Piatetski-Shapiro primes from almost primes

Piatetski-Shapiro primes from almost primes Roger C. Baker Department of Mathematics, Brigham Young University Provo, UT 84602 USA baker@math.byu.edu William D. Banks Department of Mathematics, University of Missouri Columbia, MO 65211 USA bankswd@missouri.edu Zhenyu V. Guo Department of Mathematics, University of Missouri Columbia, MO 65211 USA zgbmf@mail.missouri.edu Aaron M. Yeager Department of Mathematics, University of Missouri Columbia, MO 65211 USA amydm6@mail.missouri.edu Abstract Let be the floor function. In this paper, we show that for any fixed c (1, 77 76 ) 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

1 Introduction The Piatetski-Shapiro 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 Piatetski-Shapiro, 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 r-almost 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 Piatetski-Shapiro 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

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 1 3.1 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

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 < 8635 8568. Then, for any fixed c (1, 8635) and α ( 8, 1 136c ), the inequality (3) 8568 63 157 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 := 0.124820 and g := 63 8. 4

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). 8568 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

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

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 Heath-Brown [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. 1367 1368] 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

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

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

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 126 126 BA5 ( 1, 1 ) in the notation of 2 2 Graham [2]) we have µ 3 = 803 927 < 136 157, 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). 8568 10.

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

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 > 31 441 40 this is H 1 M 1/4 d 1/4 x 1 3ε N 1/4 5γ/4. 12

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 > 51 441 60 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 ) 1/174 +. 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 179 1 N 114+32γ 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 179 1 N 114+32γ L 23 d 32) 1/174 x 1 3ε N 1 γ (30) is equivalent to H 179 1 N 60+206γ L 23 d 32 x 174 522ε. 13

Using the first two inequalities and the upper bound on L in (29), it suffices to have D 147 x 147 380c/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 380 441 > 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 441 5 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 380 441 > 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 380 441 > 1 2. Combining the bounds (30), (31), (32), (33), (34), (35) and (36), we obtain (28), and Theorem 1 is proved. 14

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