INSULATED PRIMES ABHIMANYU KUMAR∗ AND ANURAAG SAXENA Abstract. The degree of insulation D(p) of a prime p is defined as the largest interval around the prime p in which no other prime is present. Based on this, the n-th prime pn is said to be insulated if and only if its degree of insulation is higher than its surrounding primes. Thus, a new special sequence emerges which is 7, 13, 23, 37, 53, 67, 89, 103, 113, 131, 139, 157, 173, 181, 193, 211, 233, 277, 293, and so on. Therefore, this paper thoroughly investigates its properties and explores the connection of D(p) with the gap between consecutive primes. Heuristically, it is shown that n-th insulated prime grows almost linearly, which is a property in contrast to the effort to obtain a general formula for n-th prime. Finally, the paper leaves the readers with a captivating open problem. 1. Introduction Since centuries, prime numbers have both puzzled and fascinated researchers. Several interesting subsets of primes have been identified and analyzed in literature [3]. For example, primes that are present in a specific form such as Proth prime, Mersenne prime, and Fermat prime [7]. Sequences of primes are also defined if they obey certain property, for instance, Chen prime [1], Lucky prime [4], Wilson prime [2], Ramanujan prime [11], etc. which have captured huge attention due to associated unsolved problems. Similarly, this paper defines a new special subset of primes denoted by I, called the insulated primes, which is defined based on conditions related to the gaps between consecutive primes. Sequences such as twin primes, cousin primes, sexy primes, etc. are also based on the gaps between consecutive primes. Other sequence associated with gaps is the sequence of least prime of distance n from nearest prime (n = 1; even) called the lonely/isolated primes which is given as 2, 5, 23, 53, 409, 293, 211, 1847, 3137, 2179, 3967, 23719, 16033, 40387, 44417, and so on [9]. Another interesting sequence is prime islands which is the least prime whose adjacent primes are exactly 2n apart given as 3, 5, 7, 29, 23, 53, 89, 223, 113, 331, 631, 211, 1381, 1129, 1637, 4759, 2579, 3433, and so on [10]. arXiv:2011.14210v1 [math.NT] 28 Nov 2020 In order to pursue the said objective, we begin by defining degree of insulation and then proceed to the insulated primes. Definition 1.1. Degree of insulation D(p) of a prime p is defined as the maximum value of the set X = fm : π(p − m) = π(p + m) − 1; m 2 Ng, where π(n) is the prime counting function. Definition 1.2. The n-th prime pn is said to be insulated if and only if D(pn−1) < D(pn) and D(pn+1) < D(pn). 2010 Mathematics Subject Classification. 11A41, 11K31. Key words and phrases. Special prime sequences; Prime gaps. 1 2 ABHIMANYU KUMAR∗ AND ANURAAG SAXENA The degree of insulation essentially gives the largest bracket around p such that p is the only prime in that spread, thus insulating p from neighboring primes. That is why, the set I is named insulated primes. Let us begin with an illustration using the prime triplet (19; 23; 29). Consider number 23 which is solved as π(23 − 1) =? π(23 + 1) − 1 ) 8 =? 9 − 1 ) 8 = 8 π(23 − 2) =? π(23 + 2) − 1 ) 8 =? 9 − 1 ) 8 = 8 π(23 − 3) =? π(23 + 3) − 1 ) 8 =? 9 − 1 ) 8 = 8 π(23 − 4) =? π(23 + 4) − 1 ) 8 =? 9 − 1 ) 8 = 8 π(23 − 5) =? π(23 + 5) − 1 ) 7 =? 9 − 1 ) 7 6= 8; so D(23) = 4. Now, for 19 and 29, we have π(19 − 2) =? π(19 + 2) − 1 ) 7 =? 8 − 1 ) 7 = 7 and π(29 − 1) =? π(29 + 1) − 1 ) 9 =? 10 − 1 ) 9 = 9; so D(19) = 2 and D(29) = 1, hence 23 is an insulated prime. As the determination of insulated primes requires the calculation of D(p), so the above procedure is coded in python and the resulting line plot of the values of D(p) vs primes is shown in Figure 1. Figure 1. Plot of D(p) for primes less than 1000 The sequence of insulated primes can be understood as the set of local maximas in the D(p) plot. Thus, the sequence I is 7, 13, 23, 37, 53, 67, 89, 103, 113, 131, 139, 157, 173, 181, 193, 211, 233, 277, 293, 337, 359, 389, 409, 421, 449, 479, 491, 509, 547, 577, 607, 631, 653, 691, 709, 751, 761, 797, 811, 823, 839, 863, 887, 919, 953, 983, and so on. Figure 2 is the plot of n-th insulated prime in vs n. INSULATED PRIMES 3 Figure 2. Plot of in versus n for primes less than 1000 Note that this sequence is neither present on the Online Encyclopedia of Integer Sequences (OEIS) nor its explored earlier, hence, this paper aims to investigate it. 2. Analysis of degree of insulation Certain patterns can be observed from the graphs shown earlier, but these need to be rigorously proven, which is done below along with other critical results. Theorem 2.1. Primes just adjacent to an insulated prime can never be insulated. Proof. Let pn be insulated prime, then D(pn−1) < D(pn) and D(pn+1) < D(pn) by definition. Now, for pn−1 to be insulated prime, the conditions D(pn−2) < D(pn−1) and D(pn) < D(pn−1) must hold. Clearly, the latter condition is contradictory to condition of pn, therefore, pn−1 is not an insulated prime. For pn+1 to be an insulated prime, the conditions D(pn) < D(pn+1) and D(pn+2) < D(pn+1) must hold. But the prior condition is not feasible, therefore, pn+1 is also not an insulated prime. Hence, proved. Theorem 2.2. If α2 = X then (α + r) 2= X for all r ≥ 0. Proof. For some r ≥ 0, we have π(p+α+r) ≥ π(p+α) and π(p−α) ≥ π(p−α−r) because π(n) is an increasing function. This implies π(p + α + r) − π(p − α − r) ≥ π(p + α) − π(p − α). As X contains all the possible candidates for being D(p) and α2 = X, then π(p + α) − π(p − α) > 1 because there exists at least one prime p such that p − α ≤ p ≤ p + α. Using this gives π(p + α + r) − π(p − α − r) ≥ π(p + α) − π(p − α) > 1 which implies π(p + α + r) − π(p − α − r) 6= 1. Thus, α + r is also not a possible candidate for D(p). Corollary 2.3. For prime p, if α2 = X then D(p) < α. Proof. As (α + r) 2= X that is α + r is not a candidate for being D(p) for all r ≥ 0. Now, if non-zero D(p) exists then D(p) must be less than α. 4 ABHIMANYU KUMAR∗ AND ANURAAG SAXENA Theorem 2.4. For a prime p, every m 2 X obeys (p − m)c (1) log(p + m) log(p − m) < m log ((p − m)c(p + m)) + p log p + m ln 113 provided p ≥ 17, where c = 30 113 . Proof. From [8], we have x cx < π(x) < log x log x which gives c(p + m) p − m π(p + m) − π(p − m) < − : log(p + m) log(p − m) For m to belong to set X, the left hand side of the inequality must be one. On solving the expression and rearranging the terms, we get the desired result. Theorem 2.5. For any N > N0, there exists an n 2 [N + 1; 2N + 1] such that D(pn) ≈ gn where gn = pn+1 − pn. Proof. By definition of degree of insulation, we also have D(pn) ≈ min(pn+1 − pn; pn − pn−1)(2) pn+1 − pn (3) = (pn − pn−1) × min 1; pn − pn−1 The ratio of consecutive prime gaps appears in the above expression. Using Theo- rem 3.8 in [6], for n 2 [N + 1; 2N + 1], we have pn+1 − pn c0 log N log2 N log4 N (4) > 2 pn − pn−1 (log3 N) where c0 is a constant and logt N is the t-fold iterated logarithmic function. As N is sufficiently large, so it turns out the right side of the inequality is greater than one. Thus, D(pn) ≈ pn − pn−1 = gn. For a particular prime p, the natural procedure for finding D(p) is inspired from the definition itself, which was illustrated in the previous section for number 23. The process began from m = 1 and kept on checking until the relation is violated. So, it either requires to compute prime counting function or to determine the surrounding primes of the given prime. For extremely large primes, the second approach is highly difficult. However, the prior approach can be improved in view of the above theorems to reduce the effort of checking at all the values, as they allow to use bracketing techniques for faster numerical computation of D(p). Additionally, reader may refer to Appendix A where a faster algorithm is presented for large p. In addition to the above results, which confirm the connection of D(p) with gaps between primes, the definition of D(p) also allows us to directly show the following theorem. Theorem 2.6. For a prime p, we have 1) p − (D(p) + 1) or p + (D(p) + 1) is prime if D(p) is odd, and 2) p − D(p) or p + D(p) is prime if D(p) is even.