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 = {m : π(p − m) = π(p + m) − 1, m ∈ N}, 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 α∈ / X then (α + r) ∈/ 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 α∈ / 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 α∈ / X then D(p) < α.

Proof. As (α + r) ∈/ 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 ∈ 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 ∈ [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 ∈ [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. Corollary 2.7. If D(p) = 1 then either p − 2 or p + 2 is a prime. INSULATED PRIMES 5

Therefore, counting the number of primes with D(p) = 1 gives the count of twin- prime pairs. Let fk be the probability of finding a prime p < 1000 with D(p) = k, then Figure 3 is a line plot of fk with k for primes upto 1000.

Figure 3. Plot of fk for 1 ≤ k ≤ 11 with primes p < 1000

The quantity fk is simply the fraction of primes with D(p) = k over total primes below 1000. The graph depicts that probability falls with increasing k. The prob- ability of prime with D(p) = 1 is highest. Note that D(p) = 1 corresponds to twin primes. For primes upto 105, the pattern is slightly deviated as evident from Figure 4. It is expected to behave nicely for extremely set of primes.

5 Figure 4. Plot of fk for 1 ≤ k ≤ 30 with primes p < 10 If the above trend follows, that is, the number of twin primes are larger than number of primes with other gaps, then its convenient to show the infinitude of 6 ABHIMANYU KUMAR∗ AND ANURAAG SAXENA number of twin-primes. There is huge literature in this direction as well, but this problem lies out of the scope of this paper.

3. Formula for insulated primes In addition to the results listed above, one must also be curious about an approx- imate/asymptotic formula for the insulated primes. Figure 2 indicated an almost linear growth, which was further strengthened by evaluating larger insulated primes.

5 Figure 5. Comparison of in plots for primes less than 10

Using MATLAB command cftool for the Curve fitting toolbox, a variety of curves (with different settings) were tested. It was observed that the equation y = 19.18n1.093 is an extremely good fit as indicated in Figure 5.

6 Figure 6. Comparison of in plots for primes less than 10

The equation performs even better when tested for primes upto one million as shown in Figure 6. It has an R-square value of unity which is extremely good. This INSULATED PRIMES 7

1.093 allows to conclude that in ∼ 19.18n is heuristically an accurate fit. However, this result needs to be rigorously proven, but this analysis is sufficient to convince that insulated primes are definitely well-behaved in comparison to primes or other prime subsets.

4. Conclusion and Future scope This paper defines and explores the properties of a new sequence called the insulated primes. The sequence is 7, 13, 23, 37, 53, 67, 89, 103, 113, 131, 139, 157, 173, 181, 193, 211, 233, 277, 293, and so on. The insulated primes are defined using the concept of degree of insulation, which is also shown to have deep connection with gaps between consecutive primes. A detailed mathematical analysis is conducted which unravels several characteristics. Based on this analysis, a faster algorithm for evaluation of D(p) was suggested as well as a quick discussion on Twin-Prime conjecture was also conducted. Finally, heuristics show that n-th insulated prime in obeys a power law. A strong form of this result is the conjecture that in is varies almost linearly for n → ∞. The authors encourage inquisitive readers to ponder upon the ideas presented in the paper. An extension of the insulated primes is the highly insulated primes. The set of highly insulated primes IH ⊂ I ⊂ P can be defined by treating insulation as an operation on the set I. This is done by applying the definition of insulation on in, that is, in is highly insulated prime if D(in−1) < D(in) and D(in+1) < D(in). This is demonstrated by investigating the triplet of insulated primes (in−1, in, in+1) = (13, 23, 37). As D(13) = 2, D(23) = 4 and D(37) = 3, so 23 is also an highly insulated prime. Similarly, other highly insulated primes are 23, 53, 89, 211, 293, and so on. The interested readers may investigate this sequence.

References [1] J. R. Chen, On the representation of a large even integer as the sum of a prime and the product of at most two primes, Kexue Tongbao 17 (1966), 385–386. [2] K. Goldberg, A table of wilson quotients and the third wilson prime, J. London Math. Soc. 28 (1953), 252–256. [3] R. K. Guy, Unsolved problems in number theory, Springer Verlag, New York, 2004. [4] D. Hawkins and W. E. Briggs, The lucky number theorem, Mathematics Magazine 31 (1957), 81–84. [5] James Maynard, Sums of two squares in short intervals, Analytic Number Theory (C. Pomer- ance and M. Rassias, eds.), Springer, 2015, pp. 253–273. [6] J´anosPintz, On the ratio of consecutive gaps between primes, Analytic number theory (C. Pomerance and M. Rassias, eds.), Springer, 2015, pp. 285–304. [7] R. M. Robinson, Mersenne and fermat numbers, Proceedings of the American Mathematical Society 5 (1954), 842–846. [8] J. B. Rosser and L. Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois Journal of Mathematics 6 (1962), no. 1, 64–94. [9] N. J. A. Sloane, A023188 in: On-line encyclopedia of integer sequences, https://oeis.org/A023188. [10] , A046931 in: On-line encyclopedia of integer sequences, https://oeis.org/A046931. [11] J. Sondow, Ramanujan primes and bertrand’s postulate, Amer. Math. Monthly 116 (2009), 630–635. 8 ABHIMANYU KUMAR∗ AND ANURAAG SAXENA

Appendix A. Modified algorithm for D(p) when p is large For large X and y ∈ (0,X], the prime number theorem shows that on average for x ∈ [X, 2X], we have y π(x + y) − π(x) ≈ . log x Substitute x + y = p + m and x = p − m then 2m π(p + m) − π(p − m) ≈ log(p − m) for 2m ∈ (0,X] and p ∈ [X + m, 2X + m]. For D(p), the left hand side must be unity, therefore we get 2m ≈ log(p − m). Let W(.) is Lambert’s W function then m can be explicitly written as m = p − 0.5 W(2e2p). For large primes p, evaluation of e2p is an tough task, so using the asymptotic relation for Lambert’s W function, we get m ∼ p − 0.5(log 2e2p − 2p 1  2p  log log 2e ) which simplifies to give m ∼ 2 log log 2 + 1 . The above step provides a better way to determine m rather than aimlessly beginning from 1 and linearly searching as it was illustrated for prime 23. However, the derivation of the above step imposes certain conditions which can be relaxed by not treating the obtained m as the final answer. This is done by considering m as just a better candidate to be D(p) which is then substituted in π(p+m)−π(p−m)−1 to cross-check. If its value is zero, then assign m0 = m and m1 = p, else set m0 = 1 and m1 = m. Now, apply bisection method to search D(p) in the interval [m0, m1]. This procedure immensely reduces the number of times one needs to evaluate π(x). An additional note here is that using sharper results on gaps between primes, one can choose a better m1 than just assigning to p in the first case.

Electrical and Instrumentation Engineering Department, Thapar Institute of Engi- neering and Technology, Patiala–147004, Punjab, India Email address: [email protected]

Electronics and Communication Engineering Department, Thapar Institute of Engi- neering and Technology, Patiala–147004, Punjab, India Email address: [email protected]