Elem. der Math, 64 (2009), p. 45-52.

Formulas generating prime numbers under Cramér’s conjecture Bakir FARHI [email protected] Abstract Under Cramer’s conjecture concerning the prime numbers, we prove that for any ξ > 1, there exists a real number A = A(ξ) > 1 for which the formula ξ bAn c (where b.c denotes the integer part function) gives a prime number for any positive integer n. Under the same conjecture, we also prove that for any ε > 0, there exists a real number B = B(ε) > 0 for which the formula bB · n!2+ε c gives a prime number for any sufficiently large positive integer n.

MSC: 11A41. Keywords: Prime numbers; Cramér’s conjecture.

1

Introduction

Throughout this article, we let bxc denote the integer part of a giving real number x, we let (pn )n∈N denote the sequence of all prime numbers and we set ∆pn := pn+1 − pn for any n ∈ N. Further, if A is a subset of R and x is a real number, we let A + x denote the subset of R defined by: A + x := {a + x|a ∈ A}. n In [5], Mills proved the existence of an absolute constant A > 1 for which bA3 c is a prime number for any positive integer n and in [7], Wright proved the existence α of an absolute constant α > 0 for which the infinite sequence bαc, b2α c, b22 c, . . . is composed of prime numbers. Let us describe the method used by these two authors. Both authors start from an upper bound of ∆pn in terms of pn . Such upper bound allows to construct an increasing function h (more or less elementary, according to the used upper bound of ∆pn ) such that between any two consecutive terms of the sequence (h(n))n , there is at least one prime number. Setting fn := h◦· · ·◦h (where h is applied n times), they deduce from the last fact, the existence of a real constant A for which the sequence (bfn (A)c)n is consisted of prime numbers. In this method, Wright uses the upper bound ∆pn ≤ pn which is nothing else than 5 +ε Bertrand’s postulate and Mills uses Ingham’s upper bound: ∆pn ≤ pn8 (holding for any sufficiently large n compared to a given ε > 0, see [4]). The functions h which 1

derive from these upper bounds are h(x) = 2x for Wright and h(x) = x3 for Mills. The results of [5] and [7] then follow. Notice that the more the upper bound of ∆pn is refined, the more the function h is small and the more the obtained sequence of prime numbers grows slowly (remark for instance that the sequence of Mills grows slower than the Wright’s one). From this fact, in order to obtain a sequence of prime numbers which grows more slow again, we must use upper bounds more refined for ∆pn . But up to now even the powerful 1/2 Riemann hypothesis gives only the estimate ∆pn = O(pn log pn ). A famous conjecture (which is little too strong than this last estimate) states that between two consecutive squares, there is always a prime number (see e.g., [2]). So, according to this conjecture, the function h(x) = x2 is admissible in the method described above, n consequently there exists a constant B > 1 for which bB 2 c is a prime number for any positive integer n. So that admitting that conjecture, we obtain a sequence of prime numbers growing slower than the Mills’one. By leaning on heuristic and probabilistic arguments, Cramér [1] was leaded to conjecture that we have ∆pn = O(log2 pn ); further it’s known that ∆pn = O(log pn ) cannot hold (see [6]). Thus, by taking in the method described above h(x) = c log2 x (c > 0), we obtain (via Cramér’s conjecture) sequences of prime numbers having explicit form and growing much slower than the Mills’one. The inconvenient in this application is that the explicit form bfn (A)c in question is not elementary, because fn doesn’t have (in this case) a simple expression in terms of n. To cope with this problem, we were leaded to generalize Mills’method by considering instead of one function h, a sequence of functions (hm )m and in this situation fn is rather the composition of n functions h0 , . . . , hn−1 . This fundamentally allows to give for fn any desired form, then if we set hn := fn+1 ◦ fn−1 , we have only to check whether it’s true that for any n and any x sufficiently large (compared to n), the interval [hn (x), hn (x + 1) − 1[ contains at least one prime number or not. In the affirmative case, we will deduce the existence of a real A for which the formula bfn (A)c gives a prime number for any positive integer n (see Theorem 1 and its proof). Under a conjecture less strong than the Cramér’s one, we derive from this generalization two new types of explicit formulas giving prime numbers. We also give other applications of our main result (outside the subject of prime numbers) and we conclude this article by some open questions related to the obtained results.

2

Results

The basic result of this paper is the following: Theorem 1 Let I =]a, b[ (with a, b ∈ R, a < b) be an open interval of R, n0 be a non-negative integer and (fn )n≥n0 be a sequence of real functions which are differentiable and increasing on I. 2

f0

We assume that the functions n+1 (n ≥ n0 ) are nondecreasing on I and that for all fn0 x ∈ I, the numerical sequence (fn (x))n≥n0 is increasing. We also assume that there exists a real function g, nondecreasing on its domain of definition and satisfying: g ◦ fn+1 (x) ≤

0 fn+1 (x) fn0

(∀n ≥ n0 and ∀x ∈ I).

(1)

Then for any sequence of integers (un )n∈N , satisfying: lim sup un = +∞, n→+∞

un+1 − un ≤ g(un ) − 1

(for all n sufficiently large)

(2)

and whose one at least of the terms un (n satisfying (2)) belongs to the set fn0 (I) ∩ (fn0 (I) − 1), there exists a real A ∈ I for which the sequence (bfn (A)c)n≥n0 is an increasing subsequence of (un )n . Proof. By shifting if necessary the sequence of functions (fn )n≥n0 , we may assume that n0 = 0. Also, by shifting if necessary the sequence (un )n , we may assume that we have (more generally than (2)): un+1 − un ≤ g(un ) − 1

(20 )

(for all n ∈ N).

We begin this proof by some remarks and preliminary notations which allow to understand better the situation of the theorem. Giving n ∈ N, since the function fn is assumed differentiable (so continuous) and increasing on I =]a, b[, then it is a bijection from I into fn (I) =] lim fn (x), lim fn (x)[= x→a

x→b

]λn , µn [, where λn := lim fn (x) and µn := lim fn (x) (λn and µn belong to R). x→a

x→b

Now, let us introduce the following functions:

hn : ]λn , µn [ −→ ]λn+1 , µn+1 [ hn := fn+1 ◦ fn−1

(∀n ∈ N).

Giving n ∈ N, since (from the hypothesis of the theorem), the functions fn and fn+1 are differentiable and increasing on I, then the function hn is differentiable and increasing on ]λn , µn [. On the other hand, the hypothesis of the theorem concerning the increase of the numerical sequences (fn (x))n (x ∈ I) amounts to: hn (x) > x

(∀n ∈ N and ∀x ∈]λn , µn [).

(3)

Next, let us show that for any n ∈ N, the function hn is convex on ]λn , µn [. To do so, we verify that the derivative h0n of each function hn (n ∈ N) is nondecreasing on the interval ]λn , µn [. For any n ∈ N, we have: µ 0 ¶ 0 fn+1 ◦ fn−1 fn+1 0 −1 0 −1 0 0 −1 = ◦ fn−1 . hn = (fn+1 ◦ fn ) = (fn ) · fn+1 ◦ fn = 0 −1 fn ◦ fn fn0 3

f0

Since (from the hypothesis of the theorem), the function n+1 is nondecreasing on fn0 −1 I and the function fn is increasing on fn (I) =]λn , µn [ (because fn is increasing on I), then (as a composite of two nondecreasing functions), the function h0n is nondecreasing on ]λn , µn [. So the function hn is effectively convex on ]λn , µn [ as we claimed it to be. The rest of the proof is splitted into three steps. 1st Step: We are going to show that we have: g ◦ hn (y) ≤ hn (y + 1) − hn (y)

(∀n ∈ N and ∀y ∈]λn , µn − 1[).

(4)

(We will see further that the interval ]λn , µn − 1[ is never empty). Let n ∈ N and y ∈]λn , µn − 1[ fixed and set x := fn−1 (y). The convexity of hn on ]λn , µn [, proved above, implies that we have: hn (u) ≥ h0n (t)(u − t) + hn (t)

(for all t, u ∈]λn , µn [).

By taking in this last inequality t = y and u = y + 1, we obtain: hn (y + 1) − hn (y) ≥ h0n (y) µ 0 ¶ 0 fn+1 fn+1 0 = (x) (because hn = 0 ◦ fn−1 and x = fn−1 (y)) 0 fn fn ≥ g ◦ fn+1 (x) (from the hypothesis (1) of the theorem) = g ◦ fn+1 ◦ fn−1 (y) = g ◦ hn (y). The relation (4) follows. 2nd Step: We are going to construct an increasing sequence (kn )n∈N of non-negative integers such that the subsequence of (un )n with general term vn = ukn satisfies:    vn ∈]λn , µn − 1[ and (∀n ∈ N).   h (v ) ≤ v n n n+1 < hn (vn + 1) − 1 We proceed by induction as follows: • We pick k0 ∈ N such that uk0 ∈ f0 (I) ∩ (f0 (I) − 1) =]λ0 , µ0 − 1[. Notice that the existence of such integer k0 is an hypothesis of the theorem. • If, for some n ∈ N, an integer kn ∈ N is chosen such that ukn ∈]λn , µn − 1[, let: Xn := {k ∈ N | k > kn and uk ≥ hn (ukn )} . From the hypothesis lim supn→+∞ un = +∞, this subset Xn of N is nonempty, it thus admits a smallest element which we take as the choice of kn+1 . So, we have: kn+1 > kn , ukn+1 ≥ hn (ukn ) and kn+1 − 1 6∈ Xn . 4

We claim that the facts “kn+1 > kn ” and “kn+1 − 1 6∈ Xn ” imply: ukn+1 −1 < hn (ukn ).

(5)

Indeed: either kn+1 = kn + 1 in which case we have ukn+1 −1 = ukn < hn (ukn ) (from (3)); or kn+1 > kn + 1, that is kn+1 − 1 > kn , but since kn+1 − 1 6∈ Xn , we are forced to have (also in this case) ukn+1 −1 < hn (ukn ) as required. It follows that we have: ukn+1 ≤ ukn+1 −1 + g(ukn+1 −1 ) − 1 (from (20 )) < hn (ukn ) + g ◦ hn (ukn ) − 1 (by using (5) and the non-decrease of g) ≤ hn (ukn + 1) − 1 (from (4)). Hence: ukn+1 < hn (ukn + 1) − 1. We thus have: hn (ukn ) ≤ ukn+1 < hn (ukn + 1) − 1. Finally, since the function hn takes its values in ]λn+1 , µn+1 [, then the last double inequality does show that ukn+1 ∈]λn+1 , µn+1 − 1[. This ensures the good functioning of this induction process which gives the required sequence (kn )n . Note also that the resulted subsequence (vn )n of (un )n is increasing because for any n ∈ N, we have: vn+1 ≥ hn (vn ) > vn (according to (3)). 3rd step (conclusion): To conclude this proof, we will show the existence of a real A ∈ I for which we have for any n ∈ N : vn = bfn (A)c. To do so, we introduce two real sequences (xn )n and (yn )n , with terms in I, which we define by: xn := fn−1 (vn ) and yn := fn−1 (vn + 1)

(∀n ∈ N).

Since the functions fn are increasing, we have xn < yn for all n ∈ N. On the other hand, we have for any n ∈ N: −1 −1 xn = fn−1 (vn ) = fn+1 ◦ hn (vn ) ≤ fn+1 (vn+1 ) = xn+1

and −1 −1 yn = fn−1 (vn + 1) = fn+1 ◦ hn (vn + 1) > fn+1 (vn+1 + 1) = yn+1 . −1 (Where in these last relations, we have just use the facts that fn+1 is increasing and hn (vn ) ≤ vn+1 < hn (vn + 1) − 1). This shows that the sequences (xn )n and (yn )n are respectively nondecreasing and decreasing. The intervals [xn , yn ] (n ∈ N) are thus nested inclosed intervals of R; consequently their intersection is nonempty (according to Cantor’s intersection theorem). Pick A an arbitrary real belonging to this intersection, that is: xn ≤ A ≤ yn for all n ∈ N (in particular A ∈ I). Actually A even satisfies: xn ≤ A < y n (∀n ∈ N),

5

because if A = ym for some m ∈ N, we will have (from the decreasing of the sequence (yn )n ): A > ym+1 , contradicting the inequality A ≤ ym+1 . It follows from the increase of the functions fn that we have: fn (xn ) ≤ fn (A) < fn (yn )

(∀n ∈ N),

that is: vn ≤ fn (A) < vn + 1

(∀n ∈ N).

Then (since vn is an integer for all n ∈ N): bfn (A)c = vn

(∀n ∈ N).

The proof is complete.

¥

Remarks. Mills’theorem [5] can be find again by applying Theorem 1 for I =]1, +∞[, n n0 = 0, fn (x) = x3 (∀n ∈ N and ∀x ∈ I), g(x) = x2/3 (∀x > 0) and (un )n the sequence of the prime numbers. In this application, we verify the relation (1) of Theorem 1 by a simple calculus and we deduce the relation (2) of the same theorem from Ingham’s estimate noted in the introduction. The remaining hypothesis of Theorem 1 are immediately verified. Wright’s theorem [7] can be also find again by applying Theorem 1 for I =]0, +∞[, n0 = 0, (fn )n∈N the sequence of functions defined on I by: f0 = IdI and fn+1 = 2fn (∀n ∈ N), g(x) = (log 2)x (∀x ∈ R) and (un )n the sequence of the prime numbers. To verify the relation (1) of Theorem 1, we have just to remark that for any n ∈ N: 0 fn+1 = (log 2)fn+1 . As for the relation (2), it is (in this application) a consequence of fn0 the prime numbers theorem, but it can be obtained by using Chebyshev’s elementary arguments (see e.g., [3]). The remaining hypothesis of Theorem 1 are immediately verified. N. B. It is interesting to remark that in the two above applications of Theorem 1, the sequence of functions (hn )n introduced in the proof of the latter (hn := fn+1 ◦ fn−1 ) is constant. Indeed, for the first application, we have hn (x) = x3 (∀n ∈ N) and for the second one, we have hn (x) = 2x (∀n ∈ N). As explained in the introduction, the fact to be able to take (hn )n not constant is the crucial point of our approach. In the following, we are going to give some applications of Theorem 1 in which this sequence (hn )n is not constant. If we admit the following Conjecture (which is less strong than the Cramér’s one [1]), we obtain two new types of explicit sequences of prime numbers which growing much slower than the ones of Mills and Wright. Conjecture 2 There exists an absolute constant k > 1 such that: ¢ ¡ ∆pn = O (log pn )k . Under this Conjecture, we obtain by applying Theorem 1 the two following Corollaries:

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Corollary 3 Under Conjecture 2, there exists for all real number ξ > 1, a real number ξ A = A(ξ) > 1 for which the sequence (bAn c)n≥1 is an increasing sequence of prime numbers. Proof. Let ξ > 1 be fixed, k > 1 be an admissible constant in Conjecture 2 and a > 1 be a real chosen so that: (log x)k+1 ≤ x1/2 (n + 1)k+1 ≤ a

1 ξ−1 n 2

(∀x > a)

(6)

(∀n ≥ 1).

(7) 2(k+1)

k+1

x) (a exists because limx→+∞ (logx1/2 = 0 and limn→+∞ (n + 1) nξ−1 = 1). ξ We apply Theorem 1 for I =]a, +∞[, n0 = 1, fn (x) = xn (∀n ≥ 1, ∀x ∈ I), g(x) = (log x)k+1 (∀x > 1) and (un )n the sequence of the prime numbers. Let us verify the hypothesis of Theorem 1: The functions fn are clearly increasing and differentiable on I. We have fn0 (x) = ξ ξ ξ f0 nξ xn −1 (∀n ≥ 1, ∀x ∈ I), then: n+1 (x) = ( n+1 )ξ x(n+1) −n (∀n ≥ 1, ∀x ∈ I), f0 n n

f0

which shows that the functions n+1 (n ≥ 1) are nondecreasing on I. Further, if fn0 x ∈ I is fixed, the numerical sequence (fn (x))n≥1 is clearly increasing. Now, we have for any integer n ≥ 1 and for any x ∈ I: g ◦ fn+1 (x) = (n + 1)ξ(k+1) (log x)k+1 1

ξ−1

≤ a 2 ξn ≤ x

x1/2

(from (6) and (7))

ξnξ−1

(since x > a and ξnξ−1 > 1) ξ

≤ x(n+1) −n f0 ≤ n+1 (x). fn0

ξ

(since ξnξ−1 ≤ (n + 1)ξ − nξ )

The relation (1) of Theorem 1 follows. Next, the relation (2) of Theorem 1 (related to this application) follows immediately from Conjecture 2 (admitted in this context). Finally, fn0 (I) ∩ (fn0 (I) − 1) =]a, +∞[ does contains prime numbers as large as we please. The hypothesis of Theorem 1 are thus all satisfied, and consequently we can apply this latter to the present situation. Corollary 3 follows from this application. ¥ Corollary 4 Assume that Conjecture 2 is true and let k > 1 be an admissible constant in that conjecture. Then, for any positive real number ε, there exists an integer n0 = n0 (ε, k) ≥ 1 and a positive real number B = B(ε, k) such that the sequence (bB · n!k+ε c)n≥n0 is an increasing sequence of prime numbers. Proof. let ε > 0 fixed. From Conjecture 2 (admitted with the constant k > 1), there exists ck > 0 for which we have: pn+1 − pn ≤ ck (log pn )k 7

(∀n ∈ N).

(8)

We apply Theorem 1 for I =]1, 2[, n0 ≥ 2 an integer (depending on k and ε) which we pick large enough that we have: ck ((k + ε)(n + 1) log(n + 1) + log 2)k + 1 ≤ (n + 1)k+ε

(∀n ≥ n0 ),

(9)

fn (x) = n!k+ε x (∀n ≥ n0 , ∀x ∈ I), g(x) = ck (log x)k + 1 (∀x > 1) and (un )n the sequence of the prime numbers. In this situation, we can easily verify that all the hypothesis of Theorem 1 are satisfied. We just note that the relation (1) follows from (9), the relation (2) follows from (8) and the last hypothesis of Theorem 1 concerning the sequence (un )n = (pn )n is a consequence of Bertrand’s postulate. Corollary 4 follows from this application. ¥ Apart from the context of the prime numbers, we have the following: Corollary 5 Let (un )n∈N be a sequence of integers such that: 1 ≤ lim sup(un+1 − un ) < +∞. n→+∞

Then: (1) For any positive real number λ, there exists a real number A > 1 for which the sequence (bλAn c)n≥1 is an increasing subsequence of (un )n . (2) For any real number A > lim supn→+∞ (un+1 − un ) + 1, there exists a positive real number λ for which the sequence (bλAn c)n≥1 is an increasing subsequence of (un )n . Some open problems related to the preceding study: We ask (under or without Cramér’s Conjecture) the following questions: (1) Does there exist a real number A > 1 for which bAn c is a prime number for any positive integer n? (This corresponds to the case ξ = 1 which is excluded from Corollary 3). (2) More generally than (1), does there exist a couple of real numbers (λ, A), with λ > 0, A > 1, for which bλAn c is a prime number for any positive integer n? (This is related to Corollary 5). (3) Does there exist a real number B > 1 for which bB · n!2 c is a prime number for any sufficiently large non-negative integer n? (This is related to Corollary 4).

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References [1] H. Cramér. On the order of magnitude of the difference between consecutive prime numbers. Acta Arith, 2 (1936), p. 23-46. [2] R. K. Guy. Unsolved problems in Number Theory. 2 nd ed., Springer-Verlag, New York, (1994). [3] G. H. Hardy & E. M. Wright. The theory of numbers. Oxford Univ. Press, London, 5th ed, (1979). [4] A. E. Ingham. On the difference between consecutive primes. Quarterly Journal of Mathematics, 8 (1937), p. 255-266. [5] W. H. Mills. A prime-representing function. Bull. Amer. Math. Soc, 53 (1947), p. 604. [6] E. Westzynthius. Über die Verteilung der Zahlen die zu den n ersten Primzahlen teilerfremd sind. Comm. Phys. Math. Helsingfors, (5), 25 (1931), p. 1-37. [7] E. M. Wright. A prime-representing function. Amer. Math. Monthly, 58 (1951), p. 616-618.

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