AN UPPER BOUND FOR THE PRIME GAP Ya-Ping Lu and Shu-Fang Deng
Abstract : We showed that the prime gap for a prime number p is less than or equal to the prime count of the prime number, or π(π) β€ π(π) . Definition 1: The prime gap, π(π π ) , is defined as the number of integers greater than the nth prime number p n and less than or equal to the next prime number p n+1 : π(π π ): = π < π΅ β€ π π+π } Thus, the prime gap equals to the difference between two consecutive prime numbers, π(π π ) = π π+π β π π . The smallest prime gap occurs between the first and the second prime numbers, 2 and 3, and π(π ) = π β π = 1 . For twin primes π π and π π + 2 , π(π π ) = 2 . As n increases, larger prime gaps are expected to appear and, as a matter of fact, prime gap can be arbitrarily large. For example, the π β 1 consecutive integers π! + 2 , π! + 3 , π! + 4 , β¦, π! + π are all composites [1] . If p is the largest prime number less than or equal to π! + 1 , then the prime gap π(π) β₯ π . From the prime number theorem, the number of primes less than or equal to p n is approximately π π / πππ π π , or π π ~ π πππ π π . So, on average, the prime gap between two prime numbers p n and p n+1 is πππ π π . CramΓ©r [2] showed that if the Riemann Hypothesis holds, π(π) < ππ log π . Bertrand's postulate [3] states that for π β₯ 1 , π π+1 < 2π π , or π(π π ) < π π Bertrand's postulate was proved by Chebyshev [4] and so the postulate is also called the Bertrand β Chebyshev theorem or Chebyshev's theorem.
Based on Ramanujanβs work [5] it is proved that πβπ > π π for π > π , where π = π(π π ) =π(π π ), and π π is the π - th Ramanujan prime. This means that, for π > π(π π ) , (π π ) β€ 2π πβπ β π π One can show, from the prime number theorem, that for every real number π > 0 and there is some integer m such that there is always a prime p satisfying π < π < (1 + π)π for every π > π . This shows that, for all π > π , π(π π ) < πp π . Some specific pairs of ( e , n ) are ( ), ( ), and ( ) [6] . Legendre's conjecture states that, for every π > 1 , there is a prime π , such that π < π <(π + 1) . Oppermann's conjecture [7] states that, for every integer π > 1 , there is at least one prime number between π(π β 1) and π , and at least another prime between π and π(π + 1) . If Oppermann's conjecture is true, there would be at least four prime numbers between (π π ) and (π π+1 ) for every π β₯ 2 , and the largest possible gaps between two consecutive prime numbers could be, as stated by Andrica's conjecture [8] , π(π π ) < 2βπ π + 1 which suggests that π(π π ) = πͺ(π ππ ) with π < 1/2 will suffice to prove Andrica's conjecture. However, all values of ΞΈ proved so far are larger than and the best unconditional result is π < 21/40 by R.C. Baker et al. [9] . Table 1 lists some of the prime gaps for p up to , including the first maximal prime gaps [10] , defined as the prime gaps larger than all gaps between smaller primes. We see that the prime gap π π (Column ) is smaller than or equal to the prime count of the prime number (Column 1), or π(π π ) β€ π(π π ) = π . Theorem 1 : The prime gap of a prime number p is less than or equal to the prime count of the prime number, or π(π) β€ π(π) . Proof:
Let π be the next prime number following the prime number π . From Definition 1, we have π(π) = π β π . Since π(π) β π(π) = 1 , π(π + π(π)) β π(π) = 1. As the prime counting function is non-decreasing,
Theorem 1 can be rephrased as β
There is at least one prime number in the range of (π, π + π(π)] β , or π(π + π(π)) β π(π) β₯ 1 Dusart [11] showed that the number of prime numbers less than or equal to x is bounded by (π₯) β€ π₯log π₯β1.1 if π₯ β₯ 60184 and π(π₯) β₯ π₯log π₯β1 if π₯ β₯ 5393 For π > 60184 , the number of prime numbers between p and π + π(π) is π π : = π(π + π(π)) β π(π) β₯ π (π + πlog π β 1) β πlog π β 1.1 β₯ π + πlog π β 1log (π + πlog π β 1) β 1 β πlog π β 1.1 = π [ 1 + 1log π β 1log π + πππ (1 + 1log π β 1) β 1 β 1log π β 1.1] Since, for π > 60184 and πππ(1 + π₯) < π₯ ππ π₯ > 0 , π π > π [ πππ πlog π β 1(log π β 1) + 1log π β 1 β 1log π β 1.1] = π [ πππ π(log π β 1) +1 β 1log π β 1.1] = (0.9 πππ π β 2)πlog π β 3.1 πππ π +4.2 πππ π β 2.2 = 0.9 πlog π β 7990 πππ π + 911405 + 102013645 1πππ π β 20/9 in which β πππ π + + < 0 for π > 60184 , thus π π > 0.9ππππ π > 1. It can be verified that, for π < 60184 , π(π + π(π)) β π(π) β₯ 1 (see Column 4 in Table 1). β‘ ith Theorem 1 and the lower bound of π₯log π₯β1.1 for the prime counting function by Dusart [11] , we have the following corollary: Corollary 1: The prime gap of a prime number p is less than π/(πππ π β π) for π β₯ 5 . π(π) < ππππ π β π. π Proof:
From Theorem 1 and Dusartβs upper bound [11] , we have, for π > 60184 , π(π) β€ π(π) < πlog π β 1.1. It can be verified that, π(π) < πlog πβ1.1 also holds for . β‘ Numerical verification (last column in Table 1) indicates that a slightly tighter upper bound π(π) < π + 1πππ π holds for all prime numbers. Appendix 1 is the Python code used to obtain the data given in Table 1.
Table 1. ( maximal prime gaps in C olumn 3 marked by * ) π = π(π π ) π π π(π π ) Ο(p n +n) β n (π + 1)/πππ π
1 2 1* 1 4.3 2 3 2* 1 3.6 3 5 2 1 3.7 4 7 4* 1 4.1 5 11 2 1 5.0 6 13 4 2 5.5 7 17 2 2 6.4 8 19 4 1 6.8 9 23 6* 2 7.7 10 29 2 2 8.9 11 31 6 2 9.3 12 37 4 3 10.5 13 41 2 3 11.3 14 43 4 2 11.7 15 47 6 3 12.5 16 53 6 3 13.6 17 59 2 4 14.7 18 61 6 4 15.1 19 67 4 4 16.2 20 71 2 4 16.9 21 73 6 3 17.2 2 79 4 4 18.3 23 83 6 4 19.0 24 89 8* 6 20.1 25 97 4 5 21.4 26 101 2 5 22.1 27 103 4 4 22.4 28 107 2 4 23.1 29 109 4 4 23.4 30 113 14* 4 24.1 99 523 18* 15 83.7 154 887 20* 21 130.8 189 1129 22* 25 160.8 217 1327 34* 26 184.7 1183 9551 36* 126 1042.3 1831 15683 44* 184 1623.5 2225 19609 52* 223 1984.1 3385 31397 72* 330 3032.3 14357 155921 86* 1165 13040.1 30802 360653 96* 2386 28185.6 31545 370261 112* 2439 28877.2 40933 492113 114* 3123 37547.4 103520 1349533 118* 7325 95608.1 104071 1357201 132* 7349 96112.8 149689 2010733 148* 10304 138537.5 325852 4652353 154* 21244 303028.0 1094421 17051707 180* 65621 1024018.3 1319945 20831323 210* 78221 1236136.0 2850174 47326693 220* 160910 2677972.3 6957876 122164747 222* 373308 6560632.0 10539432 189695659 234* 551956 9952066.6 10655462 191912783 248* 557801 10062250.1 20684332 387096133 250* 1044533 19575833.9 23163298 436273009 282* 1163064 21930122.7
Bertrand's postulate [3] can be proved by Theorem 1.
Theorem 2: (Bertrand β Chebyshev theorem or
Bertrandβs postulate):
For π β₯ π , the prime gap of the nth prime number p n is less than the prime number itself, π(π π ) < π π . Proof: From Theorem 1, π(π) β€ π(π) . Let p n be the nth prime number, we have π(π π ) β€ π(π π ) = π . Since n is always smaller than the nth prime number, or π < π π , the Bertrandβs postulate follows, π(π π ) β€ π < π π . β‘ References [1] Havil, J. Gamma, βExploring Euler's Constantβ, p. 1
70, Princeton, NJ: Princeton University Press, 2003. [2] H. CramΓ©r, "On the order of magnitude of the differences between consecutive prime numbers," Acta. Arith., 2 (1936) 396-403. [3]. Joseph Bertrand. MΓ©moire sur le nombre de valeurs que peut prendre une fonction quand on y permute les lettres qu'elle renferme. Journal de l'Ecole Royale Polytechnique, Cahier 30, Vol. 18 (1845), 123-140. [4]. P. Tchebychev. MΓ©moire sur les nombres premiers. Journal de mathΓ©matiques pures et appliquΓ©es, SΓ©r. 1(1852), 366-390. [5]. Ramanujan, S. (1919). "A proof of Bertrand's postulate". Journal of the Indian Mathematical Society. 11: 181 β β ] R. C. Baker, G. Harman, and J. Pintz, βThe di ο¬ erence between consecutive primes, IIβ, Proc. London Math. Soc., 83:532562, 2001. [10] N. J. A. Sloane, R. K. Guy, Sequence A005250 in The On-Line Encyclopedia of Integer Sequences (2020), published electronically at https://oeis.org. [11] P. Dusart, βEstimates of some functions over primes without RHβ, arXiv:1002.0442v1 [math.NT] 2 Feb 2010. Email address οΌ [email protected] ppendix 1 οΌ Python code from math import sqrt def prime_check(Num): of primes in (Num1,Num2] and Ο(Num2) count = 0 p_end = Num1 j = Num1 + 2 for j in range(Num1 + 2, Num2 + 1, 2): if prime_check(j) == 1: count += 1 p_end = j result = [count, p_end] return result n_max = 100000000 gap_max = 1 ct_last = 1 print(1, 2, gap_max, ct_last) list1 = [1, 3] p = 3 p_end_last = 3 for n in range(2, n_max + 1): list1 = prime_count(p_end_last, p + n) ct = ct_last ββ