aa r X i v : . [ m a t h . N T ] S e p SOME RESULTS ON A CONJECTURE OF HARDY AND LITTLEWOOD
CHRISTIAN AXLER
Abstract.
Let m and n be positive integers with m, n ≥
2. The second Hardy-Littlewood conjecturestates that the number of primes in the interval ( m, m + n ] is always less than or equal to the numberof primes in the interval [2 , n ]. Based on new explicit estimates for the prime counting function π ( x ),we give some new ranges in which this conjecture holds. Introduction
The prime counting function π ( x ) denotes the number of primes less or equal to x . In 1872, Lionnet[12] raised the question whether the inequality(1.1) π (2 n ) − π ( n ) ≤ π ( n )holds for every integer n ≥
2. This means that for each integer n ≥ n, n ] contains atmost as many prime numbers as the interval [2 , n ]. A first progress concerning this question was done byLandau [11, p. 215–216]. He used the Prime Number Theorem, i.e. π ( x ) = x log x + O (cid:18) x log x (cid:19) as x → ∞ , to show that (1.1) holds for every sufficiently large positive integer n . In 1923, Hardy andLittlewood [8] conjectured that lim sup n →∞ ( π ( x + n ) − π ( n )) ≤ π ( x )for every x ≥
2. From here it has been derived the well known conjecture (in the following denoted byHLC) that(1.2) π ( m + n ) ≤ π ( m ) + π ( n ) ∀ m, n ∈ N \ { } . Clearly, the HLC is a generalization of Lionnet’s question (1.1). Although the HLC could neither beproven nor disproved in general so far, some special cases can be shown. As a consequence of theirexplicit estimates for π ( x ), Rosser and Schoenfeld [16] stated without proof that(1.3) π (2 x ) − π ( x ) ≤ π ( x )for every real x ≥
3. A detailed proof was finally given by Kopetzky and Schwarz [10]. If we combine(1.3) with π (4) − π (2) = π (2), it turns out that Lionnet’s inequality (1.1) indeed holds for every integer n ≥
2. Erd¨os [6] reported that Ungar verified the HLC for every pair of integers ( m, n ) satisfying2 ≤ min( m, n ) ≤
41. One year later, Schinzel and Sierpi´nski [18] could show that the inequality is fulfilledfor every pair of integers ( m, n ) with 2 ≤ min( m, n ) ≤ ≤ min( m, n ) ≤ m, n ) satisfying(1.4) 2 ≤ min( m, n ) ≤ . The next result is due to Panaitopol [14, Theorem 1]. He showed that the HLC is true for every pair ofintegers ( m, n ) satisfying m, n ≥ m ≤ n ≤ m. In [4, Proposition 3], Dusart improved the result of Panaitopol by showing that the HLC is true for everypair of positive real numbers ( x, y ) satisfying x, y ≥ x ≤ y ≤ x. Using explicit estimates for the prime counting function π ( x ), we find the following improvement. Date : October 1, 2019.2010
Mathematics Subject Classification.
Primary 11N05; Secondary 11A99.
Key words and phrases. distribution of prime numbers, Hardy-Littlewood conjecture, Riemann hypothesis.
Theorem 1.1.
Let m and n be integers satisfying m, n ≥ and m/ ≤ n ≤ m . Then we have π ( m + n ) ≤ π ( m ) + π ( n ) . In 1975, Udrescu [21] found the following generalization. Under the assumption that n satisfies εm ≤ n ≤ m , where ε is a real number with 0 < ε ≤
1, he showed that the HLC holds for every sufficiently largepositive integer m . Dusart [4] showed that Udrescu’s result holds for every integer m ≥ e . / log(1+ ε ) . Wegive the following improvement. Theorem 1.2.
Udrescu’s result holds for every integer m ≥ e √ . / log(1+ ε ) . In [15], Panaitopol [15] used explicit estimates for the prime counting function π ( x ) to get that theHLC is true for all positive integers m, n ≥ π ( m ) ≤ n ≤ m. Since π ( x ) ∼ x/ log x as x → ∞ , the last result yields an improvement of Theorem 1.1 for all sufficientlylarge values of m . In this paper, we find the following refinement of (1.6). Theorem 1.3.
Let c = 0 . . Then we have π ( m + n ) ≤ π ( m ) + π ( n ) for allintegers m ≥ n ≥ with n ≥ c m/ log m . In the case where m + n ≤ , we can use some recent results concerning the distance of π ( x ) andthe logarithmic integral li( x ), which is defined for every real x > x ) = Z x d t log t = lim ε → (cid:26)Z − ε d t log t + Z x ε d t log t (cid:27) , to get the following result. Theorem 1.4.
Let c = 2(1 − log 2) = 0 . . . . . Then we have π ( m + n ) ≤ π ( m ) + π ( n ) for all integers m ≥ n ≥ satisfying m + n ≤ and n ≥ √ m (cid:18) − c log m + c (cid:19) . Finally, we find the following result which depends on the correctness of the Riemann hypothesis.
Theorem 1.5.
Let c = 1 / (4 π ) . If the Riemann hypothesis is true, then π ( m + n ) ≤ π ( m ) + π ( n ) forall integers m ≥ n ≥ satisfying n ≥ c √ m log m log( m log m ) . On a result of Segal
In 1962, Segal [20, Theorem I] obtained the following inequality condition involving only prime numberswhich is equivalent to the HLC. Here, as usual, p r denotes the r th prime number. Lemma 2.1 (Segal) . The
HLC is true if and only if (2.1) p k ≥ p k − q + p q +1 − for all integers k, q satisfying k ≥ and ≤ q ≤ ( k − / . Then, Segal [20, Theorem II] used this equivalence to get the following result.
Lemma 2.2 (Segal) . If the
HLC is false for some positive integer m + n , then the smallest such valueof m + n is the smallest value of p k for which (2.1) is false. He used a computer to see that the inequality (2.1) holds for every positive integer k ≤ p k ≤
101 081. Now it follows from Lemma 2.2 that the HLC holds for all integers m, n ≥ m + n ≤
101 081. In 2001, Panaitopol [14] improved Lemma 2.1 by showing the following
Lemma 2.3 (Panaitopol) . The
HLC is true if and only if the inequality (2.1) holds for all integers k, q satisfying k ≥ and ≤ q ≤ ( k − / . Using Lemmata 2.2 and 2.3 and a computer, Panaitopol [14] found that the HLC is true for all integers m, n ≥ m + n ≤ p . Extending this computation, we get the following Proposition 2.4.
Let N = 1 . × . Then the HLC holds for all integers m, n ≥ satisfying m + n ≤
39 708 229 123 = p N . OME RESULTS ON A CONJECTURE OF HARDY AND LITTLEWOOD 3 A Proof of Theorem 1.1
First, we set(3.1) f c ( t ) = t log t − − c/ log t . By [1, Corollary 3.5], we have π ( t ) ≥ f ( t ) for every t ≥ b a real number with b ∈ (1 ,
2) andlet B a positive real number so that π ( t ) ≤ f b ( t ) for every x ≥ B . Further, let r and s be positive realnumbers with r ≥ s ≥
1. We set(3.2) λ b ( r, s ) = ( b − r + 1) − log( s + 1) log s r log(1 + r ) + 2 log( s + 1) + 12 (cid:18) log r − log (cid:18) r (cid:19)(cid:19) and η b ( r, s ) = r log r − log s − (1 + log( s + 1) log s ) log(1 + r ) − br log sr log(1 + r ) + log( s + 1) + log (cid:18) r (cid:19) log s. Then we get the following result.
Proposition 3.1.
Let r and s be real numbers with r ≥ s ≥ . Let χ b ( r, s ) = λ b ( r, s ) + η b ( r, s ) and ϕ b ( r, s ) = sgn( χ b ( r, s )) + 12 · χ b ( r, s ) . Then we have π ( x + y ) ≤ π ( x ) + π ( y ) for every pair of real numbers ( x, y ) satisfying x ≥ y ≥ , (3.3) x ≥ max (cid:26) exp( λ b ( r, s ) + p ϕ b ( r, s )) , r, B /r (cid:27) , and x/r ≤ y ≤ x/s .Proof. By (1.5), we can assume that r ≥ s ≥ h ( x, y ) = π ( x ) + π ( y ) − π ( x + y ). We need to showthat h ( x, y ) ≥
0. First, we note that(3.4) log (cid:18) xy (cid:19) − b log( x + y ) + 1log y ≥ log 110 − ≥ . Since log( x/y ) ≥ log s , we have(3.5) 1log y − − y ≥ log s log y log x − − x . From (3.3), it follows that (log x − λ b ( r, s )) ≥ λ b ( r, s ) + η b ( r, s ). Substituting the definition of η b ( r, s )into the last inequality, we see thatlog x − λ b ( r, s ) log x − log (cid:18) r (cid:19) log r ≥ r log r − log s − (1 + log( s + 1) log s ) log(1 + r ) − br log sr log(1 + r ) + log( s + 1) . Let κ ( r, s ) = r log(1 + 1 /r ) + log( s + 1). Then we can use (3.2) to get that the last inequality is equivalentto κ ( r, s ) log (cid:16) x + xr (cid:17) log xr − b ( r + 1) log xs + r log xr ≥ ( b −
1) log s − (1 + log( s + 1) log s ) log (cid:16) x + xr (cid:17) . Since x/r ≤ y ≤ x/s , we get κ ( r, s ) r − b (1 + r )log( x + y ) + 1log( x + y ) ≥ ( b −
1) log sr log( x + y ) log y − r log y − log( s + 1) log sr log y . Hence κ ( r, s ) r − b (1 + r )log( x + y ) + 1log x ≥ b log sr log( x + y ) log y − log sr log y − r log y − log( s + 1) log sr log y . Now we substitute the definitoin of κ ( r, s ) to obtain the inequalitylog (cid:18) r (cid:19) − b log( x + y ) + 1log x + 1 r (cid:18) log (cid:18) xy (cid:19) − b log( x + y ) + 1log y (cid:19) (cid:18) s log y (cid:19) ≥ . Therefore,(3.6) log (cid:16) yx (cid:17) − b log( x + y ) + 1log x + 1 r (cid:18) log( s + 1) − b log( x + y ) + 1log y (cid:19) (cid:18) s log y (cid:19) ≥ . CHRISTIAN AXLER
Next, we note that y ≥ x/r ≥ x + y ≥ x (1+1 /r ) ≥ B . Hence h ( x, y ) ≥ f ( x )+ f ( y ) − f b ( x + y ),where f c ( t ) is defined as in (3.1). Setting g c ( t ) = log t − − c/ log t , we see that h ( x, y ) ≥ x log(1 + yx ) − b log( x + y ) + x g ( x ) g b ( x + y ) ! + y log(1 + xy ) − b log( x + y ) + y g ( y ) g b ( x + y ) ! . Now we can use (3.4) and (3.5) to get the inequality h ( x, y ) ≥ x log(1 + yx ) − b log( x + y ) + x + r (log(1 + xy ) − b log( x + y ) + y )(1 + log s log y ) g ( x ) g b ( x + y ) ! . Finally it suffices to apply the inequality (3.6). (cid:3)
Now we use Propositions 2.4 and 3.1 to give the following proof of Theorem 1.1.
Proof of Theorem 1.1.
We set b = 1 .
15. By [2, Corollary 1], we can choose B = 38 284 442 297. Inaddition, we substitute the following explicit values for r and s into Proposition 3.1 to get π ( x + y ) ≤ π ( x ) + π ( y ) for every x ≥ x and x/r ≤ y ≤ x/s , where x is equal to the least integer greater than orequal to the right-hand side of (3.3): r . . . . s . . . . . x
38 284 409 814 38 284 393 330 38 284 407 670 38 284 394 575 38 284 419 151 r . . . . . s . . . . . x
38 284 398 522 38 284 417 850 38 284 399 116 38 284 426 596 38 284 405 535 r . . . . . s . . . . x
38 284 440 640 38 284 412 784 38 284 406 728 38 284 305 355 38 083 977 941In particular, we see that π ( m + n ) ≤ π ( m )+ π ( n ) for every m ≥
38 284 440 640 and m/ ≤ n ≤ m/ m ≤
38 284 440 640 and m/ ≤ n ≤ m/ m + n ≤ (1 + 1 / m ≤
39 708 229 123 and theresult follows from Proposition 2.4. The remaining case where m, n ≥ m/ ≤ n ≤ m is a directconsequence of (1.4) and (1.5). (cid:3) A Proof of Theorem 1.2
In this section, we use explicit estimates for the prime counting function π ( x ) to give the followingproof of Theorem 1.2. Proof of Theorem 1.2. If ε ∈ [1 / , ε ∈ (0 , / m, n ≥ εm ≤ n ≤ m and m ≥ e √ . / log(1+ ε ) . Then m ≥
168 527 259 431. Hence,log(1 + ε ) ≥ . m ≥ . m + 1 . m . Hence log( m + n ) − − m + n ) − . m − . m ≥ log m − − m − . m − . m . Now we can use [2, Corollary 3] to see that(4.1) m log( m + n ) − − m + n ) − . ( m + n ) − . ( m + n ) ≤ π ( m ) . Since log 2 > . / log m + 14 . / log m , we get(4.2) log( m + n ) − − m + n ) − . ( m + n ) − . ( m + n ) ≥ log n − − n . OME RESULTS ON A CONJECTURE OF HARDY AND LITTLEWOOD 5
Note that the function f ( x ) = xe √ . / log(1+ x ) is decreasing on the interval (0 , / n ≥ εm ≥ f (1 / ≥
86 424 235. If we combine the inequality (4.2) with Corollary 3.5 of [1], it turnsout that the inequality(4.3) n log( m + n ) − − m + n ) − . ( m + n ) − . ( m + n ) ≤ π ( n )holds. By [2, Corollary 1], we have π ( m + n ) ≤ m + n log( m + n ) − − m + n ) − . ( m + n ) − . ( m + n ) and it suffices to apply (4.1) and (4.3). (cid:3) A Proof of Theorem 1.3
Let k be a positive integer and ε be positive real number. By Panaitopol [13], there exist positive realnumbers a , . . . , a k and two positive real numbers α k and β k = β k ( ε ) so that(5.1) π ( x ) ≥ x log x − − P kj =1 a j log j x ( x ≥ α k )and(5.2) π ( x ) ≤ x log x − − P kj =1 a j log j x − ε log k x ( x ≥ β k ) . Further, let γ k = γ k ( ε ) be the smallest positive integer so thatlog 2 ≥ ε log k x + k X j =1 a j log j x for every x ≥ γ k . Then we obtain the following result. Proposition 5.1.
Let k be a positive integer and ε, c be positive real numbers with c > ε . Then π ( x + y ) ≤ π ( x ) + π ( y ) for all real numbers x, y ≥ with x ≥ max { α k , β k , γ k , exp( k p c / (2( c − ε ))) } and max (cid:26) , cx log k x (cid:27) ≤ y ≤ x. Proof.
Since x ≥ exp( k p c / (2( c − ε ))), we have c log k x − c k x ≥ ε log k x . Using the inequality log(1 + t ) ≥ t − t /
2, which holds for every t ≥
0, we see thatlog( x + y ) − log x ≥ log (cid:18) c log k x (cid:19) ≥ ε log k x . If we combine the last inequality with (5.1), it turns out that(5.3) x log( x + y ) − − P kj =1 a j log j ( x + y ) − ε log k ( x + y ) ≤ x log x − − P kj =1 a j log j x ≤ π ( x ) . On the other hand, we have y ≤ x and x ≥ γ k . Hence(5.4) y log( x + y ) − − P kj =1 a j log j ( x + y ) − ε log k ( x + y ) ≤ y log y − . By Dusart [3, p. 55], we have π ( t ) ≥ t/ (log t −
1) for every t ≥ y log( x + y ) − − P kj =1 a j log j ( x + y ) − ε log k ( x + y ) ≤ π ( y ) . By (5.2), we have π ( x + y ) ≤ x + y log( x + y ) − − P kj =1 a j log j ( x + y ) − ε log k ( x + y ) and it suffices to apply (5.3) and (5.5). (cid:3) CHRISTIAN AXLER
Proof of Theorem 1.3.
Let k = 2. We set a = 1 and a = 2 .
85. By [2, Corollary 3], we can choose α = 38 099 531. Further, we set ε = 0 . x ≥ β = 14 000 264 036 190 262. A simple calculation shows that γ = 23. Now let c = c . Substituting these values into Proposition 3.1, it turns out that the inequality π ( m + n ) ≤ π ( m ) + π ( n ) holds for all integers m ≥ n ≥ m ≥
14 000 264 036 190 263 and n ≥ cm/ log m . If m ≤
14 000 264 036 190 262, the claim follows from Theorem 1.1. (cid:3) A Proof of Theorem 1.4
First, we note some results of Dusart [5] concerning the distance of π ( x ) and li( x ). Proposition 6.1 (Dusart) . For every real x with ≤ x ≤ , we have (6.1) π ( x ) ≤ li( x ) , and for every real x satisfying ≤ x ≤ , we have (6.2) li( x ) − √ x log x ≤ π ( x ) . Proof.
See [5, Lemma 2.2]. (cid:3)
Now we use this result to find the following proof of Theorem 1.4.
Proof of Theorem 1.4.
By Theorems 1.1 and 1.3, it suffices to consider the case where n satisfies2 √ m ≤ n ≤ m × min (cid:26) , c log m (cid:27) , where c is given as in Theorem 1.3. If m ≤
39 687 876 365, we get m + n ≤ (1 + / m ≤
39 708 229 123and the result follows from Proposition 2.4. So we can assume that m ≥
39 687 876 366. Using (6.1), we seethat π ( m + n ) ≤ li( m + n ). Now we can use the mean value theorem to see that π ( m + n ) ≤ li( m )+ n/ log m .Applying (6.2) to this inequality, we get π ( m + n ) ≤ π ( m ) + 2 √ m log m + n log m , which is equivalent to(6.3) π ( m + n ) ≤ π ( m ) + 2 √ m log m + n log n − − n (log( m/n ) + 1)log m (log n − . Since m ≥
39 687 876 366, we have n ≥
887 293.So we can apply the inequality including π ( x ) given in [3,p. 55] to (6.3) and get(6.4) π ( m + n ) ≤ π ( m ) + π ( n ) + 2 √ m log m − n (log( m/n ) + 1)log m (log n − . In order to prove the theorem, we consider the following three cases.
Case √ m log m/ log log m ≤ n ≤ c m/ log m .In this first case, the inequality (6.4) implies that(6.5) π ( m + n ) ≤ π ( m ) + π ( n ) + 2 √ m log m − n (2 log log m − c )log m (log m − m + c ) , where c = log( c ) − − . . . . . The assumption n ≥ √ m log m/ log log m implies that n (2 log log m − c )log m (log m − m + c ) ≥ √ m log m . Applying this to (6.5), we obtain the inequality π ( m + n ) ≤ π ( m ) + π ( n ). Case
2. 2 √ m (1 + 4 log log m/ log m ) ≤ n ≤ √ m log m/ log log m .Here, the inequality (6.4) implies that(6.6) π ( m + n ) ≤ π ( m ) + π ( n ) + 2 √ m log m − n (log( √ m log log m/ log m ) + 1)log m (log( √ m log m/ log log m ) − , We have n ≥ √ m (cid:18) m log m (cid:19) ≥ √ m × log( √ m log m/ log log m ) − √ m log log m/ log m ) + 1 . Applying this to (6.4), we see that the inequality π ( m + n ) ≤ π ( m ) + π ( n ) holds. OME RESULTS ON A CONJECTURE OF HARDY AND LITTLEWOOD 7
Case
3. 2 √ m ≤ n ≤ √ m (1 + 4 log log m/ log m ).Let r ( x ) = 1 + 4 log log x/ log x . In this latter case, a simple calculation shows that n ≥ √ m ≥ √ m × log(2 √ mr ( m )) − √ m/ (2 r ( m ))) + 1 ≥ √ m × log n − m/n ) + 1 . Now we apply this to (6.4) to get the required inequality.
Case
4. 2 √ m (1 − c / (log m + c )) ≤ n ≤ √ m .In this latter case, we have n ≥ √ m (cid:18) − c log m + c (cid:19) = 2 √ m × log(2 √ m ) − √ m/
2) + 1 ≥ √ m × log n − m/n ) + 1 . Finally, we apply this to (6.4) to arrive at the end of the proof. (cid:3) A Proof of Theorem 1.5
Under the assumption that the Riemann hypothesis is true, Schoenfeld [19, Corollary 1] showed that | π ( x ) − li( x ) | ≤ √ x π log x for every x ≥ Proposition 7.1 (Dusart) . If the Riemann hypothesis is true, then | π ( x ) − li( x ) | ≤ √ x π log (cid:18) x log x (cid:19) for every real x ≥ . We use Proposition 7.1 to find the following proof of Theorem 1.5.
Proof of Theorem 1.5. If m ≤ × , then m + n ≤ m ≤ and the result follows directly fromTheorem 1.4. So it suffices to consider the case where m ≥ × . In order to prove the theorem, weconsider the following three cases. Case n ≥ c √ m log m .If n ≥ c m/ log m , where c is given as in Theorem 1.3, the result follows directly from Theorem 1.3.Hence we can assume that c √ m log m ≤ n ≤ c m/ log m . By Proposition 7.1, we have π ( m + n ) ≤ li( m + n ) + f ( m + n ), where f ( t ) = (1 / (8 π )) √ t log( t/ log t ). Now we use the mean value theorem to get π ( m + n ) ≤ li( m ) + n log m + f ( m ) + n π √ m (cid:18) log (cid:18) m log m (cid:19) + 2 (cid:19) . Next we apply Proposition 7.1 to obtain the inequality π ( m + n ) ≤ π ( m ) + n log m + 2 f ( m ) + n π √ m (cid:18) log (cid:18) m log m (cid:19) + 2 (cid:19) , which is equivalent to π ( m + n ) ≤ π ( m ) + 2 f ( m ) + n π √ m (cid:18) log (cid:18) m log m (cid:19) + 2 (cid:19) + n log n − − n log( m/n ) + 1log m (log n − . Since m ≥ × , we have n ≥ c √ m log m ≥
52 511 298 895 885. So we can apply the inequalityincluding π ( x ) given in [3, p. 55] to the last inequality and get(7.1) π ( m + n ) ≤ π ( m ) + π ( n ) + 2 f ( m ) + n π √ m (cid:18) log (cid:18) m log m (cid:19) + 2 (cid:19) − n log( m/n ) + 1log m (log n − . Since c √ m log m ≤ n ≤ c m/ log m , we see that π ( m + n ) ≤ π ( m ) + π ( n ) + 2 f ( m ) + c √ m π log m (cid:18) log (cid:18) m log m (cid:19) + 2 (cid:19) − n m − c log m (log m − m + c ) , where c = log( c ) − − . . . . . Now we substitute the definition of f ( t ) to get π ( m + n ) ≤ π ( m ) + π ( n ) + g ( m, n ), where g ( m, n ) = √ m π (cid:18)(cid:18) c m (cid:19) log (cid:18) m log m (cid:19) + c m (cid:19) − n m − c log m (log m − m + c ) . CHRISTIAN AXLER
Clearly, it suffices to show that g ( m, n ) ≤
0. This inequality is equivalent to(7.2) n ≥ √ m π (cid:18)(cid:18) c m (cid:19) log (cid:18) m log m (cid:19) + c m (cid:19) log m (log m − m + c )2 log log m − c . Since − log log m + c m log (cid:18) m log m (cid:19) + c m ≤ , the inequality n ≥ c √ m log m implies (7.2) and we get π ( m + n ) ≤ π ( m ) + π ( n ). Case c √ m log m log( m log m ) ≤ n ≤ c √ m log m .From (7.1), it follows that the inequality π ( m + n ) ≤ π ( m ) + π ( n ) holds if(7.3) n ≥ (cid:18) c √ m log (cid:18) m log m (cid:19) + c π (cid:18) log (cid:18) m log m (cid:19) + 2 (cid:19) log m (cid:19) log m (log( c √ m log m ) − √ m/ ( c log m )) + 1 . We have c π (cid:18) log (cid:18) m log m (cid:19) + 2 (cid:19) log m ≤ c √ m log log m and log( c √ m log m ) − √ m/ ( c log m )) + 1 ≤ m log m . So if n fulfills the inequality n ≥ c √ m log m log( m log m ), we get the inequality (7.3). Hence we have π ( m + n ) ≤ π ( m ) + π ( n ). Case c √ m log m log( m log m ) ≤ n ≤ c √ m log m log( m log m ).We use (7.1) to see that the inequality π ( m + n ) ≤ π ( m ) + π ( n ) holds if(7.4) n ≥ (cid:18) c √ m log (cid:18) m log m (cid:19) + c π (cid:18) log (cid:18) m log m (cid:19) + 2 (cid:19) log m log( m log m ) (cid:19) h ( m ) log m, where h ( m ) = log( c √ m log m log( m log m )) − √ m/ ( c log m log( m log m ))) + 1 . Note that c π (cid:18) log (cid:18) m log m (cid:19) + 2 (cid:19) log m log( m log m ) ≤ c √ m log log m and h ( m ) ≤ m/ log m . So the inequality n ≥ c √ m log m log( m log m ) implies the inequality(7.4) and we arrive at the end of the proof. (cid:3) Appendix: The Incompatibility of the HLC and the Prime k -tuples Conjecture To formulate the Prime k -tuples Conjecture, we first introduce the following definition. Definition. A k -tuple of distinct integers b , . . . , b k is admissible if for each prime p , there is somecongruence class mod p which contains none of the b i . Prime k -tuples Conjecture. Let b , . . . , b k be an admissible k -tuple of integers. Then there existinfinitely many positive integers n for which all of the values n + b , . . . , n + b k are prime.Remark. The Prime k -tuples Conjecture is a special case of Schinzel’s Hypothesis H [18, p. 188].In order to show that the HLC and the Prime k -tuples Conjecture are incompatible, Hensley andRichards [9] used the following function which was introduced by Schinzel and Sierpi´nski [18, p. 201]. Definition.
Let the function ρ ∗ : N → N be defined by ρ ∗ ( m ) = max n ∈ N |{ k ∈ N | n < k ≤ m + n, gcd( k, m !) = 1 }| . This function describes the maximum number of positive integers in each interval ( n, m + n ] that arerelatively prime to all positive integers less than or equal to m .Under the assumption that the Prime k -tuples Conjecture is true, Schinzel and Sierpi´nski [18, pp.204–205] found the identity(8.1) ρ ∗ ( m ) = lim sup n →∞ ( π ( m + n ) − π ( n )) . OME RESULTS ON A CONJECTURE OF HARDY AND LITTLEWOOD 9
Hensley and Richards [9, p. 380] proved that for every real number ε there exists a m ( ε ) so that ρ ∗ ( m ) − π ( m ) ≥ (log 2 − ε ) × m log m for every m ≥ m ( ε ). In particular, the last inequality gives(8.2) lim m →∞ ( ρ ∗ ( m ) − π ( m )) = ∞ . So if the Prime k -tuples Conjecture is true, we can combine (8.1) and (8.2) to see that for every sufficientlylarge values of m there exist infinitly many positive integers n so that the inequality π ( m + n ) > π ( m ) + π ( n )holds which contadicts the HLC. Acknowledgement
I would like to express my great appreciation to Thomas Leßmann for writing a C++ program in orderto verify Proposition 2.4. I would also like to thank Pierre Dusart whose PhD thesis inspired me to dealwith this subject. Furthermore, I thank R. for being a never-ending inspiration.
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