On generalized Ramanujan primes Christian Axler January 29, 2014 Abstract 4 InthispaperweestablishseveralresultsconcerningthegeneralizedRamanujanprimes. Forn∈N 1 and k ∈ R>1 we give estimates for the nth k-Ramanujan prime which lead both to generalizations 0 and to improvements of the results presently in the literature. Moreover, we obtain results about 2 the distribution of k-Ramanujan primes. In addition, we find explicit formulae for certain nth k- n Ramanujan primes. As an application, we prove that a conjecture of Mitra, Paul and Sarkar [9] a concerningthenumberofprimesincertainintervalsholdsforeverysufficientlylargepositiveinteger. J 8 2 1 Introduction ] T Ramanujan primes, named after the Indian mathematician Srinivasa Ramanujan, were introduced by N Sondow [17] in 2005 and have their origin in Bertrand’s postulate. h. Bertrand’s Postulate. For each n∈N there is a prime number p with n<p≤2n. t a Bertrand’spostulatewasproved,forinstance,byTchebychev[19]andbyErdo¨s[6]. In1919,Ramanujan m [13] proved an extension of Bertrand’s postulate by showing that [ x π(x)−π ≥1(respectively2,3,4,5,...) 1 2 v (cid:16) (cid:17) 9 for every 7 x≥2(respectively11,17,29,41,...). 1 7 Motivated by the fact π(x)−π(x/2) → ∞ as x → ∞ by the Prime Number Theorem (PNT), Sondow . [17] defined the number R ∈N for each n∈N as the smallest positive integer such that the inequality 1 n 0 π(x)−π(x/2)≥n holds for every x≥Rn. He calledthe number Rn the nth Ramanujan prime, because 4 R ∈P for every n∈N, where P denotes the set of prime numbers. n 1 This can be generalizedas follows. Let k ∈(1,∞). Again, the PNT implies that π(x)−π(x/k)→∞ : v as x→∞ and Shevelev [16] introduced the nth k-Ramanujan prime as follows. i X Definition. Let k >1 be real. For every n∈N, let r a R(k) =min{m∈N|π(x)−π(x/k)≥nfor everyx≥m}. n This number is prime and it is called the nth k-Ramanujan prime. Since R(2) =R for every n∈N, the n n numbers R(k) are also called generalized Ramanujan primes. n In 2009, Sondow [17] showed that R ∼p (n→∞), (1) n 2n where p denotes the nth prime number. Further, he proved that n R >p (2) n 2n for every n ≥ 2. In 2011, Amersi, Beckwith, Miller, Ronan and Sondow [1] generalized the asymptotic formula (1) to k-Ramanujan primes by showing that R(k) ∼p (n→∞). (3) n ⌈kn/(k−1)⌉ 1 Inviewof (3),onemayaskwhethertheinequality(2)canalsobegeneralizedtok-Ramanujanprimes. We provethatthisisindeedthecase. Infact,wederivefurtherinequalitiesconcerningthenthk-Ramanujan prime,byconstructingexplicitconstantsn ,n ,n ,n ∈Ndependingonaseriesofparametersincluding 0 1 2 3 k (see (10), (24), Theorem 4.4, Theorem 5.3, respectively), such that the following theorems hold. Theorem A. Let t∈Z with t>−⌈k/(k−1)⌉. Then for every n≥n , 0 R(k) >p . (4) n ⌈kn/(k−1)⌉+t Another problem which arises is to find a minimal bound m=m(k,t) such that the inequality (4) holds for all n≥m. For the case t=0, we introduce the following Definition. For k >1 let N(k)=min{m∈N|R(k) >p for everyn≥m}. n ⌈kn/(k−1)⌉ In Section 3.2 we prove the following theorem giving an explicit formula for N(k). Theorem B. If k ≥745.8, then N(k)=π(3k)−1. Theorem A is supplemented by the following upper bound for nth k-Ramanujan prime. Theorem C. Let ε ≥0, ε ≥0 and ε +ε 6=0. Then for every n≥n , 1 2 1 2 1 R(k) ≤(1+ε )p . n 1 ⌈(1+ε2)kn/(k−1)⌉ By [1], there exists a positive constant β =β (k) such that for every sufficiently large n, 1 1 |R(k)−p |<β nloglogn. (5) n ⌊kn/(k−1)⌋ 1 In TheoremA, we actually obtain a lower bound for R(k)−p improving the lower bound given n ⌈kn/(k−1)⌉ in (5). The next theorem yields an improvement of the upper bound. Theorem D. There exists a positive constant γ, depending on a series of parameters including k, such that for every n≥n , 2 R(k)−p <γn. n ⌈kn/(k−1)⌉ Letπ (x) be the number ofk-Ramanujanprimesless thanorequalto x. Using PNT,Amersi, Beckwith, k Miller,RonanandSondow[1] provedthat there existsa positive constantβ =β (k) suchthatfor every 2 2 sufficiently large n, k−1 π (n) β loglogn k 2 − ≤ . (6) k π(n) logn (cid:12) (cid:12) (cid:12) (cid:12) In Section 4 we prove the following t(cid:12)wo theorems wh(cid:12)ich lead to an improvement of the lower and upper (cid:12) (cid:12) bound in (6). Theorem E. If x≥R(k) with N(k) defined above, then N(k) π (x) k−1 k < . π(x) k Theorem F. There exists a positive constant c, depending on a series of parameters including k, such that for every x≥n , 3 k−1 π (x) c k − < . k π(x) logx In 2009, Mitra, Paul and Sarkar [9] stated a conjecture concerning the number of primes in certain intervals, namely that π(mn)−π(n)≥m−1 for all m,n∈N with n≥⌈1.1log(2.5m)⌉. In Section 5 we confirm the conjecture for large m. Theorem G. If m is sufficiently large and n≥⌈1.1log(2.5m)⌉, then π(mn)−π(n)≥m−1. 2 2 Some simple properties of k-Ramanujan primes We begin with Proposition 2.1. The following three properties hold for R(k). n (i) Let k ,k ∈R with k >k >1. Then R(k1) ≥R(k2). 1 2 2 1 n n (ii) R(k) ≥p for every n∈N and every k >1. n n (iii) For each k, the sequence (R(k)) is strictly increasing. n n Proof. The assertions follow directly from the definition of R(k). n Proposition 2.2. Let k >1 and let n∈N so that R(k) =p . Then R(k) =p for every m≤n. n n m m Proof. The assertion follows from the fact that R(k) ∈P and Proposition 2.1. n For the next property we need the following Lemma 2.3. Let m,n∈N. Then π(m)+π(n)≤π(mn). Proof. By [7], we have π(m)+π(n) < π(m+n) for every m,n ∈ N with m,n ≥ 2 and max{m,n}≥ 6. Now it is easy to check that the desired inequality holds in the remaining cases. Proposition 2.4. If k ≥2, then R(k) =p . π(k) π(k) Proof. Let t=⌊k⌋ and x≥k. Let m∈N be such that mk ≤x<(m+1)k. Using Lemma 2.3, we get x π(x)−π ≥π(mt)−π(m)≥π(k), k (cid:16) (cid:17) i.e. R(k) ≤k <p . Using Proposition 2.1(ii), we obtain the required equality. π(k) π(k)+1 Corollary 2.5. If k ≥2, then R(k) =p for every n=1,...,π(k). n n Proof. The claim follows from Proposition 2.2 and Proposition 2.4. For 1<k <2 we can give more information on R(k). n Proposition 2.6. We have: (i) If 1<k<5/3, then R(k) >p for every n∈N. n n (ii) If 5/3≤k <2, then R(k) =p if and only if n=1. n n Proof. (i) If 1 < k < 3/2, we set x = 2k and obtain π(x)−π(x/k) = 0, i.e. R(k) > 2k > p . It remains 1 1 to use Proposition 2.1(iii). If 3/2≤k<5/3, we set x=3k and proceed as before. (5/3) (k) (ii) Let n = 1 and 5/3 ≤ k < 2. By Proposition 2.1(ii) we get p ≤ R = p , i.e. R = p . Let 1 1 1 1 1 n≥2. Then R(k) ≥R(2) >p and we use Proposition 2.1(iii) as in the previous case. 2 2 2 The following property will be useful in Section 4. Proposition 2.7. For every n and k, R(k) π(R(k))−π n =n. n k ! Proof. This easily follows from the definition of R(k). n 3 Finally, we formulate an interesting property of the k-Ramanujan primes. Proposition 2.8. If p∈P\{2}, then for every n∈N R(k) 6=kp−1. n Proof. It suffices to consider the case kp∈N. Assume R(k) =kp−1 for some n∈N. Since kp−1>2, n we obtain kp6∈P. Let r ∈R with 0≤r <1. Using Proposition 2.7, we get kp+r R(k) π(kp+r)−π =π(R(k))−π n −1=n−1, k n k (cid:18) (cid:19) ! which contradicts the definition of R(k). n 3 Estimates for the nth k-Ramanujan prime From here on, we use the following notation. Let m ∈N and s,a ,...,a ∈R with s≥0. We define 1 1 m1 m1 a j A(x)= logjx j=1 X and Y =Y (a ,...,a ) so that s s 1 m1 x π(x)> +s (7) logx−1−A(x) for every x≥Y . Further, for m ∈N and b ,...,b ∈R we define s 2 1 m2 ≥0 m2 b j B(x)= logjx j=1 X and X =X (b ,...,b ) so that 0 0 1 m2 x π(x)< (8) logx−1−B(x) for every x≥X . In addition, let X =X (k,a ,...,a ,b ,...,b ) be such that 0 1 1 1 m1 1 m2 logk−B(kx)+A(x)≥0 (9) for every x≥X . 1 Remark. It is clear that B(x)>A(x) for every x≥max{X ,X }. Hence b >a . 0 1 1 1 3.1 A lower bound for the nth k-Ramanujan prime The theorem below implies that the inequality (2) can be generalized, in view of (3), to k-Ramanujan primes. Theorem 3.1. Let t∈Z with t>−⌈k/(k−1)⌉ and let r=(t+1)(k−1)/k. Then R(k) >p n ⌈kn/(k−1)⌉+t for every n∈N with k−1 n≥n = (π(X )−t+1), (10) 0 2 k where X =X (k,t,m ,m ,a ,...,a ,b ,...,b )=max{X ,kX ,kY }. 2 2 1 2 1 m1 1 m2 0 1 r 4 Proof. Let x≥X /k. Then the inequality (9) is equivalent to 2 x x ≥ logx−1−A(x) log(kx)−1−B(kx) and we get x x π(kx) π(x)> +r ≥ +r > +r. (11) logx−1−A(x) logkx−1−B(kx) k By setting x=p /k in (11), we obtain ⌈kn/(k−1)⌉+t 1 1 kn π p > +t +r. k ⌈kn/(k−1)⌉+t k k−1 (cid:18) (cid:19) (cid:18)(cid:24) (cid:25) (cid:19) Hence, 1 π(p )−π p <n, ⌈kn/(k−1)⌉+t k ⌈kn/(k−1)⌉+t (cid:18) (cid:19) and we apply the definition of R(k). n Corollary 3.2. We have liminf(R(k)−p )=∞. n ⌈kn/(k−1)⌉ n→∞ Proof. FromTheorem3.1, it followsthat for everyt∈N there is anN ∈N suchthat for every n≥N , 0 0 R(k)−p ≥p −p ≥2t. n ⌈kn/(k−1)⌉ ⌈kn/(k−1)⌉+t ⌈kn/(k−1)⌉ This proves our corollary. Remark. In 2013, Sondow [18] raised the question whether the sequence (R − p ) is unbounded. n 2n n Corollary 3.2 implies that this is indeed the case. Corollary 3.3. If n≥max{2,(k−1)π(X )/k}, then 2 R(k)−p ≥6. n ⌈kn/(k−1)⌉ Proof. We set t=1 in Theorem 3.1. Then for every n≥(k−1)π(X )/k we obtain 2 R(k) ≥p ≥p +2≥p +4. n ⌈kn/(k−1)⌉+2 ⌈kn/(k−1)⌉+1 ⌈kn/(k−1)⌉ Since there is no prime triple of the form (p,p+2,p+4) for p>3, we are done. To find an explicit value for X in the case t=0, we need the following 2 Lemma 3.4. If x≥470077, then x π(x)> +1. logx−1− 1 logx Proof. We consider the function g(x)=2.65x−log4x. Then g(x)>0 for every x≥e7 and we get x x > +1 logx−1− 1 − 2.65 logx−1− 1 logx log2x logx for every x ≥ e7. By Corollary 3.11 of [2], the inequality π(x) > x/(logx−1−1/logx)+1 holds for every x ≥ 38168363. We set h(x) = x/(logx−1−1/logx). Then h′(x) ≥ 0 for every x ≥ 12.8 and we check with a computer that π(p )≥h(p )+1 for every π(470077)≤i≤π(38168363). i i+1 5 Proposition 3.5. Let X =X (k)=max{470077k,kr(k)}, where 3 3 1 3.83 r(k)= exp max −1,0 . k s (cid:26)logk (cid:27)! Then R(k) >p n ⌈kn/(k−1)⌉ for every k−1 n≥ (π(X )+1). 3 k Proof. We choose t = 0 in Theorem 3.1. Then r = (k−1)/k. We set A(x) = 1/logx and Y = 470077. r By Lemma 3.4, we get that the inequality (7) holds for every x≥Y . By choosing b =1, b =3.83 and r 1 2 X =9.25, we can use the third inequality in Corollary 3.9 of [2]. Let x≥r(k). Then it is easy to show 0 that the inequality (9) holds. Now our proposition follows from Theorem 3.1. Corollary 3.6. If n≥4, then R −p ≥6. n 2n Proof. We set t = 1 and k = 2 in Theorem 3.1. Then r = 1. From Corollary 3.3 and from the proof of Proposition3.5,iffollowsthatR −p ≥6foralln≥π(X (2))/2=π(max{940154,2r(2)})/2=37098. n 2n 3 We check with a computer that the inequality R −p ≥6 also holds for every 4≤x≤37097. n 2n Remark. Since R −p =4 and R −p =4, Corollary3.6 gives a positive answer to the questionraised 2 4 3 6 by Sondow [18], whether min{R −p |n≥2}=4. n 2n 3.2 An explicit formula for N(k) In the introduction we defined N(k) to be the smallest positive integer so that R(k) >p n ⌈kn/(k−1)⌉ for every n≥N(k). By Proposition 3.5, we get k−1 N(k)≤ (π(max{470077k,kr(k)})+1) k (cid:24) (cid:25) for every k >1. We can significantly improve this inequality in the following case. Theorem 3.7. If k ≥745.8, then N(k)≤π(3k)−1. Proof. We have R(k) >3k. Since 3k≥p , we obtain by Proposition 2.2 that π(3k)−1 π(3k) R(k) >p (12) n n+1 for every n≥π(3k)−1. We set A(x)=−7.1/logx, s =1 and Y =3. Then, as in the proof of Lemma 1 3.4,weobtainthattheinequality(7)holdsforeveryx≥Y . BysettingB(x)=1.17/logxandX =5.43 1 0 and using Corollary 3.9 of [2], we see that the inequality (8) holds. Let 1 8.27 2 1 8.27 r(k)=exp 7.1+ logk− − logk− . s 4 logk 2 logk  (cid:18) (cid:19) (cid:18) (cid:19) e   It is easy to see that x≥r(k) implies the inequality (9). By Theorem 3.1, we obtain k−1 e N(k)≤ (π(X4)+1) , (13) k (cid:24) (cid:25) where X = X (k) = max{5.43,3k,kr(k)}. Since r(k) is decreasing, from r(745.8) ≤ 2.999966 we get 4 4 that r(k) ≤ 3. Hence, X = 3k for every k ≥ 745.8. Since π(3k)+1≤ k for every k ≥ 745.8, we obtain 4 N(k)≤π(3k)+1 by (13). Finally, we apply (12). e e e e 6 Next, we find a lower bound for N(k). Proposition 3.8. For every k >1, N(k)>π(k). Proof. First, let k ≥2. Using Proposition 2.4, we get R(k) <p . (14) π(k) ⌈kπ(k)/(k−1)⌉ Hence, N(k)>π(k) for every k ≥2. The asserted inequality clearly holds for every 1<k<2. In order to prove a sharper lower bound for N(k), see Theorem 3.11, we need the following lemma. Lemma 3.9. Let r,s∈R with r >s>0. If t≥s/r·R(r/s), then π(r) rt π(r)+π(t)≤π . s (cid:18) (cid:19) Proof. Since rt/s≥R(r/s), the claim follows from the definition of R(k). π(r) n Proposition 3.10. If m,n∈N with m,n≥5 and max{m,n}≥18, then mn π(m)+π(n)≤π . 3 (cid:16) (cid:17) Proof. Without loss of generality, let m ≥ n. First, we consider the case m ≥ n ≥ 20. By [20], we have π(x)<8x/(5logx) for every x>1. Using an estimate from [15] for π(x), we get mn mn 8m 8m π −π(m)≥ − ≥ >π(n). 3 3log(mn/3) 5logm 5logm (cid:16) (cid:17) So the proposition is proved, when m≥n≥20. Now let m≥18 and min{m,20}≥n≥5. We have: r 20 19 18 . (r/3) ⌈3/r·R ⌉ 5 5 4 π(r) We apply Lemma 3.9 with s=3, r =m and t=n. Theorem 3.11. For every k >1, N(k)≥π(3k)−1. Proof. For every 1<k <5/3 the claim is obviously true. For every 5/3≤k <7/3, we have R(k) ≤R(5/3) <p ; π(3k)−2 1 ⌈k(π(3k)−2)/(k−1)⌉ i.e., N(k)>π(3k)−2. Similarly, for every p /3≤k <p /3, where i=4,...,8, we check that i i+1 R(k) ≤p . π(3k)−2 ⌈k(π(3k)−2)/(k−1)⌉ Hence our theorem is proved for every 1<k <19/3. Now, let k ≥19/3. For p ≤x<3k and for π(3k)−1 3k≤x<5k it is easy to see that π(x)−π(x/k)≥π(3k)−2. So let x≥5k and let m∈N be such that m≥5 and mk ≤x<(m+1)k. Since 3k ≥19, we use Proposition 3.10 to get the inequality x m(3k) π(x)−π ≥π −π(m)≥π(3k). k 3 (cid:16) (cid:17) (cid:18) (cid:19) Hence, altogether we have R(k) ≤p ≤p (15) π(3k)−2 π(3k)−1 ⌈k(π(3k)−2)/(k−1)⌉ and therefore N(k)>π(3k)−2. Remark. The proof of Theorem 3.11 yields R(k) ≤p for every k ≥19/3. π(3k) π(5k) From Theorem 3.7 and Theorem 3.11, we obtain the following explicit formula for N(k). Corollary 3.12. If k ≥745.8, then N(k)=π(3k)−1. 7 3.3 An explicit formula for N (k) 0 By replacing “>” with “≥” in the definition of N(k), we get the following Definition. For k >1, let N (k)=min{m∈N|R(k) ≥p for everyn≥m}. 0 n ⌈kn/(k−1)⌉ Since N (k)>π(k) for every 1<k <2, it follows from (14) that 0 N (k)>π(k) 0 is fulfilled for every k >1. In the following case we obtain a sharper lower bound for N (k). 0 Theorem 3.13. If k ≥11/3, then N (k)≥π(2k). 0 Proof. First, we show that x π(x)−π ≥π(2k)−1 (16) k (cid:16) (cid:17) for every x≥p . For p ≤x<2k and for 2k ≤x<3k, the inequality (16) is obviously true. π(2k)−1 π(2k)−1 Let 3k ≤ x < 5k. Since π(3t)−π(2t) ≥ 1 for every t ≥ 11/3 = 1/3·R(3/2), it follows π(x)−π(x/k) ≥ 1 π(2k)−1. So let x ≥5k and let l ∈N be such that l ≥ 5 and lk ≤x <(l+1)k. Similarly to the proof of Proposition 3.10, we get that mn π(m)+π(n)≤π (17) 2 (cid:16) (cid:17) for every m,n≥4 with max{m,n}≥6. Since 2k>7, using (17) we obtain the inequality x 2lk π(x)−π ≥π −π(l)≥π(2k). (18) k 2 (cid:16) (cid:17) (cid:18) (cid:19) Hence, we proved that the inequality (16) holds for every x≥p . So, π(2k)−1 R(k) ≤p <p , (19) π(2k)−1 π(2k)−1 ⌈k(π(2k)−1)/(k−1)⌉ which gives the required inequality. Using (19), we get an improvement of Corollary 2.5. Corollary 3.14. If k ≥11/3, then R(k) =p for every 1≤n≤π(2k)−1. n n Proof. Follows from Proposition(2.1)(ii), the left inequality in (19) and Proposition2.2. To prove an upper bound for N (k), the following proposition will be useful. 0 Proposition 3.15. If k ≥29/3, then R(k) =p . π(2k) π(2k)+1 Proof. Since π(2k)−π(2k/k)<π(2k) and 2k<p , we have R(k) ≥p for every k >1. To π(2k)+1 π(2k) π(2k)+1 prove R(k) ≤p , it suffices to show that π(2k) π(2k)+1 x π(x)−π ≥π(2k) (20) k (cid:16) (cid:17) for every x ≥ p . It is clear that (20) is true for every p ≤ x < 3k. Let 3k ≤ x < 5k. We π(2k)+1 π(2k)+1 have π(3t)−π(2t) ≥ 2 for every t ≥ 29/3 = 1/3·R(3/2) and thus (20) holds. By (18) we already have 2 that the inequality (20) also holds for every x≥5k. We can show more than in Corollary 3.14 for the following case. 8 Corollary 3.16. If k ≥29/3, then: (i) R(k) =p if and only if 1≤n≤π(2k)−1. n n (ii) R(k) =p if and only if π(2k)≤n≤π(3k)−2. n n+1 Proof. (i) From Proposition 3.15, we get R(k) >p for every n≥π(2k) and then use Corollary 3.14. n n (ii) Let π(2k)≤n≤π(3k)−2. By (i) and (15), we obtain R(k) =p . Now let R(k) =p . Since n n+1 n n+1 p π(3k) π(p )−π ≤π(3k)−π(3)<π(3k)−1, π(3k) k (cid:16) (cid:17) we obtain R(k) >p . (21) π(3k)−1 π(3k) Hence, n≤π(3k)−2. By (i), we get n≥π(2k). Remark. Similarly to the proof of (21), we obtain in general that for every real r ≥2/k, R(k) >p . π(rk)−π(r)+1 π(rk) Remark. Corollary 3.16 implies that for every k ≥ 29/3 the prime numbers p and p are not π(2k) π(3k)−1 k-Ramanujan primes. Corollary 3.17. For each m∈N there exists k =k(m)≥29/3 such that p ,...,p are all π(2k)+1 π(2k)+m k-Ramanujan primes. Proof. By PNT, we obtain π(3k)−2−π(2k)→∞ as k →∞. Then use Corollary 3.16(ii). The next lemma provides an upper bound for N (k). 0 Lemma 3.18. Let X =X (k)=max{X ,kX ,kY }. Then the inequality 5 5 0 1 0 R(k) ≥p n ⌈kn/(k−1)⌉ holds for every k−1 n≥ (π(X )+2). 5 k Proof. We just set t=−1 in Theorem 3.1. The inequality in Theorem 3.13 becomes an equality in the following case. Theorem 3.19. If k ≥143.7, then N (k)=π(2k). 0 Proof. We set A(x) = −3.3/logx, s = 0 and Y = 2. Similarly to the proof of Corollary 3.11 from [2], 0 we get that (7) is fulfilled for every x ≥ Y . By setting B(x) = 1.17/logx and X = 5.43 and using 0 0 Corollary 3.9 from [2], we see that the inequality (8) is fulfilled for every x≥X . Let 0 2 1 4.47 1 4.47 z(k)=exp 3.3+ logk− − logk− . s 4 logk 2 logk  (cid:18) (cid:19) (cid:18) (cid:19)   It is easy to show that x ≥ z(k) is equivalent to the inequality (9). By setting X = X (k) = 6 6 max{2k,5.43,kz(k)}and using Lemma 3.18, we get that k−1 N (k)≤ (π(X )+2) . (22) 0 6 k (cid:24) (cid:25) In the proof of Theorem 3.7, we showed that r(k)≤3 for every k ≥745.8. Analogously, we get z(k)≤2 for every k ≥ 143.7. Hence we obtain X = 2k and therefore N (k) ≤ π(2k)+2 for every k ≥ 143.7. 6 0 Proposition 3.15 and Theorem 3.13 finish the proof. e 9 3.4 An upper bound for the nth k-Ramanujan prime After finding a lower bound for the nth k-Ramanujan prime, we find an upper bound by using the following two propositions, where Υ (x) is defined by k x 1 1 logk−A(x)+B(x/k) Υ (x)=Υ (x)= 1− − . k k,a1,...,am1,b1,...,bm2 logx−1−A(x) k k log(x/k)−1−B(x/k) (cid:18) (cid:19) Proposition 3.20. If x≥max{Y ,kX }, then 0 0 x π(x)−π >Υ (x). k k (cid:16) (cid:17) Proof. We have x x x/k π(x)−π > − k logx−1−A(x) log(x/k)−1−B(x/k) (cid:16) (cid:17) and see that the term on the right hand side is equal to Υ (x). k Proposition 3.21. For every sufficiently large x, the derivative Υ′(x)>0. k Proof. We set x F(x)=F (x)= a1,...,am1 logx−1−A(x) and 1 1 logk−A(x)+B(x/k) G(x)=G (x)=1− − . k,a1,...,am1,b1,...,bm2 k k log(x/k)−1−B(x/k) It is clear that there exists an X = X (k,m ,m ,a ,...,a ,b ,...,b ) such that G(x) > 0 for 7 7 1 2 1 m1 1 m2 every x ≥ X . Since k > 1, we obtain B(x/k) > B(x) > A(x) as well as log(x/k) ≤ logx and 7 log(x/k)−1−B(x/k)>0 for every x≥max{Y ,kX }. It follows that 0 0 1 m2 j·b m1 i·a G′(x)> j − i (23) k(log(x/k)−1−B(x/k))  logj+1x logi+1x j=1 i=1 X X   for every x ≥ max{Y ,kX }. Since b > a , there is an X = X (m ,m ,a ,...,a ,b ,...,b ) such 0 0 1 1 8 8 1 2 1 m1 1 m2 that m2 j·b m1 i·a j i − ≥0 logj+1x logi+1x j=1 i=1 X X for every x ≥ X . Hence, G′(x) > 0 for every x ≥ max{Y ,kX ,X }. We have F(x) > 0 for every 8 0 0 8 x≥Y . Further, there exists an X =X (m ,a ,...,a ) such that 0 9 9 1 1 m1 m1 i·a i logx−2−A(x)− >0 logi+1x i=1 X for every x≥X . Therefore, 9 1 m1 i·a F′(x)= logx−2−A(x)− i >0 (logx−1−A(x))2 logi+1x! i=1 X for every x ≥ max{Y ,X }. So, for every x ≥ max{Y ,kX ,X ,X ,X }, we get Υ′(x) = F′(x)G(x)+ 0 9 0 0 7 8 9 k F(x)G′(x)>0. Now, let m = m = 1. By Proposition 3.21, there exists an X = X (k,a ,b ) such that Υ′(x) > 0 1 2 10 10 1 1 k for every x≥X . Let X =X (b )∈N be such that 10 11 11 1 p ≥n(logp −1−b /logp ) n n 1 n 10