Primitive Root Bias for Twin Primes

EXPERIMENTAL MATHEMATICS 2019, VOL. 28, NO. 2, 151–160 https://doi.org/./..

Stephan Ramon Garciaa,Elvis Kahoroa,and Florian Lucab,c,d aDepartment of Mathematics, Pomona College, Claremont, CA, USA; bSchool of Mathematics, University of the Witwatersrand, Wits, Johannesburg, South Africa; cMax Planck Institute for Mathematics, Bonn, Germany; dDepartment of Mathematics, Faculty of Sciences, University of Ostrava, Ostrava, Czech Republic

ABSTRACT KEYWORDS Numerical evidence suggests that for only about 2% of pairs p, p 2oftwinprimes,p 2 has more prime; twin prime; primitive primitive roots than does p.If this occurs,we say that p is exceptional+ (there are only two+exceptional root; Bateman–Horn conjecture; Twin Prime pairs with 5 ! p ! 10,000).Assuming the Bateman–Horn conjecture,we prove that at least 0.459% of twin prime pairs are exceptional and at least 65.13% are not exceptional.We also conjecture a precise Conjecture; Brun Sieve formula for the proportion of exceptional twin primes. 2010 AMS SUBJECT CLASSIFICATION A; A; N; N

1. Introduction primitive roots as p 2; that is, ϕ(p 1) ϕ(p 1). + − " + Let n be a positive integer. An integer coprime to n is a If this occurs, then p is unexceptional.Thepreceding primitive root modulo n if it generates the multiplicative inequality holds for all twin primes p, p 2with5 p + ! ! group (Z/nZ)× of units modulo n.Afamous resultof 10,000, except for the pairs 2381, 2383 and 3851, 3853. Gauss states that n possessesprimitiverootsifandonly If p, p 2areprimeswithp 5andϕ(p 1)< + " − if n is 2, 4, an odd prime power, or twice an odd prime ϕ(p 1),thenp is exceptional.Wedonotregardp 3 + = power. If a primitive root modulo n exists, then n has as exceptional for technical reasons. Let πe(x) denote the precisely ϕ(ϕ(n)) of them, in which ϕ denotes the Euler number of exceptional primes p ! x;thatis, totient function. If p is prime, then ϕ(p) p 1and = − π ( ) hence p has exactly ϕ(p 1) primitive roots. e x # p ! x : − = If p and p 2areprime, then p and p 2are twin p and p 2areprimeandϕ(p 1)<ϕ(p 1) . + + #+ − + primes. The Twin Prime Conjecture asserts that there are Computational evidence suggests that approximately 2%$ infnitelymanytwinprimes.Whileitremains unproved, of twin primes are exceptional; see Table 2.Wemakethe recent years have seen an explosion of closely-related following conjecture. work [Castryck et al. 14, Zhang 14, Maynard 15]. Let π2(x) denote the number of primes p at most x for which Conjecture 1. Apositiveproportionofthetwinprimes p 2 is prime. The frst Hardy–Littlewood conjecture are exceptional. That is, limx πe(x)/π2(x) exists and + →∞ asserts that is positive. x dt π2(x) 2C2 (1–1) We are able to prove Conjecture 1,ifweassumethe ∼ (log t)2 !2 Bateman–Horn conjecture (stated below). Our main the- in which orem is the following. p(p 2) Theorem 1. Assume that the Bateman–Horn conjecture C2 − 0.660161815 .... (1–2) = (p 1)2 = holds. p!3 − " (a) The set of twin prime pairs p, p 2 for which + is the twin primes constant [Hardy and Littlewood 04]. ϕ(p 1)<ϕ(p 1) has lower density (as a subset − + A simpler expression that is asymptotically equivalent to of twin primes) at least 0.459%. 2 (1–1)is2C2x/(log x) . (b) The set of twin prime pairs p, p 2 for which + Acasualinspection(see Table 1)suggeststhatif p and ϕ(p 1) " ϕ(p 1) has lower density (as a subset p 2 are primes and p 5, then p has at least as many − + + " of twin primes) at least 65.13%. This article has been republished with minor changes. These changes do not impact the academic content of the article. CONTACT Stephan Ramon Garcia [email protected] Department of Mathematics, Pamona College,  N. College Ave., Claremont, CA , USA. Color versions of one or more of the ﬁgures in the article can be found online at www.tandfonline.com/uexm. © Taylor & Francis Group, LLC 152 S.R. GARCIA ET AL.

Table . For twin primes p, p 2 with 5 p 2000,thediﬀer- in which + ! ! ence δ(p) ϕ(p 1) ϕ(p 1) is nonnegative. That is, p has m at least as many= primitive− − roots+ as does p 2. 1 N f (p)/p + D deg fi and C − , = = (1 1/p)m p ϕ(p 1)ϕ(p 1)δ(p) p ϕ(p 1)ϕ(p 1)δ(p) i 1 p − + − + "= " −  where N f (p) is the number of solutions to f (n)         ( ) ≡         0 mod p [Bateman and Horn 62].         If f1(t) t and f2(t) t 2, then f (t) t(t 2),         = = + = + N f (2) 1, and N f (p) 2forp " 3. In this case,         = =         Bateman–Horn predicts (1–1), the frst Hardy–         Littlewood conjecture, which in turn implies the Twin                 Prime Conjecture.         Although weaker than the Bateman–Horn conjecture,                 the Brun sieve [Tenenbaum 15, Thm. 3, Section I.4.2]         has the undeniable advantage of being proven. It says that                 there exists a constant B that depends only on m and D         such that         BC x dt BC x         ( ) ( ( ))         P x ! m 1 o 1 m D 2 (log t) = + D (log x)         !         for sufciently large x.Inparticular,                 Kx         π ( ) 2 x ! 2         (log x)                 for some constant K and sufciently large x. The best         known K in the estimate above is K 4.5[Wu 04].         = 3. An heuristic argument Computations suggest that the value of the limit in Conjecture 1 is approximately 2%; see Figures 1 and 2.A We give an heuristic argument which suggests that ϕ(p 1) ϕ(p 1) for an overwhelming proportion value for the limiting ratio is proposed in Section 5. − " + of twin primes p, p 2. It also identifes specifc con- It is also worth pointing out that this bias is specifc + ditions under which ϕ(p 1)<ϕ(p 1) might occur. to the twin primes since the set of primes p for which − + ϕ(p 1) ϕ(p 1) is positive (respectively, negative) This informal reasoning can be made rigorous under the − − + has density 50% as a subset of the primes [Garcia and assumption of the Bateman–Horn conjecture (see Section Luca]. That is, if we remove the assumption that p 2is 4). + also prime, then the bias completely disappears. Although Observe that each pair of twin primes, aside from 3,5, is of the form 6n 1, 6n 1. Thus, if p, p 2aretwin only tangentially related to the present discussion, it is − + + primes with p 3, then 2 (p 1) and 6 (p 1).Weuse worth mentioning the exciting preprint [Lemke Oliver " | − | + and Soundararajan]whichconcernsapeculiarandunex- this in the following lemma to obtain an equivalent char- pected bias in the primes. acterization of (un)exceptionality. Lemma 2. If p and p 2 are prime and p " 5,then 2. The Bateman–Horn conjecture + ϕ(p 1) ϕ(p 1) ϕ(p 1) " ϕ(p 1) − " + . − + ⇐⇒ p 1 p 1 The proof of Theorem 1 is deferred until Section 4.We − + frst require a few words about the Bateman–Horn con- (3–1) jecture. Let f1, f2,..., fm be a collection of distinct irre- Proof. The forward implication is straightforward arith- ducible polynomials with positive leading coefcients. metic, so we focus on the reverse. If the inequality on the An integer n is prime generating for this collection if right-hand side of (3–1) holds, then each f1(n), f2(n), . . . , fm(n) is prime. Let P(x) denote 0 p(ϕ(p 1) ϕ(p 1)) ϕ(p 1) ϕ(p 1) the number of prime-generating integers at most x and ! − − + + − + + suppose that f f f f does not vanish identically ! p(ϕ(p 1) ϕ(p 1)) 1 (p 1) 1 (p 1) = 1 2 ··· m − − + + 2 − + 3 + modulo any prime. The Bateman–Horn conjecture is < p(ϕ(p 1) ϕ(p 1)) 5 p − − + + 6 C x dt since 2 (p 1) and 6 (p 1).Fortheprecedingtohold, P(x) , | − | + ∼ D (log t)m the integer ϕ(p 1) ϕ(p 1) must be nonnegative. # !2 − − + EXPERIMENTAL MATHEMATICS 153

Table . The ﬁrst exceptional p.Hereδ(p) ϕ(p 1) ϕ(p 1). = − − +

p δ(p)π2(p)πe(p)πe(p)/π2(p) p δ(p)π2(p)πe(p)πe(p)/π2(p)     . ,    .  −    . , −    . , −   . , −    . , −    . , −   . , −   . , −   . , −    . , −    . , −   . , −   . , −    . , −   . , −   . , −    . , −   . , −    . , −   . , −   . , −   . , −   . , −    . , −    . , −    . , −   . , −   . , −   . , −    . , −   . , −    . , −    . , −   . , −   . , −    . , −    . , −   . , −   . , −   . , −    . , −    . , −   . , −    . , −    . , −    . , −    . , −   . , −   . , −    . , −    . , −   . , −    . , −    . , −    . , −   . , −   . , −    . , −,   . , −   . , −    . , −    . , −    . , −    . , −   . , −    . , −,   . , −   . , −    . , −    . , −    . , −   . , −,   . , −    . , −    . , −   . , −    . , −    . , −,   . , −    . , −,   . , −   . , −    . , −    . , −   . , −   . , −,   . , −   . , − ,   . , −    . , −    . , −   . , −    . , −    . , −   . , −   . , −   . , −    . , −,   . − −

In light of (3–1)andtheformula(inwhichq is prime) (p 1) and the quantities in (3–2)become + ϕ(n) 1 24 1 1 1 1 , 1 and 1 . n = − q 77 − q 3 − q q n q (p 1) % & q (p 1) % & | % & | − | + " q"!13 q"!13 it follows that p is exceptional if and only if p 2isprime + Since and 24 1 1 1 1 1 0.3117 < and 2 5 7 11 770, 1 < 1 (3–2) 77 ≈ 3 · · · = 2 − q 3 − q q (p 1) % & q (p 1) % & |"− |"+ one expects (3–2)toholdoccasionallyifp 770n q!5 q!5 = + 1. Dirichlet’s theorem on primes in arithmetic progres- because 2 (p 1),3! (p 1) and 6 (p 1).Thecondi- sions ensures that p 2 770n 3isprime1/ϕ(770) | − − | + + = + = tion (3–2)canoccurifp 1isdivisiblebyonlysmall 1/240 0.4167% of the time. Thus, we expect a small − = primes. For example, if 5, 7, 11 (p 1), then 5, 7, 11 ! proportion of twin prime pairs to satisfy (3–2). For | − 154 S.R. GARCIA ET AL.

Figure . Numerical evidence suggests that limx πe(x)/π2(x) exists and is slightly larger than 2%.Thehorizontalaxisdenotesthe →∞ number of exceptional twin prime pairs. The vertical axis represents the ratio πe/π2. example, among the frst 100 exceptional pairs (see under the assumption that the Bateman–Horn conjecture Table 2), the following values of p have the form 770n 1: holds. + 3851, 20021, 26951, 47741, 50051, 52361, 70841, 87011, 4. Proof of Theorem 1 98561, 117041, 165551, 167861, 197891, 225611, 237161, 241781, 274121, 278741, 301841, 315701, 322631, Assume that the Bateman–Horn conjecture holds. We 345731, 354971, 357281, 361901, 371141, 410411, frst prove statement (a) of Theorem 1.Inwhatfollows, 424271, 438131, 440441, 470471. p, q, r denote prime numbers. This accounts for 31% of the frst 100 exceptional Proof of (a). Consider twin primes p, p 2suchthat pairs. We now make this heuristic argument rigorous, + 5, 7, 11 (p 1). Let π ′ (x) be the number of such p x. | − 2 !

Figure . Plots in the xy-plane of ordered pairs (p,ϕ(p 1)) (in red) and (p 2,ϕ(p 1)) (in cyan) for twin primes p, p 2.There − + + + are no exceptional pairs visible in Figure a;thatis,ϕ(p 1) " ϕ(p 1) in each case. The exceptional pairs , and , are visible in Figure b.Asmatteringofexceptionalpairsemergeasmoretwinprimesareconsidered.− + EXPERIMENTAL MATHEMATICS 155

Step 1.Since5 7 11 385, the desired primes are pre- in which k0 is the smallest positive integer with k0 · · = 1 ≡ cisely those of the form 2(385)− (mod r). Let b 385k 1. Then − r = 0 + n 385k 1 ! x such that n 2 385k 3isprime. n 385rℓ b and n 2 385rℓ (b 2), = + + = + = + r + = + r + (4–3) In the Bateman–Horn conjecture, let are both prime, n ! x,and f1(t) 385t 1, f2(t) 385t 3, and f f1 f2. = + = + = x br ℓ ! − . Then 385r 1ifp 2, In the Bateman–Horn conjecture, let = 2ifp 3, N f (p) ⎧ = (4–1) f1(t) 385rt br, f2(t) 385rt (br 2), = ⎪0ifp 5, 7, 11, = + = + + ⎨⎪ = and f f f . 2ifp " 13. = 1 2 ⎪ Since p x,wemusthave⎩ k (x 1)/385. For suf- Then N f (p) is as in (4–1) except for p r,inwhichcase ! ! − = ciently large x, the Bateman–Horn conjecture predicts N f (r) 0. Indeed, = that the number of such k is f1(t) br 385k0 1 1 (mod r) (x 1)/385 1 N (p)/p ≡ ≡ + ≡− − f π2′ (x) (1 o(1)) − and f (t) b 2 1 (mod r) = + (log((x 1)/385))2 (1 1/p)2 2 ≡ r + ≡ − "p!2 % − & 2x 1 N f (p)/p for all t.Asx , the Bateman–Horn conjecture pre- (1 o(1)) − →∞ = + 385(log x)2 (1 1/p)2 dicts that the number of such ℓ is % & "p!3 % − & 2x (x br )/(385r) (1 o(1)) π ′, (x) (1 o(1)) − = + 385(log x)2 2 r = + (log((x b )/(385r)))2 % & − r 1 1 2/p 1 N (p)/p − − f × ( / )2 ( / )2 2 , , 1 1 p 1 1 p × (1 1/p) p 5 7 11 % − & p!13 % − & p!2 % − & =" or"p 3 " = x 1 N f (p)/p 2x 1 2/p (1 o(1)) − (1 o(1)) − = + 385r(log x)2 (1 1/p)2 = + 385(log x)2 (1 1/p)2 p!2 % − & % & p!3 % − & " " ( )/ 1 2x 1 N f p p 1 2/p − (1 o(1)) − × − = + 385r(log x)2 (1 1/p)2 p 5,7,11 p!3 % − & =" + , " 2x 2x 1 (1 o(1)) (1 o(1)) = + ( )2 = + 385r(log x)2 (1 1/p)2 385 log x p 5,7,11,r % & % & "p!3 =" − p(p 2) 5 7 11 1 2/p − · · − × ( )2 ( )( )( ) × (1 1/p)2 p 1 5 2 7 2 11 2 p 5,7,11,r % & % − & − − − =" − 2C2x (1 o(1)) 2x p(p 2) ( )2 (1 o(1)) − = + 135 log x = + 385r(log x)2 (p 1)2 % & p!3 % − & π2(x) " (1 o(1)) 5 7 11 r = + 135 · · · × (5 2)(7 2)(11 2)(r 2) > 0.00740740 π2(x). (4–2) − − − − 2C2x (1 o(1)) π ( ) = + 135(r 2)(log x)2 Step 2.Fixaprimer " 13. Let 2′,r x be the number − of primes p x such that p, p 2areprime, π2(x) ! + (1 o(1)) . (4–4) 5, 7, 11 (p 1),andr (p 1). The desired = + 135(r 2) | − | + − primes are precisely those of the form Step 3.Supposethatp is counted by π2′ (x);thatis, suppose that p, p 2areprimeandthat n 385k 1 ! x such that + = + 5, 7, 11 (p 1). Then 6 (p 1),5, 7, 11 ! ( ). | − | + n 2 385k 3isprimeandr 385k 2 (p 1),and + = + | + + In particular, k must be of the form ϕ(p 1) 1 24 − ! 1 . p 1 − q = 77 k k0 rℓ, q 2,5,7,11 % & = + − = " 156 S.R. GARCIA ET AL.

If the pair p is unexceptional, then Lemma 2 ensures 0.02549678 ! (1 o(1)) π2(x) that + 135 < 0.000188865 π (x). 1 1 ϕ(p 1) ϕ(p 1) 24 2 1 + ! − ! . 3 − r = p 1 p 1 77 < ( )3 r (p 1) % & (b) If z r ! log x ,weusetheBrunsieveand |"+ + − r!13 manipulations similar to those used to obtain Consequently, (4–4)tofndanabsoluteconstantK such that 1 77 1 " , K(x/(135r)) + r 1 72 π ′ ( ) r (p 1) 2,r x ! 2 | + % − & (log(x/(135r))) r"!13 in which r is prime. Let for sufciently large x.Sincer ! (log x)3, 1 ( /( )) ( 1/2) ( )/ F(p) log 1 . log x 135r " log x " log x 2 = + r 1 r (p 1) | + % − & 14 r-!13 holds if x " 10 . Then (1–1) ensures that

4Kx 5Kπ2(x) Step 4. We want to count the twin primes pairs π ′, (x) ! ! 2 r 135r(log x)2 135(r 2) p, p 2withp x, F(p) log(77/72),and − + ! " 5, 7, 11 (p 1).Todothis,wesumupF(p) over for sufciently large x.Nowwefxz such | − 9 all twin primes p counted by π ′ (x) and change that 5K/(135(z 2)) < 10− .Sincelog(1 2 − + the order of summation to obtain t) 0, for sufciently large x we A(x) F(p) obtain = p counted- by 1 π ′ (x) A (x) π ′ (x) log 1 2 2 = 2,r + 3 r 1 1 z r 2 + r 1 1 -r z − % − & ! π2′,r(x) log 1 5Kπ (x) 1 + r 1 < 2 13"r"z % − & ( )( ) - 135 r>z r 2 r 1 1 - − − π ′, (x) log 1 5Kπ2(x) 1 1 + 2 r + r 1 z r 2 − r 1 -r z % − − & 1 5Kπ (x) π2′,r (x) log 1 2 + + r 1 ! (log x)3

π2(x) ber is at most π ′, (x) (1 o(1)) 2 r = + 135(r 2) x x π , (x) 1 1. − 2 r ! + ! + uniformly for r [13, z]asx . For suf- 385r 385r ∈ →∞ 2 3 fciently large x we have1 Since log(1 t) 0, the terms of the series are O(1/r2 ) x 1 1 + ! and hence it converges rapidly enough for reliable numerical evaluation. ! ( ) + 385 3 r r 1 3 r 1 Mathematica provides the value ..... r>(-log x) − (log x-)

x 1 1 x dt since 3 ! (p 1) and 6 (p 1). If we let ! − | + 385 r 1 − r + (log x)3 2 t r>(log x)3 % − & ! − 1 - G(p) log 1 , x t x = + r 1 = r (p 1) % & ! log t |-− − 385((log x)3 1) + t (log x)3 2 r!5 − % 6 = − & 2x 6 ( ) ( / ) log x 6 then G p " log 3 2 holds for all exceptional primes p. ! ( )3 385 log x + Let πe(x) denote the number of exceptional primes p ! x. 3 1 (log x) 2C2x Then 2 = 385C2 log x + 2C2x (log x) % & πe(x) log(3/2) G(p) 3 ! 1 (log x) p counted (1 o(1)) π (x) -π ( ) = + + 2 by 2 x %385C2 log x 2C2x & 9 1 < 10− π (x). log 1 2 = + r 1 p counted r!5 % − & by-π (x) r -(p 1) Step 5.Returningto(4–5)andusingtheprecedingthree 2 | − 1 estimates, we have ! log 1 1 + r 1 5"r"x % − & p counted by π2 (x) - p 1-(mod r) A(x) A (x) A (x) A (x) ≡ = 1 + 2 + 3 9 9 1 1 < 0.000188865 π (x) 10− π (x) 10− π (x) ! (1 o(1))π2(x) log 1 2 + 2 + 2 + (r 2) + r 1 r!5 − % − & < 0.000188866 π2(x). - < 0.14137 π2(x),

for sufciently large x. which shows that there are at least Step 6. Let (x) be the set of primes p counted by U 0.14137 π ′ (x) that are unexceptional; that is, ϕ(p 1)/ π ( ) π ( ) π ( ) > . π ( ) 2 − 2 x e x " 2 x 1 0 6513 2 x (p 1) ϕ(p 1)/(p 1) by Lemma 2.Aswe − − log(3/2) − " + + % & have seen, if p (x), then F(p) log(77/72). ∈ U " unexceptional primes at most x. # Thus,

0 # (x) log(77/72) F(p) A(x) 5. Conjectured density ! U ! ! p (x) ∈-U Below we conjecture a value for the density of the excep- 0.000188866 π (x), ! 2 tional primes relative to the twin primes. In what follows, we let P(n) denote the largest prime factor of n and let from which we deduce that p(n) denote the smallest. We let µ denote the Möbius 0.000179 function and remind the reader that µ2(n) 1ifandonly # (x) π (x)<0.00281306 π (x). = U ! ( / ) 2 2 if n 1orn is the product of distinct primes. %log 77 72 & = Conjecture 2. The density of the exceptional twin primes The primes p counted by π2′ (x) which are not in (x) are exceptional; that is ϕ(p 1)/(p 1)<ϕ(p is U − − + 1)/(p 1).By(4–2)andtheprecedingcalculation,for πe(x) q 4 1 + lim lim − . ε large x there are at least x π2(x) = 0 q 2 p 4 →∞ → 1 a,b p ab 5"q" ε % − &% | % − && " µ2-(ab) 1 " = 1 π ( ) ( )> . . π ( ) 5"p(ab)"P(ab)" ε 2′ x # x (0 00740740 0 00281306) 2 x ϕ(a) ϕ(b) − U − 2a " 3b > 0.00459 π2(x) (5–1) such primes. This completes the proof of statement (a) A few remarks about the imposing expression (5–1) from Theorem 1. # are in order. First of all, for each fxed ε>0, the sum involves only fnitely many pairs a, b.Indeed,thecondi- Proof of (b). This is similar to the preceding, although it is tion µ2(ab) 1 ensures that ab is a product of distinct much simpler. As before, p, q, r denote primes. If p, p 2 = + prime factors. The restriction 5 ! p(ab) ! P(ab) ! 1 are prime and p is exceptional, then ε implies that only fnitely many prime factors are available 1 1 ϕ(p 1) ϕ(p 1) 1 to form a and b.Inprinciple,theright-handsideof 1 − ! + ! (5–1)canbeevaluatedtoarbitraryaccuracybytaking 2 − r = p 1 p 1 3 r (p 1) | − % & − + ε sufciently small. Unfortunately, the number of terms "r!5 158 S.R. GARCIA ET AL. involved in the sum grows rapidly as ε shrinks and we are that qt 1andqt 3 are prime. The number of them is, + + unable to obtain a reliable numerical estimate from (5–1). by the Brun sieve, As a brief “sanity check,” we also remark that the limit x π2(x q, 1) . in (5–1), if it exists, is at most 1. Without the condition ; ≪ (q 1)(log x)2 − ϕ(a) ϕ(b) ! , The Prime Number Theorem and Abel summation reveal 2a 3b that the inner sum in (5–1)is x 1 επ2(x) ε ( ) . 1 1 F p 2 2 1 ω(n) ≪ (log x) (q 1) ≪ log( ) 2 p x > 1 ε p 4 = p 4 " q ε − a,b p ab n p n p, p -2 prime - | % − & µ2( ) | % − & µ2-(ab) 1 " -n 1 " + = = 1 1 5"p(n)"P(n)" ε 5"p(ab)"P(ab)" ε If we let 2 1 = + p 4 ε p : p, p 2primeandFε (p)>ε, 1 % & A ={ + } 5""p" ε − p 2 then − , = 1 1 p 4 − 5"p" ε % − & 1 " # ε (x)ε Fε (p) ε log π (x), A ! ≪ ε 2 which precisely ofsets the frst product in (5–1). p"x % & p, p -2 prime 4 5 To proceed, we need to generalize the functions F and + ε> 1 1 G that appeared in the proof of Theorem 1. Let 0and which gives # ε (x) O((log( ε ))− π2(x)). # defne A = 1 1 To justify our conjecture, we look at the ε -part of Fε (p) log 1 and p2 1. We frst let ε 0.5. We note that 2 (p 1), = + r 1 − ! | − r (p 1) % − & ( ) ( ) |-+1 2 p 1 and 3 p 1 for all twin primes p " 5. For r! ε | + | + two coprime square-free numbers a, b with 5 ! p(ab) ! 1 P(ab) 1 ,wesaythatthetwinprimep is of 1 -type (a, b) Gε (p) log 1 . ! ε ε = + r 1 r (p 1) % − & if |-−1 r! ε p 1 2α qαq qγq and − = Particular instances of these functions have appeared in q a > 1 "| q"ε the proof of Theorem 1 with ε 1/5forFε (called F)and = p 1 2β 3γ qβq qδq ε 1/13 for Gε (called G), respectively. + = = q b > 1 "| q"ε Lemma 3. For ε>0,thenumberoftwinprimesp! x 1 1 for some positive α,β,γ,αq and βq for q ab and non- such that Fε (p)>εis O((log( ))− π2(x)).Thesamecon- ε γ ,δ 1 | clusion holds with Fε replaced by Gε. negative q q for q " ε .Thatis,theprimefactorsof p 1thatare 1 are exactly the ones dividing 2a and − ! ε Proof. The argument is essentially already in the proof of the prime factors of p 1thatare 1 are exactly the ones + ! ε Theorem 1.WedoitonlyforFε (p) since the argument dividing 6b. for Gε (p) is similar. We sum Fε (p) for p x with p, p 2 ε ( , ) ! + Given and a b , let prime and use the fact that log(1 t) ! t to obtain + c , q. 1 a b = ( ) 5 q 1 Fε p ! """ ε q 1 q ab p x p x q (p 1) ! " " | − − p, p -2 prime p, p -2 prime -q> 1 + + ε Note that 1 1 ϕ(p 1) 1 ϕ(a) 1 = q 1 − 1 and > 1 p, p 2 prime q ε − + p 1 = 2 a − q - p 1-(mod q) − q (p 1) % & ≡ |"−1 q> ε π2(x, q, 1) , ϕ( ) ϕ( ) = q 1 p 1 1 b 1 > 1 + 1 . -q ε − p 1 = 3 b − q + q (p 1) % & |"+1 in which π (x q, 1) denotes the number of primes p q> ε 2 ; ! x with p, p 2primeandp 1 (mod q).Bytheusual + ≡ Since argument, the number of twin primes p, p 2withp ! + 2y y 1 x and p 1 (mod q) equals the number of t x/q such e− < 1 y < e− for y < , ≡ ! − 2 EXPERIMENTAL MATHEMATICS 159 it follows that up to x,thesameprogramcanbeappliedtoprovethatthe additional assumption holds under the Bateman–Horn 2ε Fε (p) 1 1 4ε ε set of twin primes p ! x such that ( ) hold for all twin primes p ! x except the ones in ε x ,a ϕ( ) ϕ( ) 1 1 A p 1 p 1 set of cardinality O((log( )− π (x)). Consequently, − < + ε 2 p 1 p 1 − + ϕ(a) ϕ(p 1) (1 4ε) ! − is within a set of cardinality O(h(ε)π2(x)) from the set of − 2a p 1 − primes for which 1 1 holds for all but O((log( ε ))− π2(x)) twin primes p ! x. ϕ(a) ϕ(b) < . (5–4) Thus, the inequality 2a 3b ϕ(p 1) ϕ(p 1) With this assumption, we proceed as in [Garcia and Luca, − ! + 1 p 1 p 1 Section 2.11]. Fix , a, b,andc ca,b.Wealsofx a residue − + ε = class for p modulo c which is not 0, 1, 2 .Inthiscase implies that { ± − } we need to count natural numbers of the form ϕ(a) 1 ϕ(b) ! (1 4ε)− . abct κ, 2a − 3b + Let us consider twin primes for which in which κ is fxed such that abct κ ! x, ϕ(b) ϕ(a) 1 ϕ(b) + < <(1 4ε)− (5–2) abct κ and abct κ 2areprime, − r + + + 3b 2a 3b abct κ 1aredivisiblebyallprimesina and r + − occurs. Since r coprime to cb, ϕ(a) ϕ(p 1) abct κ 1 is divisible by all primes in b (and − (1 O(ε)) and + + 2a = p 1 + coprime to ca). − r ϕ(b) ϕ(p 1) Observe that κ is uniquely determined modulo abc + (1 O(ε)) 3b = p 1 + once it is determined modulo c. By the Bateman–Horn + conjecture, this number is 1 1 for all p ! x with O((log( ))− π2(x)) exceptions, it fol- ε 1 lows that twin primes p ! x for which (5–2)holdshave (1 o(1))π2(x) . + (p 2) the additional property that p abc "| − ϕ(p 1) ϕ(p 1) We next sum this over all q 4 progressions modulo q − + O(ε). (5–3) − − = for which abct κ is not congruent modulo q to some 6 p 1 p 1 6 + 6 − + 6 member of 0, 1, 2 and for all q c getting an amount Let ε be6 the set of twin primes for6 which (5–3)holds.We { ± − } | B 6 6 of make the following additional assumption. 1 q 4 (1 o(1))π2(x) − Additional assumption.Thenumberoftwinprimesp ! + p 2 q 2 p ab % & p c % & x for which (5–3)holdsisO(h(ε)π2(x)) for some func- "| − "| − tion h(y) with h(y) 0asy 0. q 4 1 → → (1 o(1)) − . = + q 2 q 4 Assumption (5–3)hasbeenshowntoholdwhenp 1 % & q ab % & 5""q" ε − "| − is only a prime [Garcia and Luca]. That is, the num- We now sum up over all pairs a, b with ber of primes p ! x such that (5–3)holdsisatmost O(h(ε)π(x)), where h(ε) tends to zero when ε 0. In ϕ(a) ϕ(b) → < , fact, this was a crucial step in showing that ϕ(p 1) 2a 3b − − ϕ(p 1) has no bias if only p is assumed to be prime. + which yields a proportion of Proving this for primes uses the Turan–Kubilius the- q 4 1 orem about the number of prime factors q ! y of p 1 (1 o(1)) − µ2(ab) ± + 1 q 2 , p 4 when p is prime as the parameter y tends to infnity and 5"q" ε % − & a b p ab % − & " ( -) ( ) 1 "| also Sperner’s theorem from combinatorics. With some 5"p ab "P ab " ε ϕ(a) < ϕ(b) nontrivial efort, which involves proving frst a Turan– 2a 3b Kubilius estimate for the number of distinct primes q ! of π2(x) with a number of exceptions p ! x of counting 1/ε of p 1andp 1 when p ranges over twin primes function O(h(ε)π (x)).ThissupportsConjecture2. − + 2 160 S.R. GARCIA ET AL.

6. Comments follows that the set of such primes has density zero in the twin primes. We did not need the full strength of the Bateman–Horn conjecture, just the case r 2andD 1 for certain spe- = = cifc pairs of linear polynomials f1(t) and f2(t). Under Acknowledgments this conjecture, we have seen that ϕ(p 1) ! ϕ(p 1) − + We thank Tomás Silva for independently computing the ratio for a substantial majority of twin prime pairs p, p 2. + πe(x)/π2(x) for large x.Wealsothanktheanonymousreferee There are a few twin primes p, p 2forwhich + for suggesting the approach of Section 5. ϕ(p 1) ϕ(p 1). (6–1) − = + Funding For only such p ! 100,000,000 are SRG supported by NSF grant DMS-1265973, a David L. Hirsch 5, 11, 71, 2591, 208,391, 16,692,551, 48,502,931, III and Susan H. Hirsch Research Initiation Grant, and the 92,012,201, 249,206,231, 419,445,251, 496,978,301. Budapest Semesters in Mathematics (BSM) Director’s Mathe- matician in Residence (DMiR) program. The following result highlights the rarity of these twin FL was supported in part by grants CPRR160325161141 and primes. an A-rated researcher award both from the NRF of South Africa and by grant no. 17-02804S of the Czech Granting Agency. Theorem 4. The number of primes p xwithp 2 prime ! + and ϕ(p 1) ϕ(p 1) is O(x/ exp((log x)1/3). − = + References Proof. Suppose that j and j k have the same prime fac- + [Bateman and Horn 62] P. T. Bateman and R. A. Horn. “A tors, let g ( j, j k),andsupposethat = + Heuristic Asymptotic Formula Concerning the Distribution j j k of Prime Numbers.” Math. Comput. 16 (1962), 363–367. r 1and + r 1 (6–2) [Castryck et al. 14] Wouter Castryck, Étienne Fouvry, Gergely g + g + Harcos, Emmanuel Kowalski, Philippe Michel, Paul Nelson, Eytan Paldi, János Pintz, Andrew V. Sutherland, Terence are primes that do not divide j. Then Tao, and Xiao-Feng Xie. “New Equidistribution Estimates j k of Zhang Type.” Algebra Number Theory 8: 9 (2014), 2067– n j + r 1 (6–3) 2199. = + % g & [Garcia and Luca] S. R. Garcia and F. Luca. “On the Difer- satisfes ϕ(n) ϕ(n k) [Graham et al. 99, Thm. 1]. ence in Values of the Euler Totient Function near Prime = + Arguments.” Irregularities in the Distribution of Prime For k fxed, the number of solutions n x to ϕ(n) ! = Numbers Research Inspired by Maier’s Matrix Method. ϕ(n k) which are not of the form (6–3) is less than https://arxiv.org/abs/1706.00392 + / x/ exp((log x)1 3) for sufciently large x [Graham et al. [Graham et al. 99] S. W. Graham, J. J. Holt, and Carl Pomer- ance. “On the Solutions to ϕ(n) ϕ(n k).” In Num- 99, Thm. 2]. = + We are interested in the case k 2andn p 1, in ber Theory in Progress, Vol. 2 (Zakopane-Ko´scielisko, 1997), = = − edited by K. Györy, H. Iwaniec, and J. Urbanowicz, pp. 867– which p, p 2areprime.If j and j 2havethesame + + 882. Berlin: de Gruyter, 1999. prime factors, then they are both powers of 2. Thus, j 2 = [Hardy and Littlewood 04] G. H. Hardy and J. E. Little- and j k 4, so g 2. From (6–2)weseethatr is such + = = wood. “Some Problems of ‘Partitio Numerorum’; III: On the that Expression of a Number as a Sum of Primes.” Acta Math. 114: 3 (2004), 215–273. r 1and2r 1 [Lemke Oliver and Soundararajan] R. Lemke Oliver and K. + + Soundararajan. “Unexpected Biases in the Distribution of are prime. Then n 2(2r 1) p 1, from which it = + = − Consecutive Primes.” Proc. Natl. Acad. Sci. 113: 31, E4446– follows that p 4r 3andp 2 4r 5areprime. E4454. (http://arxiv.org/abs/1603.03720). = + + = + Consequently, [Maynard 15] J. Maynard. “Small Gaps between Primes.” Ann. of Math. (2) 181: 1 (2015), 383–413. r 1, 2r 1, 4r 3, and 4r 5, [Tenenbaum 15] G. Tenenbaum. Introduction to Analytic and + + + + Probabilistic Number Theory,volume163ofGraduate Stud- are prime. However, this occurs only for r 2 since oth- ies in Mathematics.Thirdedition.Providence,RI:Ameri- = erwise one of the preceding is a multiple of 3 that is larger can Mathematical Society, 2015. Translated from the 2008 French edition by Patrick D. F. Ion. than 3. # [Wu 04] J. Wu. “Chen’sDouble Sieve, Goldbach’sConjecture and the Twin Prime Problem.” Acta Arith. 114: 3 (2004), 215– In particular, the number of primes p ! x for which 273. p 2isprimeandϕ(p 1) ϕ(p 1) is o(x/(log x)2). + − = + [Zhang 14] Y. Zhang. “Bounded Gaps between Primes.” Ann. Assuming the frst Hardy–Littlewood conjecture, it Math. (2) 179: 3 (2014), 1121–1174.