#A61 INTEGERS 14 (2014) SHORT EFFECTIVE INTERVALS CONTAINING PRIMES

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1 #A6 INTEGERS 4 (24) SHORT EFFECTIVE INTERVALS CONTAINING PRIMES Habiba Kadiri Department of Mathematics and Computer Science, University of Lethbridge, Lethbridge, Canada habiba.kadiri@uleth.ca Allysa Lumley Department of Mathematics and Computer Science, University of Lethbridge, Lethbridge, Canada allysa.lumley@uleth.ca Received: /9/4, Revised: 7/23/4, Accepted: 9/25/4, Published: /3/4 Abstract We prove that if x is large enough, namely x x, then there exists a prime between x( ) and x, where is an e ective constant computed in terms of x.. Introduction In this article, we address the problem of finding short intervals containing primes. In 845 Bertrand conjectured that for any integer n > 3, there always exists at least one prime number p with n < p < 2n 2. This was proven by Chebyshev in 85, using elementary methods. Since then other intervals of the form (kn, (k + )n) have been investigated. We refer the reader to [] for k = 2 and to [2] for k = 3. Assuming that x is arbitrarily large, the length of intervals containing primes can be drastically reduced. To date, the record is held by Baker, Harman, and Pintz [2] as they prove that there is at least one prime between x and x + x.525+". This is an impressive result since under the Riemann Hypothesis the exponent.525 can only be reduced to.5. On the other hand, maximal gaps for the first primes have been checked numerically up to 4 8 by Oliveira e Silva et al. [4]. In particular, they find that the largest prime gap before this limit is 476 and occurs at = e The purpose of this article is to obtain an e ective result of the form: for all x x, there exists > such that the interval (x( ), x) contains at least one prime. In 976 Schoenfeld s Theorem 2 of [8] gave this for x = and = In 23 Ramaré and Saouter improved on Schoenfeld s method by using a smoothing argument. They also extended the computations to many other values for x ([6, Theorem 2 and

2 INTEGERS: 4 (24) 2 Table ]). In [9], the first author generalized this theorem to primes in arithmetic progression and applied this to Waring s seven cubes problem. Here, our theorem improves [6] by making use of a new explicit zero-density for the zeros of the Riemann zeta function: Theorem.. Let x 4 8 be a fixed constant and let x > x. Then there exists at least one prime p such that ( )x < p < x, where is a constant depending on x and is given in Table 2. In Section 2, we prove a general theorem (Theorem 2.7) which provides conditions for intervals of the form (( )x, x) to contain a prime. In Section 3, we apply this theorem to compute explicit values for. We present an example of the numerical improvement this theorem allows, for instance when x = e 59, Ramaré and Saouter [6] found that the interval gap was given by = In [5, page 74], Helfgott mentioned an improvement of Ramaré using Platt s latest verification of the Riemann Hypothesis [5]: = Our Theorem. leads to = We now mention an application to the verification of the Ternary Goldbach conjecture. This conjecture was known to be true for su ciently large integers (by Vinogradov), and Liu and Wang [] prove it for all integers n e 3. On the other hand, the conjecture was verified for the first values of n. In [6, Corollary ], Ramaré and Saouter verified it for n apple Very recently, Oliveira e Silva et. al. [4, Theorem 2.] extended this limit to n apple In [5, Proposition A..], Helfgott applied the above result on short intervals containing primes ( = ) and found n apple This allowed him to complete his proof [5, 6] of the Ternary Goldbach conjecture for the remaining integers. Here our main theorem gives: Corollary.2. Every odd number larger than 5 and smaller than is the sum of at most three primes = As of today, Helfgott and Platt [7] have announced a verification up to Proof of Theorem. We recall the definition of the classical Chebyshev functions: (x) = log p, (x) = ( if n = p k for some k 2 N, (n), with (n) = otherwise. papplex napplex For each x, we want to find the largest > such that, for all x > x, there exists a prime between x( ) and x. This happens as soon as (x) (x( )) >.

3 INTEGERS: 4 (24) Introduction of Parameters We list here the parameters we will be using throughout the proof. m integer with m 2, apple u apple., = mu and apple apple., apple a apple /2, = ( + a)( + ( a)) e u, e 38, x = e u ( + ( a)) x = e u ( + ( a)), y = ( + a) = x. () 2.2. Smoothing the Di erence (x) (y) We follow here the smoothing argument of Ramaré and Saouter [6]. Let f be a positive function integrable on (, ). We denote kfk = (f, a) = Z Z a and I,u, = kfk f(t)dt, (2) Z f(t)dt + Z a f(t)dt, (3) ( (e u ( + t)) (( + t))) f(t)dt. (4) Note that for all a apple t apple a, (e u ( + t)) (( + t)) apple (x) (y). We integrate with the positive weight f and obtain: Z a a Z ( (e u ( + t)) (( + t))) f(t)dt apple ( (x) (y)) a a f(t)dt. (5) We extend the left integral to the interval (, ) and use a Brun-Titchmarsh inequality to control the primes on the extremities (, a) and ( a, ) of the interval (see [6, page 6, line -5] or [3, Theorem 2]): Z t2(,a)[( ( (e u ( + t)) a,) Note that [6] uses the slightly larger bound (( + t))) f(t)dt apple 2( + )(e u log(e u ) ) log((e u (f, a). (6) )) 2.4u log (f, a). log(u)

4 INTEGERS: 4 (24) 4 Combining (5) and (6) gives I,u, apple ( (x) (y)) R a a f(t)dt +2(+ )(e u ) log(eu ( + )) kfk log((e u )) (f, a) kfk. (7) Thus (x) (y) > when I,u, 2( + )(e u ) log(eu ( + )) (f, a) log((e u >. (8) )) kfk It remains to establish a lower bound for I,u,. To do so, we first approximate (x) with (x). This will allow us to translate our problem in terms of the zeros of the zeta function. We use approximations proven by Costa in [3, Theorem 5]: Lemma 2.. Let x e 38. Then.999 p x + 3p x < (x) (x) <. p x + 3p x. (9) We have that for all < t <, ( (e u ( + t)) (e u ( + t))) ( (( + t)) (( + t))) < p p +.e u/ /6 ( + ) /6 (e u/3 ) <! p, () where we can take, under our assumptions (), We denote J,u, = kfk It follows from () that Z! = () ( (e u ( + t)) (( + t))) f(t)dt. (2) I,u, J,u,! p. (3) Note that [6] used older approximations from [8], which lead to! =.325. To summarize, we want to find conditions on m,, u, a so that J,u,! p 2( + )(e u ) log(eu ( + )) (f, a) log((e u >. (4) )) kfk

5 INTEGERS: 4 (24) 5 We are now left with evaluating J,u,, which we shall do by relating it to the zeros of the Riemann zeta function through an explicit formula An Explicit Inequality for J,u, Lemma 2.2. [6, Lemma 4] Let 2 apple b apple c, and let g be a continuously di erentiable function on [b, c]. We have Z c b (u)g(u)du = Z c b ug(u) % Z c u % + b Z c b % g(u)du log 2 2 log( u 2 ) g(u)du. (5) We apply this identity to g(t) = f e u t, b = e u, c = e u ( + ), and g(t) = f t, b =, c = ( + ), respectively. It follows that J,u, = (eu ) kfk 2kfk Z Z ( + t)f(t)dt kfk % Z (e u% ) % ( + t) % f(t) dt % log (e u ( + t)) 2 log ( ( + t)) 2 f(t)dt. Observe that the last term is at least u 2. We obtain J,u, (e u ) R ( + t)f(t)dt kfk % R (e u% ) ( + t)% f(t)dt (e u Re% )% kfk u 2(e u ) 2. (6) We obtain some small savings by directly computing the first term, whereas [6, Equation (3)] uses the following bound in (6) instead: R ( + t)f(t)dt kfk u e u. Let s be a complex number. We let G m,,u (s) be the summand

6 INTEGERS: 4 (24) 6 and we rewrite inequality (6) as J,u, (e u ) G m,,u() R G m,,u (s) = (eus ) ( + t)s f(t)dt (e u, (7) )s kfk % G m,,u (%) Re% u 2(e u ) 2. (8) Since the right term increases with, we can replace with for that this is also the case for For simplicity we let! (e u ) p 2( + ( + )) (f, a) )log(eu log((e u. )) kfk. Note = m,,u, = % G m,,u (%) Re%. (9) The following Proposition gives a first inequality in terms of the zeros of the Riemann zeta function and conditions on m, u,, a (and thus ) so that (x) (x( )) > : Proposition 2.3. Let m, u,, a,, satisfy (). If and u G m,,u () m,,u, 2(e u ) 2! (e u ) /2 2 (f, a)( + ) log(e u ( + )) kfk log( (e u >, (2) )) then there exists a prime number between x( ) and x. We are now going to make this lemma more explicit by providing computable bounds for the sum over the zeros m,,u, Evaluating G m,,u Let f be an m-admissible function over [, ]. We recall the properties it entitles according to the definition of [6]: f is an m-times di erentiable function, f (k) () = f (k) () = for apple k apple m, f, f is not identically.

7 INTEGERS: 4 (24) 7 For k =,..., m and for s = + i a complex number with > and apple apple, we let R F k,m, = ( + t)+k f (k) (t) dt. (2) kfk We provide finer estimates than [6] for G m,,u. Observe that and e us s e us s Z = Z u e xs dx apple Z u e x dx = eu, (22) apple eu +, (23) ( + t) s f(t)dt apple k k F k,m,. (24) We easily deduce bounds for G m,,u (s) by combining (22) and (24) with respectively k =, k =, and k = m: G m,,u (s) apple F,m, e u (e u ), (25) G m,,u (s) apple F,m, e u (e u ), (26) and G m,,u (s) apple F m,m, e u (e u ) m m. (27) Lastly by combining (23) and (24) with k = m we obtain e u + G m,,u (s) apple F m,m, (e u ) m. (28) m Zeros of the Riemann Zeta Function We denote each zero of the zeta funtion by % = + i, the number of zeros in the rectangle < <, < < T, by N(T ), and the number of those zeros in the rectangle < <, < < T, by N(, T ). We assume that we have the following information. Theorem A numerical verification of the Riemann Hypothesis: There exists H > 2 such that if ( + i ) = at apple apple and apple apple H, then = /2. 2. A direct computation of some finite sums over the first zeros:

8 INTEGERS: 4 (24) 8 Let < T < H and S > satisfy apple N = N(T ), (29) and < applet =/2 apple S. (3) < applet =/2 3. A zero-free region: There exists R >, a constant, such that ( + it) does not vanish in the region and t 2. (3) R log t 4. An estimate for N(T ): There exist positive constants a, a 2, a 3 such that, for all T 2, N(T ) P (T ) apple R(T ), where P (T ) = T 2 log T T , R(T ) = a log T + a 2 log log T + a 3. (32) 5. An upper bound for N(, T ): Let 3/5 < <. Then there exist constants c, c 2, c 3 such that, for all T H, N(, T ) apple c T + c 2 log T + c 3. (33) Note that [6] did not use any information of the type (3), (3), or (33). Instead they used (29), the fact that all nontrivial zeros satisfied <, and the classical bound (32) for N(T ) as given in [7, Theorem 9]. Our improvement will mainly come from using a new zero-density of the form of (33) Evaluating the Sum Over the Zeros m,,u, We assume Theorem 2.4. We split the sum m,,u, vertically at heights = (so as to use the symmetry with respect to the x-axis) and consider G m,,u ( + i ) = G m,,u ( + i ) + G m,,u ( i ). We then split at = H (so as to take advantage of the fact that all zeros below this horizontal line satisfy = /2), and again at = T and = T (where T will be

9 INTEGERS: 4 (24) 9 chosen between T and H), and consider: = Gm,,u (/2 + i ) /2, (34) < applet = Gm,,u (/2 + i ) /2, (35) T < applet and 2 = G m,,u (/2 + i ) /2. (36) T < appleh For the remaining zeros (those with > H), we make use of the symmetry with respect to the critical line, and we split at = for some fixed > /2 (we will consider 9/ apple apple 99/ for our computations). We denote 3 = >H =/2 + >H G m,,u (/2 + i ) /2 /2< apple Gm,,u ( + i ) + G m,,u ( + i ), (37) 4 = Gm,,u ( + i ) + G m,,u ( + i ). (38) >H < < As a conclusion, we have m,,u, = (39) We state here some preliminary results (see [4, Equations (2.8), (2.9), (2.2), (2.2), (2.26)]). Lemma 2.5. Let T, H, R, be as in Theorem 2.4. Let m 2, >, and T between T and H. We define S (T ) = 2 + q(t ) log T p T T 2R(T ) log, (4) T 2 T! S 2 (m, T ) = 2 + q(t + m log T 2 + m log H 2 ) m 2 T m m 2 H m + 2R(T ) T m+, (4) + m log H 2 )! S 3 (m) = 2 + q(h) S 4 (m, ) = S 5 (, m, ) = c + m m 2 H m + c 2 log H + H c + c 2 log H + c 3 H H + + 2R(H), (42) Hm+ c 3 + c 2 m + H H m, (43) c + c 2 R (log H) 2 H 2 log ( mr log )(log H) 2 H m. (44)

10 INTEGERS: 4 (24) We assume Theorem 2.4. Then apple S (T ), (45) T < applet m+ apple S 2(m, T ), (46) T < appleh >H >H < < Moreover, if log < R m(log H) 2, then >H < < m+ apple S 3(m), (47) m+ apple S 4(m, ). (48) R log R apple S m+ 5 (, m, ) log H. (49) Lemma 2.6. Let m,, satisfy (). We assume Theorem 2.4. If log < R m(log H) 2, then m,,u, apple B (m, ) /2 + B (m,, T ) /2 + B 2 (m,, T ) /2 + B 3 (m, ) + + B 4 (, m,, ) R log(h) + B 42 (m,, ) + R log H, (5) where the B i s are respectively defined in (5), (54), (58), (6), (62), and (63). Proof. We investigate two ways to evaluate and. For, we can either combine (26) with (3), which computes P < applet, or (25) with (29), which computes P < applet. We denote with We obtain (m, ) = B (m, ) = min( (m, ), 2 (m, )), (5) 4F,m, (e u/2 + ) S and 2 (m, ) = 4F,m, (e u/2 + ) N. (52) apple B (m, ) /2. (53) For, we can either combine (26) with the bound (45) for P T < applet, or (25) with the bound (32) for N(T ) from Theorem 2.4. We denote B (m,, T ) = min( (m,, T ), 2 (m,, T )), (54)

11 INTEGERS: 4 (24) with (m,, T ) = 4F,m, (e u/2 + ) S (T ), and 2 (m,, T ) = 4F,m, e u/2 + (N(T ) N ). (55) We obtain apple B (m,, T ) /2. (56) It follows from (28) and (46) that with 2 apple B 2 (m,, T ) /2, (57) B 2 (m,, T ) = We use (28) to bound G in 3 : 3 apple 2F m,m, (e u ) m >H /2apple apple 2F m,m, (e u/2 ) m S 2(m, T ). (58) (e u + ) + (e u( ) + ) m+. Note that since log > u, then (e u + ) + (e u( ) + ) increases with /2. Moreover, we use (47) to bound the sum P >H (m+), and we obtain /2 where 3 apple B 3 (m,, ) + B 3 (m,, ), (59) B 3 (m,, ) = 2F m,m, m e u + e u S 3(m). (6) For 4 we again use (28) to bound G and the fact that + increases with. Since apple R log and > H, we obtain 4 apple 2(eu + )F m,m, (e u ) m >H < < R log R log H + + m+ >H < <. m+ We apply (48) and (49) to bound the above sums over the zeros and obtain with R log(h) 4 apple B 4 (, m,, ) + B 42 (m,, ) + R log H, (6) B 4 (, m,, ) = 2(eu + )F m,m, (e u ) m S 5(, m, ), (62) B 42 (m,, ) = 2(eu + )F m,m, (e u ) m S 4(m, ). (63) Note that G m,,u () = F,m,. Finally we apply Proposition 2.3 and Lemma 2.6.

12 INTEGERS: 4 (24) Main Theorem Theorem 2.7. Let m, u,, a,,, and x satisfy (). Let T, H, R, be as in Theorem 2.4. We assume Theorem 2.4. If and F,m, B (m, ) /2 B (m,, T ) /2 B 2 (m,, T ) /2 B 3 (m,, ) B 3 (m,, ) B 4 (, m,, ) u B 42 (m,, ) + R log H then there exists a prime number between x( ) and x. R log H 2(e u ) 2! (e u ) /2 2 (f, a)( + ) log(e u ( + )) kfk log( (e u >, (64) )) 3. Computations 3.. Introducing the Smooth Weight f We choose the same weight as [6]. That is f m (t) = (4t( t)) m if apple t apple, and otherwise. Faber and Kadiri proved in [4] that a primitive form of f m provided a near optimal weight to estimate (x). Thus we believe that the above weight should also be near optimal to evaluate (y) (x) when y is close to x. We recall [6, Lemma 6]: We now provide estimates for F k,m, as defined in (2). kf m k = 22m (m!) 2 (2m + )!, (65) kf (m) m k 2 = 22m m! p 2m +. (66) Lemma 3.. Let m 2, >, and < <.We define (m, ) = (2m + )! 2 2m (m!) 2, (m, ) = ( + )2 (2m + )! 2 2m (m!) 2, s ( + ) (m, ) = 2m+3 (2m + )! (2m + 3) m! p 2m +.

13 INTEGERS: 4 (24) 3 Then apple F,m, apple +, (67) (m, ) apple F,m, ( ) apple (m, ), (68) F m,m, ( ) apple (m, ). (69) Proof. Inequalities (67) follow trivially from the fact that apple ( + t) apple +. To bound F,m,, we note that Since f m(t) has the same sign as kf mk = Z /2 kf mk kf m k apple F,m, apple ( + )2 kf mk kf m k. f m(t)dt Z /2 2t, we have f m(t)dt = 2f m (/2) f m () f m () = 2. This together with (65) achieves (68). Lastly, for F m.m,, we apply (66) together with the Cauchy-Schwarz inequality: q R q R F m,m, ( ) apple ( + s t)2(m+) (m) dt f m (t) 2 dt ( + ) = 2m+3 kf m (m) k 2. kf m k (2m + 3) kf m k Note that, while F,m, and F,m, can be easily computed as integrals, this is not the case for F m,m,. The following observation helps us to compute F m,m, directly. We recognize in the definition of f m (m) the analogue of Rodrigues formula for the shifted Legendre polynomials: f (m) m (t) = 4 m m!p m ( 2t), where P m (x) is the m th Legendre polynomial, and m m m + k P m ( 2t) = ( ) m ( t) k. k k k= For each P m ( 2t), we denote r j,m, with j =,..., m, its m + roots. Since P m ( 2t) alternates sign between each root, we have F m,m, = R ( + t)m+ P m ( kfk 2t) dt = m Z rj+ ( ) j ( + t) m+ P m ( 2t)dt, kfk r j and GP-Pari is able to compute quickly this sum of polynomial integrals. j=

14 INTEGERS: 4 (24) Explicit Results About the Zeros of the Riemann Zeta Function We provide here the latest values for the constants appearing in Theorem 2.4: Theorem A numerical verification of the Riemann Hypothesis (Platt [5]): H = A direct computation of some finite sums over the first zeros (using A. Odlyzko s list of zeros): For T = 32 49, then N = N(T ) = 2 52, and S = A zero-free region (Kadiri [8, Theorem.]): R = An estimate for N(T ) (Rosser [7, Theorem 9]): a =.37, a 2 =.443, a 3 = An upper bound for N(, T ) (Kadiri []): For all T H, where the c i s are given in Table. N(, T ) apple c T + c 2 log T + c 3, Table : N(, T ) apple c T + c 2 log T + c 3. c c 2 c Note that [7, Theorem 9] was recently improved by T. Trudgian in [9, Corollary ] with a =., a 2 =.275, a 3 = Our results are valid with either Rosser s or Trudgian s bounds.

15 INTEGERS: 4 (24) Understanding the Contribution of the Low Lying Zeros We assume Theorem 3.2 and that m m = 5, < = 2 8, and T > t = 9 (7) (this would be consistent with the values we choose in Table 2). We observe that B (m, ) = 2 (m, ) and B (m,, T ) = 2 (m, ). where 2 and 2 are defined in (52) and (55), respectively. In other words, it turns out that we obtain a smaller bound for the sum over the small zeros ( < < T ) by using N(T ) directly instead of evaluating P < <T. This essentially comes from the fact that our choice of parameters ensures F,m, S F,m, N and F,m, S (T ) F,m, (N(T ) N ). We first prove the inequality Proof. We denote w = q(t ) log 2 w 3 = 2 + q(t (T ) ) 2 v = 2 =.59..., v 2 = S (t) N(t) = , w 2 = log(2 ) log t c. (7) t 2 + q(t ) = , + log(t ) log(2 ) + 2R(T ) = , T log(2 ) 2 v 4 = a 2 =.443, v 5 = a = and = , v 3 = a =.37, S (t) = w (log t) 2 +w 2 log t+w 3, P (t)+r(t) = v t log t+v 2 t+v 3 log t+v 4 log log t+v 5. From (4) and Theorem 3.2 (d), we have S (t) N(t) S (t) P (t) + R(t) = w (log t) 2 + w 2 log t + w 3 v t log t + v 2 t + v 3 log t + v 4 log log t + v 5. Since t > t = 9, we deduce the bound where c = w + S (t) N(t) w2 log t + w3 (log t ) 2 log t c, (72) t.758. (73) v + v3 v4 log log t t + t log t + v5 t log t

16 INTEGERS: 4 (24) 6 We now establish that max ( +, + 2, 2 + ) apple We make use of Lemma 3., to provide estimates for the F k,m, s in (72), and of the assumptions (7) on m,, T. Proof. We have 4 F,m, ( + ) ( ) = (S e u/2 + S (T )) F,m, N(T ) + > 4( + )N(T ) (2m + )! S e u/ m (m!) 2 ( + ) P (t ) + R(t ) + c log t t > 4( + )N(T ) ( ) >. e u/2 + We have S ( + 2 ) ( ) = F,m, N F,m, > 4( + )N (2m + )! S e u/ m (m!) 2 ( + ) N Finally, ( 2 + ) ( ) = > 4( + )(N(T ) N ) e u/2 + 4 e u/2 + > 4( + )N e u/2 + (574 ) >. 4 F,m, S e u/2 (T ) F,m, (N(T ) N ) +! (2m + )! 2 2m (m!) 2 ( + ) > 4( + )(N(T ) N ) e u/2 + S (t ) ( S(t)t c log t N ) (.3737 ) > Table of Computations The values for T and a given in the next table are rounded down to the last digit. We start the computations at x 4 8, that is log x

17 INTEGERS: 4 (24) 7 Table 2: For all x x, there exists a prime between x( ) and x. log x m T a log(4 8 ) Verification of the Ternary Goldbach Conjecture Proof of Corollary.2. Let N = 4 8. We follow Oliveira e Silva, Herzog and Pardi s argument in [4] where the authors computed all the prime gaps up to 4 8. From Table 2, we have that for x = e 6 and = , there exists at least one prime in the interval (x x/, x]. This one has length Then N = and we may infer that the gap between consecutive primes up to N can be no larger than N (since N / = N). The corollary follows by using all the odd primes up to N to extend the minimal Goldbach partitions of 4, 6,..., N up to N (the method of computation is explained in [4, Section ]). We also note that N + 2 = 2 + (N 29) and N + 4 = 33 + (N 39), where 2, 33, N 29, and N 39 are all prime. Thus, there is at least one way to write each odd number greater than 5 and smaller than N as the sum of at most 3 primes. Acknowledgments. The authors would like to thank Nathan Ng and Adam Felix for their comments on this article.

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