Journal of Inequalities in Pure and Applied Mathematics http://jipam.vu.edu.au/ Volume 4, Issue 1, Article 23, 2003 ERDŐS-TURÁN TYPE INEQUALITIES LAURENŢIU PANAITOPOL U NIVERSITY OF B UCHAREST FACULTY OF M ATHEMATICS 14 ACADEMIEI S T. RO–70109 B UCHAREST ROMANIA . pan@al.math.unibuc.ro Received 10 January, 2003; accepted 19 February, 2003 Communicated by J. Sándor A BSTRACT. Denoting by (rn )n≥1 the increasing sequence of the numbers pα with p prime and α ≥ 2 integer, we prove that rn+1 − 2rn + rn−1 is positive for infinitely many values of n and negative also for infinitely many values of n. We prove similar properties for rn2 − rn−1 rn+1 and 1 2 1 rn−1 − rn + rn+1 as well. Key words and phrases: Powers of prime numbers, Inequalities, Erdos-Turán theorems. 2000 Mathematics Subject Classification. 11A25, 11N05. 1. I NTRODUCTION Let (rn )n≥0 be the increasing sequence of the powers of prime numbers (pα with p prime and α ≥ 2 integer). Thus, we have r1 = 4, r2 = 8, r3 = 9, r4 = 16, etc. Properties of the sequence (rn )n≥1 were studied in [5] and [3]. Denote by pn the n-th prime number. In [1], Erdős and Turán proved that pn+1 −2pn +pn−1 is positive for infinitely many values of n and negative also for infinitely many values of n. Until now, no answer is known for the following question raised by Erdős and Turán: Do there exist infinitely many numbers n such that pn+1 − 2pn + pn−1 = 0? 2 Erdős and Turán also proved that each of the sequences (pn − pn−1 pn+1 )n≥2 and 1 1 − p2n + pn+1 has infinitely many positive terms and infinitely many negative ones. pn−1 n≥2 Denoting by (qn )n≥1 the increasing sequence of the powers of prime numbers, the author proved in [4] that the value of qn+1 − 2qn + qn−1 changes its sign infinitely many times. In the present paper, we raise similar problems for the sequence (rn )n≥1 . We need a few preliminary properties, which will be proved in the next section. ISSN (electronic): 1443-5756 c 2003 Victoria University. All rights reserved. 004-03 2 L AUREN ŢIU PANAITOPOL 2. O N THE D IFFERENCE rn+1 − rn Property 2.1. We have lim sup(rn+1 − rn ) = ∞. (2.1) n→∞ Proof. Let m ≥ 4. We show that, among the numbers √ m! + 2, m! + 3, . . . , m! + [ m], there is no term of the sequence (rn )n≥1 . √ Assume that there exists an integer a such that 2 ≤ a ≤ [ m] and m! + a = pi (2.2) where p is prime and i ≥ 2. The relation (2.2) can also be written in the form m! a + 1 = pi , whence a = pj with 1 ≤ j ≤ i. a It follows that m! m! + 1 = pi−j , hence j is not divisible by p. j p p If ep (n) is Legendre’s function, we have ep (m) = j, that is, ∞ X m (2.3) = j. s p s=1 √ √ j Since a ≤ m, it follows that p ≤ m, that is, m ≥ p2j , and then (2.3) implies that 2j 2j 2j p p p j≥ + + · · · + 2j 2 p p p 2j−1 2j−2 =p +p + ··· + p + 1 ≥ 22j−1 + 22j−2 + · · · + 2 + 1 = 22j − 1. Since for j ≥ 1 we have 22j − 1 > j, we obtained a contradiction. Since our assumption turned out to be false, it follows that for every m ≥ 4 there exists k = k(m) such that √ rk ≤ m! + 1 and rk+1 ≥ m! + [ m] + 1, √ whence rk+1 − rk ≥ [ m], and finally lim sup(rn+1 − rn ) = ∞, n→∞ and the proof ends. We now denote an = (2.4) rn+1 −rn n log2 n and recall that, in [2], H. Meier proved that lim inf n→∞ pn+1 − pn < 0.248. log n In connection with this result, we prove: Property 2.2. We have (2.5) J. Inequal. Pure and Appl. Math., 4(1) Art. 23, 2003 lim inf an < 0.496. n→∞ http://jipam.vu.edu.au/ E RD ŐS -T URÁN T YPE I NEQUALITIES 3 Proof. We consider the indices m such that pm+1 − pm < 0.248. log m Both the numbers p2m and p2m+1 occur in the sequence (rn )n≥1 , that is, p2m = rk and p2m+1 = rh , with k = k(m), h = h(m) and h ≥ k + 1. In [5], it was proved that, for m ≥ 1783, we have p2m ≥ rm > m2 log2 m. (2.6) Since pm ∼ m log m, it follows that rk ∼ k 2 log2 k. But rk = p2m , hence k log k ∼ m log m. One can show without difficulty that k(m) ∼ m. It then follows that √ √ √ √ rk+1 − rk rh − r k pm+1 − pm < = . log k log k log k Since log k ∼ log m, we get √ lim inf k→∞ Since √ rk ∼ k log k and √ √ rk+1 − rk pm+1 − pm ≤ lim inf < 0.248. m→∞ log k log m rk+1 ∼ (k + 1) log(k + 1) ∼ k log k, it follows that lim inf rk+1 − rk < 0.496, k log2 k lim inf rn+1 − rn < 0.496. n log2 n k→∞ where k = k(m). Consequently, n→∞ 3. E RD ŐS -T URÁN T YPE P ROPERTIES For k ≥ 2 we denote Rk = rk+1 − 2rk + rk−1 , and prove Property 3.1. There exist infinitely many values of n such that Rn > 0, and also infinitely many ones such that Rn < 0. Pm Proof. Denoting Sm = k=2 Rk , we have Sm = rm+1 − rm − r2 + 1. By (2.1) we have lim sup Sm = ∞, hence Rn > 0 for infinitely many values of n. m→∞ P Denoting σm = m k=2 kRk , we have rm 2 2 σm = m(rm+1 − rm ) − rm − r2 + 2r1 = m log m am − 2 2 . m log m Since rm ∼ m2 log2 m, we get by (2.5) that lim inf σm = −∞, hence Rn < 0 for infinitely m→∞ many values of n. For k ≥ 2, denoting ρk = 1 rk−1 − J. Inequal. Pure and Appl. Math., 4(1) Art. 23, 2003 2 rk + 1 , rk+1 we have http://jipam.vu.edu.au/ 4 L AUREN ŢIU PANAITOPOL Property 3.2. There exist infinitely many values of n such that ρn > 0, and also infinitely many ones such that ρn < 0. 0 Proof. For α > 3, denoting Sm (α) = m P k α ρk , we get k=2 0 Sm (α) m−1 mα (rm+1 − rm ) mα − (m − 1)α X k α − 2(k − 1)α + (k − 2)α =− − + + O(1). rm rm+1 rm rk k=2 We have rk ∼ k 2 log2 k, k α − (k − 1)α ∼ αk α−1 , k α − 2(k − 1)α + (k − 2)α ∼ α(α − 1)k α−2 , whence mα (rm+1 − rm ) mα−3 am ∼ , rm rm+1 log2 m αmα−3 mα − (m − 1)α ∼ , rm log2 m α(α − 1)k α−4 k α − 2(k − 1)α + (k − 2)α ∼ . rk log2 k Since m−1 X k=2 k α−4 (α − 3)mα−3 ∼ , log2 m log2 m it follows that mα−3 · − am − α + α(α − 1)(α − 3) . 2 log m 0 Then lim Sm (3.1) = −∞, and thus there exist infinitely many values of n such that ρn < 0. 0 Sm (α) ∼ m→∞ 0 On the other hand, we have by (2.5) that lim sup Sm (4) = ∞, which shows that there exist m→∞ infinitely many values of n such that ρn > 0. A consequence of Properties 3.1 and 3.2 is the following. Property 3.3. There exist infinitely many values of n such that rn−1 rn+1 > rn2 , and also infinitely many ones such that rn−1 rn+1 < rn2 . Proof. If rn > rn+1 +rn−1 , 2 √ then rn > rn−1 rn+1 . On the other hand, if 1 1 √ rn < 2 + < rn−1 rn+1 , rn−1 rn+1 and then the desired conclusion follows by Properties 3.1 and 3.2. J. Inequal. Pure and Appl. Math., 4(1) Art. 23, 2003 2 rn > 1 rn−1 + 1 , rn+1 then http://jipam.vu.edu.au/ E RD ŐS -T URÁN T YPE I NEQUALITIES 5 Open problem. Do there exist infinitely many values of n such that rn+1 − 2rn + rn−1 = 0? R EFERENCES [1] P. ERDŐS AND P. TURÁN, On some new question on the distribution of prime numbers, Bull. Amer. Math. Soc., 54(4) (1948), 371–378. [2] H. MEIER, Small difference between prime numbers, Michigan Math. J., 35 (1988), 324–344. [3] G. MINCU, An asymptotic expansion, (in press). [4] L. PANAITOPOL, Some of the properties of the sequence of powers of prime numbers, Rocky Mountain J. Math., 31(4) (2001), 1407–1415. [5] L. PANAITOPOL, The sequence of the powers of prime numbers revisited, Math. Reports, 5(55)(1) (2003), (in press). J. Inequal. Pure and Appl. Math., 4(1) Art. 23, 2003 http://jipam.vu.edu.au/