaler s approximation (2) we derive the inequality S S 1 ( b h e( h(n + 1) γ d 1 ) + ) b h e( hn γ d 1 ), where S 1 := Λ(n) Λ(n) h H 0< h H h H a h ( e( h(n + 1) γ d 1 ) e( hn γ d 1 ) ). Because h H b h e(th) is nonnegative for all t R, it follows that S S 1 log N 1 = log N 1 h H ( h H b h e( h(n + 1) γ d 1 ) + h H b h e( hn γ d 1 ) b h ( S(h, 0) + S(h, 1) ), (7) ) where we have put S(h, j) := e( h(n + j) γ d 1 ). We now choose H := x 1+ε Nd (8) so that the contribution from h = 0 on the right side of (7) does not exceed 2b 0 N log N 1 x 1 ε/2 d 1. On the other hand, using the exponent pair ( 1, 1 ) we derive that 2 2 S(h, j) h 1/2 N γ/2 d 1/2 + h 1 N 1 γ d (0 < h H, j = 0, 1). 5
6 We bound the contribution in (7) from integers h in the range 0 < h H by observing that b h h 1/2 N γ/2 d 1/2 H 1/2 N γ/2 d 1/2 = x 1/2+ε/2 N 1/2+γ/2 x 1 ε d 1 0< h H since and that since d D x 1 c/2 2ε x 3/2 3ε/2 N 1/2 γ/2, 0< h H b h h 1 N 1 γ d N 1 γ d x 1 ε d 1 d 2 D 2 x 2 c ε x 1 ε N γ 1. Putting everything together, we conclude that S S 1 + O(x 1 ε/2 d 1 ), and it remains only to bound S 1. Next, we use a partial summation argument from the book of Graham and Kolesnik [3]. Writing S 1 = a h Λ(n)φ h (n)e(hn γ d 1 ) with 0< h H 0< h H φ h (t) := e(h(t + 1) γ d 1 ht γ d 1 ) 1, we would like to show that h 1 Λ(n)φ h (n) e(hn γ d 1 ) x1 ε d 1 (d D). (9) Taking into account the bounds φ h (t) hn γ 1 d 1 and φ h(t) hn γ 2 d 1 (N t N 1 ), 6
7 the left side of (9) is, on integrating by parts, bounded by h 1 φ h (N 1 ) Λ(n) e(hn γ d 1 ) h H + N1 h 1 φ h(t) Λ(n) e(hn γ d 1 ) dt h H N N<n t N γ 1 d 1 Λ(n) e(hn γ d 1 ) h H N<n N 2 for some number N 2 (N, N 1 ]. Therefore, it suffices to show that the bound Λ(n) e(hn γ d 1 ) x1 ε N 1 γ (10) N<n N 2 h H holds uniformly for d D, N x c, N 2 N. To establish (10) we use the decomposition of HeathBrown [5]; it suffices to show that our type I and type II sums satisfy the uniform bounds S I := a l e(hl γ m γ d 1 ) x1 2ε N 1 γ, (11) S II := H 1 h H 2 H 1 h H 2 l L l L m M lm J a l b m e(hl γ m γ d 1 ) x1 2ε N 1 γ, (12) m M lm J in some specific ranges. Here, J is an interval in (N, N 1 ], H 1 H, H 2 H 1, LM N, and the numbers a l, b m C satisfy a l 1, b m 1. In view of [5, pp ] we need to show that (12) holds uniformly for all L in the range u L N 1/3 for some u x ε N 1/5, (13) and for such u we need to show that (11) holds uniformly for all M satisfying M N 1/2 u 1/2. Put F := H 1 N γ d 1. For the type II sum, we apply Baker [1, Thm. 2], which yields the bound S II ( T II,1 + T II,2 ) H1 Nx ε 7
8 with T II,1 := (H 1 L) 1/2 and T II,2 := ( ) k/(2+2k) F M (1+k l)/(2+2k) H 1 L for any exponent pair (k, l) provided that F H 1 L. For the type I sum, by Robert and Sargos [9, Thm. 3] we have the bound with T I,1 := S I ( T I,1 + T I,2 + T I,3 ) H1 Nx ε ( ) 1/4 F, T H 1 LM 2 I,2 := M 1/2 and T I,3 := F 1. Hence, to establish (11) and (12) it suffices to verify that max { T I,1, T I,2, T I,3, T II,1, T II,2 } x 1 3ε H 1 1 N γ. (14) From the definition of F we see that the bound T I,3 = F 1 x 1 3ε H 1 1 N γ (15) is equivalent to d x 1 3ε and thus follows from the inequality D x 1 3ε which is implied by (6). To guarantee that holds for all L u we simply define T II,1 = (H 1 L) 1/2 x 1 3ε H 1 1 N γ (16) u := x 2+6ε H 1 N 2γ. We need to check that the condition u x ε N 1/5 of (13) is met. For this, taking into account (8), it suffices to have D x 3 8ε N 4/5 2γ. The worst case occurs when N = x c, leading to the constraint D x 1 4c/5 8ε, which follows from (6). Next, if M satisfies the lower bound M N 1/2 u 1/2 = x 1 3ε H 1/2 1 N 1/2 γ, 8
9 then the upper bound holds; therefore, the bound holds provided that M 1/2 x 1/2+2ε H 1/4 1 N 1/4+γ/2 (17) T I,2 = M 1/2 x 1 3ε H 1 1 N γ (18) H 1 x 6/5 4ε N 1/5 6γ/5. Using (8) again, we see that (18) follows from the inequality D x 11/5 5ε N 4/5 6γ/5. Taking N := x c leads to the restriction D x 1 4c/5 5ε, and this is implied by (6). Next, using the definition of F and the relation LM N one sees that the bound ( ) 1/4 F T I,1 = x 1 3ε H 1 H 1 LM 2 1 N γ (19) holds whenever H 1 M 1/4 d 1/4 x 1 3ε N 1/4 5γ/4. Taking into account (8) and (17) this bound follows from the condition D x 19/7 7ε N 6/7 12γ/7. With N := x c we derive the constraint D x 1 6c/7 7ε, which is a consequence of (6). Our next goal is to establish the bound ( ) N γ k/(2+2k) T II,2 = M (1+k l)/(2+2k) x 1 3ε H1 1 N γ. (20) Ld To begin, we check that the condition F H 1 L is met, or equivalently, that d N γ L 1. Since L N 1/3 for the type II sum, it is enough to have D N γ 1/3 ε. (21) But this follows essentially from (6). Indeed, since c > 1 and γ := 1/c, the inequality (1 136c/157)(2/3 + 2γ) < 2(γ 1/3) 9
10 is easily verified, and we have which implies that D 2/3+2γ ε ( x 1 136c/157) 2/3+2γ ε ( x 2 3ε ) γ 1/3 ε, ( ) x D 1+γ 2 3ε γ 1/3 ε. D On the other hand, we can certainly assume that HN x 1 2ε N 1 γ, for otherwise (11) and (12) are trivial; therefore N 1+γ x 2 3ε d 1, and we have ( ) x D 1+γ 2 3ε γ 1/3 ε ( ) x 2 3ε γ 1/3 ε ( N 1+γ) γ 1/3 ɛ, D d which yields (21). Using the relation LM N, the lower bound M N 2/3 (which follows from L N 1/3 ) and the definition (8), we see that the bound (20) holds if d ν k x 2ν 4νε N ν γν γk+k+2(1 l)/3, where we have put ν := 2k + 2. The exponent of N is negative since k + 2(1 l)/3 < k + 1/3 < ν/2; therefore, the worst case occurs when N = x c, and it suffices to have where µ := D x 1 cµ/3 2ε, (22) ν k 2(1 l)/3 ν k = 3k + 2l + 4 k + 2 With the choice (k, l) := ( 57, 64 ) (which is BA5 ( 1, 1 ) in the notation of 2 2 Graham [2]) we have µ 3 = < , and therefore (22) follows from (6). This proves (20). Combining the bounds (15), (16), (18), (19) and (20), we obtain (14), and this completes the proof of Theorem 1 for c (1, 8635)
11 3.2 Refinement Here, we extend the ideas of 3.1 to show that for any δ > 0, the bound (5) holds for all sufficiently small ε > 0 (depending on δ) under the less stringent condition that 380c 1 D x 441 δ. (23) After this has been done, taking into account that 1 380c 441 > 8 63 whenever c < 77 76, the proof of Theorem 1 for the full range c (1, 77 ) is completed using the 76 sieve argument presented after (6). Following Rivat and Sargos [7, Lem. 2] it suffices to show that (i) The type II bound (12) holds for L in the range u 0 L u 2 0 for some u 0 [N 1/10, N 1/6 ]; (ii) For such u 0, the type I bound (11) holds whenever M N 1/2 u 1/2 0 ; (iii) For such u 0 and any numbers a m, c h C with a m 1, c h 1, the type I bound S I := c h a m e(hl γ m γ d 1 ) x 1 2ε N 1 γ h H m M l L holds whenever u 2 0 L N 1/3. Taking u := x 2+6ε H 1 N 2γ as in 3.1, we put u 0 := max{n 1/10, u}. The condition u 0 N 1/6 follows from the inequality x 2+6ε HN 2γ N 1/6, which in view of (8) is implied by D x 3 7ε N 5/6 2γ. Taking N := x c leads to D x 1 5c/6 7ε, which follows from (23). To verify condition (i) we again use the bound S II ( T II,1 + T II,2 ) H1 Nx ε, 11
12 but we now make the simple choice (k, l) := ( 1, 1 ), so that 2 2 ( ) N γ 1/6 T II,2 = M 1/3. Ld Since u 0 u the bound (16) for T II,1 remains valid in this case. As for T II,2 we have T II,2 N (γ 1)/6 d 1/6 M 1/6, and we now suppose that M Nu 2 0 = min { } N 4/5, N 1 4γ x 4 12ε H1 2. Thus, to obtain (12) we need both of the inequalities to hold: N (γ 1)/6 d 1/6 (N 4/5 ) 1/6 x 1 3ε H1 1 N γ, (24) N (γ 1)/6 d 1/6 (N 1 4γ x 4 12ε H1 2 ) 1/6 x 1 3ε H1 1 N γ. (25) Since H 1 x 1+ε Nd and d D, the first inequality (24) is a consequence of the bound D x 12/5 5ε N 21/25 7γ/5, which is satisfied since N x c and D x 1 21c/25 5ε. Similarly, the inequality (25) follows from D x 18/7 6ε N 6/7 11γ/7, which is satisfied since N x c and D x 1 6c/7 6ε. Condition (ii) also follows from 3.1 in the case that u N 1/10. When u < N 1/10 it suffices to show that max { T I,1, T I,2, T I,3 } x 1 3ε H 1 1 N γ (26) when M N 9/20. Taking into account (8) we see that the bound T I,2 = M 1/2 x 1 3ε H 1 1 N γ (27) holds provided that D x 2 4ε N 31/40 γ. The worst case N = x c leads to D x 1 31c/40 4ε, and since 380 implied by (23). We also know that (19) holds whenever > this is H 1 M 1/4 d 1/4 x 1 3ε N 1/4 5γ/4. 12
13 Taking into account (8) this bound follows from the inequality D x 8/3 3ε N 51/60 5γ/3. With N := x c we derive the constraint D x 1 51c/60 3ε, and as 380 > this a consequence of (23). Combining (15), (19) and (27) we obtain (26) as required. It remains to verify condition (iii). Rather than adapting Rivat and Sargos [7], we quote an abstraction of their method due to Wu [11]. Taking k := 5 in [11, Thm. 2] we have (in Wu s notation) a bound of the form (log x) 1 S I (X 32 H 114 M 147 N 137 ) 1/ However, in place of (X, H, M, N) we use the quadruple (H 1 N γ d 1, M, H 1, L). The triple of exponents (α, β, γ) becomes (γ, 1, γ) in our case, and it is straightforward to check that the various hypotheses of [11, Thm. 2] are satisfied. Applying the theorem, it follows that where S I ( T I,1 + T I,2 + T I,3 + T I,4 + T I,5 + T I,6 + T I,7) x ε, T I,1 := ( H N γ L 23 d 32) 1/174, TI,5 := H 1/2 1 NL 1/2, T I,2 := H 3/4 1 N 1/2+γ/4 L 1/2 d 1/4, T I,6 := H 1 N 1/2 L 1/2, T I,3 := H 5/4 1 N 1/2+γ/4 d 1/4, T I,7 := H 1/2 1 N 1 γ/2 d 1/2. T I,4 := H 1 NL 1, It suffices to show that given that max { T I,1, T I,2, T I,3, T I,4, T I,5, T I,6, T I,7} x 1 3ε N 1 γ (28) N x c, H 1 x 1+ε Nd and N 1/5 L N 1/3. (29) First, we note that the bound T I,1 = ( H N γ L 23 d 32) 1/174 x 1 3ε N 1 γ (30) is equivalent to H N γ L 23 d 32 x ε. 13
14 Using the first two inequalities and the upper bound on L in (29), it suffices to have D 147 x c/3 701ε, which is (23). Similarly, the bounds follow from the inequalities T I,2 = H 3/4 1 N 1/2+γ/4 L 1/2 d 1/4 x 1 3ε N 1 γ, (31) T I,6 = H 1 N 1/2 L 1/2 x 1 3ε N 1 γ, (32) D x 1 5c/6 15ε/2 and D x 1 2c/3 4ε, respectively, and these are easy consequences of (23) since > 5 6 > 2 3. On the other hand, using the first two inequalities and the lower bound on L in (29), we see that both bounds follow from the inequality T I,4 = H 1 NL 1 x 1 3ε N 1 γ, (33) T I,5 = H 1/2 1 NL 1/2 x 1 3ε N 1 γ, (34) D x 1 4c/5 7ε, which is implied by (23) since 380 > 4. Next, using the first two inequalities in (29) and disregarding the bounds on L, it is easy to check that follows from T I,3 = H 5/4 1 N 1/2+γ/4 d 1/4 x 1 3ε N 1 γ (35) D x 1 3c/4 17ε/4, which is implied by (23) since > 3 4. Similarly, is a consequence of the inequality T I,7 = H 1/2 1 N 1 γ/2 d 1/2 x 1 3ε N 1 γ (36) D x 1 c/2 7ε/2, which follows from (23) since > 1 2. Combining the bounds (30), (31), (32), (33), (34), (35) and (36), we obtain (28), and Theorem 1 is proved. 14
15 References [1] R. C. Baker, The squarefree divisor problem, Quart. J. Math. Oxford 45 (1994), [2] S. W. Graham, An algorithm for computing optimal exponent pairs, J. London Math. Soc. (2) 33 (1986), no. 2, [3] S. W. Graham and G. Kolesnik, Van der Corput s method of exponential sums. London Mathematical Society Lecture Note Series, 126. Cambridge University Press, Cambridge, [4] G. Greaves, Sieves in Number Theory. Results in Mathematics and Related Areas (3), 43. SpringerVerlag, Berlin, [5] D. R. HeathBrown, Prime numbers in short intervals and a generalized Vaughan identity, Canad. J. Math. 34 (1982), no. 6, [6] I. I. PiatetskiShapiro, On the distribution of prime numbers in the sequence of the form f(n), Mat. Sb. 33 (1953), [7] J. Rivat and P. Sargos, Nombres premiers de la forme n c, Canad. J. Math. 53 (2001), no. 2, [8] J. Rivat and J. Wu, Prime numbers of the form n c, Glasg. Math. J. 43 (2001), no. 2, [9] O. Robert and P. Sargos, Threedimensional exponential sums with monomials, J. Reine Angew. Math. 591 (2006), [10] J. D. Vaaler, Some extremal problems in Fourier analysis, Bull. Amer. Math. Soc. 12 (1985), [11] J. Wu, On the primitive circle problem, Monatsh. Math. 135 (2002), no. 1,