FREQUENCIES OF SUCCESSIVE PAIRS OF PRIME RESIDUES

Transcription

1 FREQUENCIES OF SUCCESSIVE PAIRS OF PRIME RESIDUES AVNER ASH, LAURA BELTIS, ROBERT GROSS, AND WARREN SINNOTT Abstract. We consider statistical roerties of the sequence of ordered airs obtained by taking the sequence of rime numbers and reducing modulo m. Using an inclusion/exclusion argument and a cut-off of an infinite roduct suggested by Pólya, we obtain a heuristic formula for the robability that a air of consecutive rime numbers of size aroximately x will be congruent to a, a+d) modulo m. We demonstrate some symmetries of our formula. We test our formula and some of its consequences against data for x in various ranges. 1. Introduction In a beautiful aer from 1959, Pólya [5] resented a heuristic formula to aroximate the number of rimes < x such that q = + d is also rime, where d is a ositive even integer. His formula is a secial case of the Bateman Horn conjecture [2], and had already been roosed by Hardy Littlewood [4]. However, his method of derivation is much more elementary. Pólya uses the standard robabilistic model of the rimes. His innovation is deciding where to cut off a certain roduct, which he determines by invoking Mertens s Theorem. In our treatment, this cut-off ste, involving Euler s constant, occurs in 3.1). Pólya does not care whether there are rimes between and q. In contrast, we ask about consecutive rime airs, and we are interested in gas not only equal to d but congruent to d modulo a given ositive integer m. This interest stems from our earlier work [1] in which we studied the statistics of the sequence of rime numbers modulo m. For this reason, we actually look at the finer count of consecutive rime airs congruent to a given a, a + d) mod m). We can thus state our roblem as follows: ) Given ositive integers a, d, and m, and a ositive number x, define Na, d, m, x) to be the number of consecutive rime airs < q such that < x, a and q a + d mod m). What are the asymtotics of Na, d, m, x) as x tends to infinity? Dirichlet s Theorem tells us that if we look only at single rimes, the number of rimes < x with a mod m) is aroximately πx)/ϕm), indeendent of a rovided that a is relatively rime to m). Here ϕ is the Euler ϕ-function. One might suose that the residues modulo m of consecutive rime airs would be indeendent in a seudo-robabilistic sense, and therefore that Na, d, m, x) would be aroximately indeendent of a and d rovided that a and a + d are relatively rime to m). However, this aears not to be the case in the numerical evidence comiled in [1] and in this aer. Date: Mathematics Subject Classification. 11N05, 11K45, 11N69. Key words and hrases. Prime airs, Bateman Horn, Hardy Littlewood, Pólya. 1

2 2 AVNER ASH, LAURA BELTIS, ROBERT GROSS, AND WARREN SINNOTT For examle, suose m = 4 and consider consecutive airs of rimes congruent to 1, 1) and 1, 3) resectively. Our data show that there is a definite tendency for the 1, 1)-airs to occur considerably less frequently than the 1, 3)-airs. We do not know if this tendency ersists as x increases. The work of Hardy and Littlewood [4] suggests an exlanation of the imbalance. If d is a ositive even integer, then Hardy and Littlewood roose that see [4,. 42, Conjecture B]: the robability that x and x + d are rime 1 2 2C 2 log x). 2 Here In articular C 2 = >2 ) ) 2 d,>2 the robability that x and x + 2 are rime 2C 2 log x). 2 We observe that the deendence on d occurs only through the factor 1 d,>2, and 2 therefore this factor should tell us the relative frequency of rimes such that + d is also rime when comared to the frequency of twin rimes. Here is a table of values for small d: d,>2 d Thus, for examle, rimes such that + 6 is also rime should occur twice as often as twin rimes. If is a rime congruent to 1 mod 4, and the next rime q is congruent to 3 mod 4, then q = + d for d {2, 6, 10, 14,... }; while if the next rime q is congruent to 1 mod 4, then q = + d for d {4, 8, 12, 16,... }. The table above suggests that, q) 1, 3) mod 4) would be more likely. However the suggestion raises further questions, since even if + d is rime, it does not mean that it is the next rime after. For examle, it is erfectly ossible for each of, + 4, and + 6 to be rime. Thus there are interactions between the various events and + d are rime that need to be taken into account if we want to redict whether, q) 1, 3) mod 4) is more likely than, q) 1, 1) mod 4). To the best of our knowledge, Problem ) is wide oen, and cannot be treated using L-functions, unlike the case of Dirichlet s Theorem. In this aer, we attemt a heuristic solution to Problem ) by alying an inclusion/exclusion argument, together with Pólya s heuristic argument, in 3. This leads to a formula for a function we call P a, d, m, x), which heuristically should be the robability that x is a rime congruent to a mod m) and the next largest rime is congruent to a + d mod m). Thus, we should have the aroximation Na, d, m, x) P a, d, m, x)x, in the sense that 1.1) lim x P a, d, m, x)x Na, d, m, x) = 1. However, 1.1) is not the aroriate conjecture to verify. For the same reasons that the aroximation πx) x/ log x is less accurate than πx) x dy/ log y, we instead comare 2

3 FREQUENCIES OF SUCCESSIVE PAIRS OF PRIME RESIDUES 3 Na, d, m, X) Na, d, m, x) with X y=x+1 P a, d, m, y). We do not attemt to rovide a measure of how good this aroximation should be, since there is no actual underlying robability distribution we are samling. Neither did Pólya in [5]. In fact, our exression for P is an infinite series, and we do not know whether this series converges. Thus, we choose an integer J 0 and truncate P at the J th term to obtain a finite exression P J a, d, m, x). Our heuristic rooses that P J a, d, m, x) should be the robability that x is a rime congruent to a mod m) and that the next rime is x+d+jm for some j = 0,..., J. Thus, if J is sufficiently large given x), our heuristic suggests that P J a, d, m, x) should be a density function for Na, d, m, x) in the sense given above. Our heuristic does not tell us exactly how to choose J. We suggest below that d + Jm 4 log x is a reasonable choice. We first investigate roerties of the heuristic formula for P J for a fixed J, and test them against some actual data. Below we resent data for only a few values of m, which suffice to convey the general icture. The function P J a, d, m, x) ossesses two exact symmetries. First, Proosition 4.2 states that P J a, d, m, x) = P J a d, d, m, x). We call this antidiagonal symmetry because of its aearance when we arrange the values of P J a, d, m, x) in a matrix indexed by a and a+d. Second, Proosition 4.1 says that when m is a ower of 2, P J a, d, 2 k, x) is indeendent of a. We test these two symmetry redictions against actual data in 5 and 6 resectively. The idea is that X y=x+1 P Ja, d, m, y) should closely aroximate Na, d, m, X) Na, d, m, x). Therefore the matrix of Na, d, m, X), indexed by a, a + d), should show aroximately these two symmetries. In addition to these symmetries, the observations at the start of 4 imly that if we combine the terms in 1/log x) 2 from P J a, d, m, x), we obtain a sum which is indeendent of a. Heuristically, we might exect this term to be dominant as x and therefore indeendence of a should aear for large x. We test this rediction numerically in 8. It would be interesting to have analytical roofs of any of these symmetries for the ratio Na, d, m, x)/πx) in the limit as x. As stated above, we do not know whether Na, d, m, x)/πx) might be indeendent of both a and d in this limit. We comare the actual values of X y=x+1 P Ja, d, m, y) and Na, d, m, X) Na, d, m, x) for x = 10 3, X = 10 6, and various small values of a, d, and m in 7. We begin our sum at 10 3 because our heuristic relies on x being large relative to a, d, and m. The alternating sums 3.7) become too large for feasible comutation when X gets much beyond For any x, P J 1, 2, 2, x) should aroximate log x) 1. We test this for J = 28 and various x in 4. We also discuss what haens when J varies. We rove an internal consistency, called vertical comatibility, in Proosition 4.3 and Corollary 4.4. This asserts that if m n and J is given, there exists J such that P J a, d, m, x) equals a certain sum of P J a, d, n, x) s. The function P J aears to stabilize surrisingly quickly as a function of J. Exerimentally, for small values of x, P J still aears to remain aroximately constant, even when J is much larger than 4 log x. See the remarks at the end of 4. This stabilization as J varies is somewhat surrising; our heuristic is based on a robabilistic icture, whereas, for examle,

4 4 AVNER ASH, LAURA BELTIS, ROBERT GROSS, AND WARREN SINNOTT we know that the next rime after is certainly less than 2. It would be interesting to have a theoretical gras of how P J varies with J. We could try to eliminate the deendence on J by looking at P a, d, m, x) := lim J P J a, d, m, x). However, we see no way of roving that the limit exists. The sum defined by the limit is certainly not absolutely convergent see 4 below). Our heuristic makes less and less sense if x is fixed and J increases. Acknowledgments. We thank Jenny Baglivo and K. Soundarajan for their hel, and the referee for helful comments. The first and second authors thank the National Science Foundation for artial funding of this research under grant DMS Notation To simlify the job of the reader, we record all of our considerable notation at the outset. Fix an integer m > 1, and a ositive integer a. Let S be a finite set of nonnegative numbers containing 0. Let be a rime. Let x > 2 be an integer. We define: 2.1) 2.2) 2.3) n = ns) = cards) n = n S) = cards mod ) 1 n ) 1 if n = 1 n otherwise ) 2.4) A n = 1 n ) n 2.5) 2.6) 2.7) 2.8) 2.9) 1 1 αs) = 1 n ) S) 1 ns) ) 1 if a + s, m) = 1 for all s S βa, m, S) = m n S) m 0 otherwise S k = {0, 1,..., k} Qa, d, m, x, j) = 1) ns) A ns) αs)βa, m, S) log x) ns) {0,d+jm} S S d+jm P J a, d, m, x) = J Qa, d, m, x, j) j=0

5 FREQUENCIES OF SUCCESSIVE PAIRS OF PRIME RESIDUES 5 The definition of n in 2.2) means the number of distinct residue classes modulo in the set S. Once is larger than the largest element in S, n S) = ns). The roduct in 2.4) is infinite but converges, because for > n, both the numerator and denominator have leading terms 1 n. The roduct in 2.5) contains only finitely many terms that do not equal 1, because once > n and > maxs), the numerator and denominator are equal. Below, we will tyically think of a as a ositive integer. However, a aears only as the first argument in βa, m, S). In turn, the definition of βa, m, S) only deends on a in testing the condition a + s, m) = 1. Thus, βa, m, S), Qa, d, m, x, j), and P J a, d, m, x) only deend on the value of a mod m). 3. The Heuristic We aly Pólya s heuristic [5] to the roblem of determining the following robabilities: 1. Given a ositive integer d, the robability that x and x + d are successive rimes. 2. Given integers a, d 1 and m 2, the robability that x and x + d are successive rimes and x a mod m). 3. Given integers a, d 1 and m 2, the robability x is rime, x a mod m) and the next rime is congruent to a + d mod m) The robability that x and x + d are successive rimes. Let S be a nonemty finite set of nonnegative integers. Let be a rime number. Then for integers x Prob x + k for k S) n, where n = n S) is defined in 2.2). Note that 1 n. Furthermore, if n =, then each residue class mod is reresented by some k S, hence x + k for some k S, so that Prob x + k for k S) is indeed 0. Also n = n := card S as soon as is sufficiently large larger than the largest element in S will suffice). In [5], Pólya roosed a way to ass from the robability that x + k for k S to the robability that x + k is rime for k S. He argues that the events x + k for k S for various rimes should be considered heuristically indeendent as long as is small comared with x, so that we may conclude that Prob x + k for k S for all < y) n <y as long as x is sufficiently larger than y. Now if x > y > x 1/2 then x + k for k S for all < y x + k is rime for k S at least if x is large enough that the size of the elements of S can be ignored. Pólya roosed that the correct value of y should be x µ, where µ = e γ, and γ = is Euler s constant. His justification for this trick of the magic µ is simly and solely that it works when S = {0}: indeed by Mertens s Theorem, 1 1 log x, <x µ

6 6 AVNER ASH, LAURA BELTIS, ROBERT GROSS, AND WARREN SINNOTT while the robability that x is rime is also 1 we roose log x, by the Prime Number Theorem. Accordingly 3.1) Probx + k is rime for each k S) n. <x µ We rewrite 3.1) as follows: n n <x µ n 3.2) n n n n n<<x µ n n n n n<<x µ n<< n n 1 n )) ) 1 n ) n 1 1 n<<x µ <x µ ) 1 n ) n 1 1 <x µ n<<x µ 1 1 ) n 1 1 ) n. The second roduct may be viewed as a finite roduct, because n = n for sufficiently large. The first two roducts are αs), as defined in 2.5). The third roduct aroaches A n, as defined in 2.4). The fourth roduct satisfies <x µ 1 1 ) n by Mertens s Theorem. Putting all this together gives: 1 log x) n. 3.3) Probx + k is rime for k S) αs) log x). n As a check, let d be a ositive integer, and take S = {0, d}. We should retrieve the Hardy Littlewood formula [4]. In fact, ) ) A 2 = = = 4 ) 1 1, 1 ) 1 >2 1 ) 1 1) 2 >2 and n S) = 1 if d, n S) = 2 otherwise. Hence, 0 if d is odd α{0, d}) = 1 1 if d is even, 2 2 d,>2 which agrees with the formula from [4] quoted in the Introduction note that our A 2 is 4 times Hardy and Littlewood s C 2 ). Let d be a ositive integer. We want to aly the formula 3.3) to find the robability that x and x + d are successive rimes. Let S d = {0, 1,..., d}. Then by inclusion/exclusion: A n

7 FREQUENCIES OF SUCCESSIVE PAIRS OF PRIME RESIDUES 7 3.4) Probx is rime and x + d is the next rime) = 1) ns) A ns) αs) log x). ns) {0,d} S S d 3.2. The robability that x and x + d are successive rimes and x a mod m). We start over with a congruence condition. Let m 2 and a 1 be integers. Then Probx a mod m)) 1 m. Suose that is a rime divisor of m: then the congruence x a mod m determines whether x + k for any k S. Indeed if a + k is not relatively rime to m for some k S then Probx a mod m) and x + S are rimes) = 0. Here x + S denotes {x + s : s S}. On the other hand, if m, the conditions x + k for k S should be indeendent of congruences modulo m and indeendent of each other), heuristically seaking. So if a + k is relatively rime to m for every k S then Probx a mod m) and x + S are rimes) 1 m <x µ m n. So 1 n if a + k, m) = 1 k S m Probx a mod m) and x + S are rimes) <x µ m 0 otherwise. Now, in the first case, when the elements of a + S are all relatively rime to m, n < for all m, so we have 1 n = 1 n 1 A n αs) m m n m n <x µ log x). n Thus <x µ m m 3.5) Probx a mod m) and x + S are rimes) βa, m, S)αS) log x), n where βa, m, S) is defined in 2.6). Let d be a ositive integer. Then by inclusion/exclusion: m A n 3.6) Probx is rime, x a mod m), and x + d is the next rime) 1) ns) 2 Probx a mod m) and x + S are rimes) {0,d} S S d 1) ns) A ns) αs)βa, m, S) log x). ns) {0,d} S S d = Qa, d, m, x, 0).

8 8 AVNER ASH, LAURA BELTIS, ROBERT GROSS, AND WARREN SINNOTT 3.3. The robability that x is rime, x a mod m), and the next rime is a + d mod m). Finally, suose that 1 d m, and that a, m) = 1 and a + d, m) = 1. Then 3.7) Probx a mod m), x rime, and next rime is a + d mod m) J P J a, d, m, x) = 1) ns) A ns) αs)βa, m, S) log x). ns) j=0 {0,d+jm} S S d+jm What value of J should be used in the comutation of P J a, d, m, x)? One ossibility is to let j range from 0 u to. However, there are various finite uer bounds which can be used. We need to use a large enough value of J to insure that the set x + S d+jm contains the next rime after x assuming that x is rime). Bertrand s Postulate guarantees that choosing J so that d + Jm x is sufficient. We can safely use a much smaller value of J, however. The average ga between rimes of size x is log x, and the standard deviation is also log x. We can be almost certain that x + S d+jm will contain the next rime if d + Jm 4 log x. It is worth recalling Pólya s remark [5,. 384, footnote]: When we consider a fixed number of rimes, the robabilities introduced can be regarded as indeendent, but they cannot be so regarded when the number of rimes increases in an arbitrary manner. Unfortunately, as a ractical matter, we are only able to choose J so that d+jm is smaller than 55; otherwise, the number of subsets of S d+jm is too large to be comutationally feasible. Thus when comuting our heuristic we need to limit ourselves to 4 log x 55, or x less than roughly We will discuss this further below. 4. Some Observations We see that: If S contains any odd integers, then αs) = 0, because n 2 S) = 2. If a, m) 1, then βa, m, S) = 0, because 0 S. Therefore, we will henceforth assume that a, m) = 1. If {0, d + jm} S and a + d, m) 1, then βa, m, S) = 0. If a a mod m), then βa, m, S) = βa, m, S). If βa, m, S) 0, then βa, m, S) is indeendent of a. Proosition 4.1. If m is a ower of 2, then αs)βa, m, S) is indeendent of a. Hence, if m = 2 k and a is odd, then P J a, d, m, x) is indeendent of a. Proof. Because a, m) = 1, we know that a is odd. If S contains any odd elements, then αs) = 0. If S contains only even numbers, then a + S will contain only odd numbers, and hence will be relatively rime to m, regardless of the value of a. It is temting to rearrange the terms in 2.9) in order of increasing ns) and let J. However, the sum of the terms with ns) = 2 is divergent. This is true in general, but it is simlest to see when m = 2 k. In that case, we must have a odd and d even in order for αs)βa, m, S) 0. The terms with ns) = 2 all have the form S = {0, d + jm}. In this situation, we can easily comute that ) ) 1 2 βa, m, S) = = 1 2 k 1 2. k 1

9 FREQUENCIES OF SUCCESSIVE PAIRS OF PRIME RESIDUES 9 On the other hand, αs) will always be larger than 1, so all of the infinitely many) terms 2 A with ns) = 2 will be larger than 2, and hence will diverge. 2 k log x) 2 Proosition 4.2 Antidiagonal symmetry ). P J a, d, m, x) = P J a d, d, m, x). Proof. We will show that for each set {0, d + jm} S S d+jm in the sum in 2.9) for P J a, d, m, x), there is a set {0, d + jm} S S d+jm with ns) = ns ), αs) = αs ), and βa, m, S) = β a d, m, S ). In articular, set S = d + jm) S. We automatically have {0, d + jm} S S d+jm. Obviously, ns) = ns ), and for any rime, n S) = n S ). This shows that αs) = αs ). Finally, suose that for every element s S, a + s, m) = 1, so βa, m, S) 0. Then a s, m) = 1, and so jm a s, m) = 1. Therefore, d + jm s) + a d), m) = 1, showing that β a d, m, S ) 0. We know that n S) = n S ), and therefore if βa, m, S) 0, we see that βa, m, S) = β a d, m, S ). Conversely, if βa, m, S) = 0, the same argument shows that β a d, m, S ) = 0. We may conclude that P J a, d, m, x) = P J a d, d, m, x). Proosition 4.3 Vertical comatibility ). Fix J. Suose that n is a ositive integer and m n. Then P J a, d, m, x) = P J a, d, n, x), where J + 1)m = J + 1)n. a a mod m) d d mod m) 1 a,d n Proof. We will first show that each set S d+jm used in the sum on the left-hand side of the equation aears in the sum on the right-hand side of the equation for a unique d. To see that, choose d d + jm mod n) with 1 d n. Write d + jm) d = j n, and then d + jm = d + j n. On the other hand, if we start with d d mod m) and j 0, it is clear that we can find a unique j so that d + jm = d + j n. We know that j J and d n + d m. Therefore, the uer bound for J is J +1)n m. m This leaves us needing to show that βa, m, S) = βa, n, S). a a mod m) 1 a n Notice that if βa, m, S) = 0, then automatically βa, n, S) = 0 for all ossible values of a a mod m). So we may assume that βa, m, S) 0. It suffices to consider the case where n = qm, with q rime. Notice that the set {a, a + m, a + 2m,..., a + q 1)m} contains the comlete set of integers a with a a mod m) and 1 a qm. In articular, there are q such integers a. There are two cases to finish the roof: 1. q m. If a + S contains only elements relatively rime to m, and a a mod m), then a + S contains only elements relatively rime to m. Hence, a + S contains only elements relatively rime to qm. Therefore, βa, m, S) = 0 βa, qm, S) = 0. Note also that if is rime, then qm m.

10 10 AVNER ASH, LAURA BELTIS, ROBERT GROSS, AND WARREN SINNOTT If βa, qm, S) 0, then we know that βa, qm, S) is indeendent of a. There are q ossible values of a, and we have βa, qm, S) = 1 qm n a S) qm = 1 qm n S) m = q 1 qm n S) m = 1 ) = βa, m, S). m n S) m 2. q m. In this case, the elements of the set {a, a + m,..., a + q 1)m} are distributed over all q of the residue classes modulo q. Suose that n q S) = c. Then there are q c ossible residue classes r mod q) so that r + S will be relatively rime to q, and there are c residue classes so that r + S will have a factor in common with q namely r s mod q) for some s S). Let r be one of the residue classes with r + S relatively rime to q. Recall as usual that if βa, qm, S) 0, then it is indeendent of a. We have βa, qm, S) = q n q S))βr, qm, S) a a mod m) 1 a qm = q n q S)) 1 qm = q n q S)) 1 qm = 1 m m qm n S) n S) ) q q n q S) = βa, m, S). m n S) Corollary 4.4. Choose an even integer m > 2, fix J and let J = J +1)m 2. Then 2 4.1) P J a, d, m, x) = P J 1, 2, 2, x). 1 a m 1 d m Proof. Proosition 4.3 reduces the left-hand side of 4.1) to P 1, 2, 2, x) + P 1, 1, 2, x) + P 2, 1, 2, x) + P 2, 2, 2, x). The last three terms in this sum are all 0. Note that if for some reason it is desirable to begin with odd m, we may relace m with 2m without changing the sum on the left-hand side of equation 4.1), again because of Proosition 4.3.

11 FREQUENCIES OF SUCCESSIVE PAIRS OF PRIME RESIDUES 11 Theoretically, P J 1, 2, 2, x) should give the robability that x is rime, which according to our heuristic is log x) 1. We can only sum u to J = 28 because the comutational time involved becomes inordinate, and we tabulate our results here: x P 28 1, 2, 2, x) log x) 1 Ratio We conjecture that the relatively oor agreement with rediction when x = 10 is caused by relacing a finite roduct over the range < x µ with the infinite roduct defining A n in the enultimate term of formula 3.2). The roduct defining A n converges relatively raidly, but there is still an areciable error introduced by including so many more terms when x = 10. The fact that the ratio is slowly tending away from 1 might show that for x 10 4, J should be larger than 28 for best results. More interestingly, the comutations for x = 10 2 and x = 10 3 converged relatively raidly and did not change significantly when more terms were included in the sum. For x = 10 2, the series attained the value for J = 13. Extending the sum over larger values of J did not change the first three significant digits. The sum took the value at J = 20 and that did not change u to our maximum value of J = 28. For x = 10 3, the summation takes values varying between and when J runs from 16 to 28. It is hard to understand this aarent stabilization of P J for varying J; our heuristic makes less and less sense as J increases for fixed x. 5. Numerical checks of antidiagonal symmetry Take for examle m = 5. We count residues mod 5 of consecutive rime airs < q with < is the 50, 000, 000 th rime number). We then record our results in a matrix with ϕm) rows and columns. The results for m = 5 are where a ij records the count of i mod 5), q j mod 5)). The airs 3, 5) and 5, 7) are thus omitted from the counts in this table.) To check antidiagonal symmetry, we take the ratio of each entry to its reflection across the antidiagonal. We obtain rounded to 6 laces): The entries are all within.0006 of 1, while the ratio of the largest to the smallest entry in this array is about 1.84.

12 12 AVNER ASH, LAURA BELTIS, ROBERT GROSS, AND WARREN SINNOTT If m = 12, we have the following table of counts of rime air residues mod 12, q mod 12), where < q are consecutive rimes with < we omit the airs 2, 3) and 3, 5) that are not relatively rime to For examle the entry records the number of airs of consecutive rimes congruent to 5, 7) mod 12). Again, our heuristic redicts that this matrix should be symmetric across the antidiagonal. Taking the ratio of each entry to its reflection across the antidiagonal, we obtain rounded to 6 laces): The entries are all within.002 of 1, while the ratio of the largest to the smallest entry in this array is about Numerical Checks of Power of 2 Prediction As noted above, when m = 2 k, our heuristic redicts that Na, d, 2 k, x) should be indeendent of a. Taking m = 16, we count residues mod 16 of consecutive rime airs < q with < : Our conjecture redicts that the numbers on the broken diagonals arallel to the main diagonal should be aroximately equal. For examle, the numbers in bold in the table above form a broken diagonal. We can test our conjecture by noting that the counts in the matrix range from to , with a ratio aroximately 2.66, while the counts in the broken diagonal in bold above range from to , with a ratio of The next broken diagonal beginning with ) varies from to , with a ratio of The remaining broken diagonals exhibit similar behavior. 7. Numerical checks of the heuristic In 5 and 6, we tested the heuristic indirectly, by observing symmetries in the heuristic and seeing if the same symmetries aeared in the counts of residues of rime airs. In this

13 FREQUENCIES OF SUCCESSIVE PAIRS OF PRIME RESIDUES 13 section we comare our heuristic formula directly with such counts. This requires comuting the heuristic numerically, which, as noted above, becomes raidly unwieldy as J increases, so that we must restrict J so that d + Jm < 55 in our calculations, and corresondingly restrict x to be at most We will use m = 4, 5, 11, and 12 as tyical examles. modulo m between 10 3 and 10 6 : m = We count rime airs reduced m = We comute 106 P J a, d, m, x) for all ossible values of a and d for each value of m above. x=10 3 Because of limited comuter time, we choose J maximal with d + Jm < 55. The results are: m = m = To gauge the success of the heuristic in redicting the counts, we take the ratios of the corresonding entries in these tables, i.e., we calculate, for each congruence class i, j) mod m), the ratio actual count)/redicted count), with the following results, rounded to two laces: m = m = By comarison, the ratio of the largest to the smallest count is 6934/ for m = 5), and 7034/ for m = 12). Here are the results for m = 4, again for the range 10 3 to 10 6 : 10 6 m = 4 P J a, d, 4, x) x=10 3 ratios

14 14 AVNER ASH, LAURA BELTIS, ROBERT GROSS, AND WARREN SINNOTT Finally, m = 11: m = P J a, d, 11, x) x= ratios The counts for m = 11 vary from 192 to 1826, with ratio 1826/ Our redicted values are for the most art are quite close to the observed counts, though the redictions for a few entries notably, and curiously, 3, 3), 4, 4), 5, 5), 6, 6), 7, 7)) are a bit low, only 85% 90% of their observed counts. This is artly exlained by the size of the entries themselves the main diagonal entries are the smallest, in either the table of counts or the table of redictions, so a discreancy there between rediction and observation shows u as a larger

15 FREQUENCIES OF SUCCESSIVE PAIRS OF PRIME RESIDUES 15 ercentage error than the same discreancy elsewhere. For examle, the observed count for 5, 5) is 41 above the redicted count, or 18% above; the observed count for 10, 9) is 42 above redicted, or 5% above; the observed count for 2, 4) is 39 below redicted, or 3% below. 8. Another rediction If we let x and fix J, we can view the dominant terms in P J a, d, m, x) as the ones with ns) = 2. In reality, the estimation is more comlex, because the value of J should increase with x.) If ns) = 2, then S = {0, d + jm}. As long as a, m) = a + d, m) = 1, these dominant terms in P J a, d, m, x) will be indeendent of a. To see if in fact this rediction aears to be true, we comute rime air counts modulo 5 and 12 for the first 10 7 robable) rimes larger than The results are m = m = This data is consistent with the rediction: for examle for m = d = 5 res. m = d = 12), art of the rediction is that the entries on the main diagonal should tend to equality; an ad hoc way to judge this is to take the ratio of the largest to the smallest terms on the diagonal for both our first set of residues, involving rimes less than 10 6, and for this set. For m = 5 resectively m = 12), the ratio for the small-rime data set is res ), while for the large-rime data set, the ratio is res ). 9. Further Oen Questions One obvious question is whether the different residue classes of rime airs occur asymtotically equally often: in other words, given any a, d, a, d with a, a + d, a, a + d relatively rime to m, does Na, d, m, x) Na, d, m, x) 1 as x? This corresonds to asking whether all of the entries in the matrices in the revious section are tending towards equality as x tends to infinity. One way to gauge the variation in our tables is to take the ratio of the largest to the smallest count. In 7, the ratio for m = 5 is aroximately 2.47 res for m = 12), while the large rime data set in 8 the ratio for m = 5 is aroximately 1.08 res. 1.07). Obviously, the terms aear to be getting closer together, but of course we can t tell if they are tending towards a limiting ratio of 1. If the different residue classes of rime airs do occur asymtotically equally often, we are in a rime race situation, as studied in [6] and [3]. For examle, take m = 4 and count rime airs congruent to 1, 1) mod 4) versus rime airs congruent to 1, 3) mod 4). Even if the counts are asymtotically equal, i.e. if lim x N1, 4, 4, x)/n1, 2, 4, x) = 1, we observe for finite chunks of rimes a decided tendency for the total of 1, 3) s to exceed

16 16 AVNER ASH, LAURA BELTIS, ROBERT GROSS, AND WARREN SINNOTT the total of 1, 1) s. We can kee score, as we search through the sequence of odd rimes, 3, 5, 7,..., and see when the 1, 1) s are ahead of the 1, 3) s and when the oosite holds. It seems quite ossible that N1, 4, 4, x) > N1, 2, 4, x) for almost all values of x. Another question concerns the aroriate value of J to use in our heuristic. The comutation mentioned at the end of 4 shows that in at least one case, P J a, d, m, x) is nearly indeendent of J for a range of values of J. A recise understanding of the deendence of P J a, d, m, x) on J would go a long way towards establishing the solidity of our heuristic. References [1] A. Ash, B. Bate, and R. Gross, Frequencies of consecutive tules of Frobenius classes, Exeriment. Math., to aear. [2] Paul T. Bateman and Roger A. Horn, A heuristic asymtotic formula concerning the distribution of rime numbers, Math. Com ), MR #6139) [3] Andrew Granville and Greg Martin, Prime number races, Amer. Math. Monthly ), no. 1, MR h:11112) [4] G.H. Hardy and J.E. Littlewood, Some roblems of Partitio Numerorum III: On the exression of a number as a sum of rimes, Acta Math ), [5] G. Pólya, Heuristic reasoning in the theory of numbers, Amer. Math. Monthly ), MR #3392) [6] Michael Rubinstein and Peter Sarnak, Chebyshev s bias, Exeriment. Math ), no. 3, MR d:11099) Deartment of Mathematics, Boston College, Chestnut Hill, MA address: ashav@bc.edu Deartment of Mathematics, Boston College, Chestnut Hill, MA address: lbeltis@gmail.com Deartment of Mathematics, Boston College, Chestnut Hill, MA address: gross@bc.edu Deartment of Mathematics, The Ohio State University, Columbus, OH address: sinnott@math.ohio-state.edu