π(x) π( x) = x<n x gcd(n,p)=1 The sum can be extended to all n x, except that now the number 1 is included in the sum, so π(x) π( x)+1 = n x

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1 Math 05 notes, week 7 C. Pomerance Sieving An imortant tool in elementary/analytic number theory is sieving. Let s begin with something familiar: the sieve of Ertatosthenes. This is usually introduced as a mechanical rocess for finding the rimes u to some number x. One starts with a list of the integers from 2 to x. The first number in the list is 2, one circles this, and then crosses off every 2nd number thereafter, namely 4, 6, etc. The lowest remaining unmarked number is 3, so this is circled, and then every 3rd number is crossed off, namely 6, 9, etc. Some numbers are crossed off multile times. This continues until one reaches the squareroot of x, and then all remaining unmarked numbers are circled. The circled numbers are the rimes to x. This we all know. But now think of this as a counting rocedure for the number of rimes in the final circling, namely the number of rimes in the interval x,x]. These are circled because they were not reviously crossed off, so that they have no rime divisor less than or equal to their squareroots, and so they must be rime. Let P denote the roduct of the rimes in [, x], so the numbers in the final circling are those n in x,x] that are corime to P. That is, πx π x =. x< gcdn,p= The sum can be extended to all n x, excet that now the number is included in the sum, so πx π x+ = gcdn,p=. Recall the arithmetic function I which takes to and every other n to 0. An imortant roerty of our basic arithmetic functions is that I = µ u. Here µ is the Möbius function and u is the function which is identically. Thus, πx π x+ = Igcdn,P = µ ugcdn,p = µd = d n,d P d P = x µd. d d P µd d n

2 Note that this is an exact formula for πx π x+. To make use of it, a natural thing to do is to relace x/d with x/d+o. This gives us πx π x+ = x µd d +O. d P d P The sum in the main term can be written as a roduct or one can notice that it is ϕp/p, while the sum in the O-term is 2 π x, since that is exactly how many divisors P has. This gives us πx π x = x +O 2 π x. Mertens theorem says that so that We thus have that x x πx = 2 e γ x = e γ log x+o, = 2 e γ logx +O. logx 2 x x logx +O +O 2 π x. logx 2 Is this an asymtotic formula for πx? Something must be wrong since the rime number theorem is suosed to be asserting that πx is asymtotically x/logx and the constant 2/e γ is not. It is about.2. The roblem is the error term O2 π x. It is ossible to show by elementary means that πy > cy/logy for some ositive constant c and all y 2. Thus, the O-term is huge! It exceeds every fixed ower of x. There are two ways to take this bad news. You can be discouraged that in the end we have a worthless formula. Or, you can be uzzled and heartened that in fact.2 is not so far away from, and erhas this aroach has some ossibilities. In the early art of the twentieth century, the Norwegian mathematician Viggo Brun found an elementary way to turn this elementary sieve into a machine to give rigorous and nontrivial inequalities, not only for the distribution of rime numbers, but also for other sievable sets. There is another thought here. If y is some fixed number at least 2 and Py is the roduct of the rimes to y, it is easy to see that +O. gcdn,py= = x y 2

3 We have just seen that this formula is not true when y = x /2, but somehow it is not far off from the truth. Somewhere between y fixed and y as big as x /2 there must be a transition oint, where the naive estimate is asymtotically true below that oint. We shall see that Brun s method can say something interesting about this roblem as well. 2 Twin rimes and heuristics In the mid nineteenth century, de Polignac conjectured that for each even number k > 0 there are infinitely many rimes with +k rime. The case of k = 2 is known as the twin rime conjecture. There is no value of k where de Polignac s conjecture is roved, and until recently we did not even know that there are infinitely many rimes with the next rime after having bounded. This last assertion was roved earlier this year by Yitang Zhang at the University of New Hamshire. So now we know that there is some k with and + k both rime infinitely often, and we even have a bound on such a number k Zhang showed that k < 70,000,000 and this has been imroved somewhat, but we still do not know a secific number k where de Polignac s conjecture holds. Why do we believe that de Polignac s conjecture should hold? The theory of rime numbers resembles in many ways an exerimental science, where we think we know what is going on and we can test this with numerical exeriments. A basic heuristic rincile is that a random number near x should be rime with robability /logx. In fact, this is a way of describing the rime number theorem. And it is interesting that this rincile was discovered exerimentally bygaussovertwocenturiesagojustbyoringoveratableofrimes. Hefoundthatindividually theyaearerraticallyamongthenaturalnumbers, butifonedrawsback abit andcountsthem in fairly long intervals, a natural law aears to emerge. Gauss counted them in intervals of 000 and he observed the /logx law. It is for this reason that he conjectured that a reasonable formula for πx is the logarithmic integral lix = x 2 dt logt. How good is it? Here s a numerical exeriment far beyond what Gauss could have done: π 0 22 = 20,467,286,689,35,906,290, li 0 22 = 20,467,286,69,248,26, We should note that it is still not roved that lix is such a fantastic aroximation to πx. The Riemann hyothesis, which is still unroved, is equivalent to the assertion that πx lix < xlogx, x 3. However, we do know that lix is a retty good aroximation to πx, the best we currently have is that x πx lix = O exlogx 3/5 ǫ 3

4 for each fixed ǫ > 0, a result of Vinogradov and Korobov from around 60 years ago. Let s try out the heuristic on twin rimes. If we think of and + 2 as indeendent events, then the likelihood that a number near x should have both and + 2 be rime should be about /logx 2. The trouble with this is that they are not indeendent events! For examle, if is rime and larger than 2, then it is odd, so that +2 is also odd, and so it has a leg u on being rime. Also, for > 3, we have that if it is rime then it is or 2 mod 3, with roughly equal chance, so that +2 is 0 or mod 3 with roughly equal chance. So, if is rime, the chance that 3 +2 is, which is not the chance that one would exect for 2 3 a random integer. Thus, the twin rime distribution should be corrected for these effects of the small rimes. The correction factor is 2C 2 := 2 >2 = 2 >2 = Thus, one might conjecture that the number π 2 x of rimes x with + 2 also rime is about x dt 2C 2 logt 2. Testing this guess: 2C π = 0,304,95,697,298, dt logt 2 = 0,304,92,554, So, as exerimental scientists we should be feeling retty smug, we ve nailed it. But as mathematicians, we still do not have a roof that there are infinitely many twin rimes! This tye of heuristic reasoning leads to many conjectures of this tye. For examle, we conjecture that there are infinitely many rimes of the form n 2 +, and we have a reasonable guess for what the distribution of such numbers n should be, but we don t have a roof of their infinitude. Part of the landscae is the rime k-tules conjecture. This is a grand generalization of the conjecture of de Polignac to k linear exressions instead of just 2. Basically what the conjecture asserts is that if k linear olynomials a i x + b i for i =,2,...,k have a chance to be simultaneously rime infinitely often, then they will be. Let us assume that the coefficients a i,b i are all integers and each a i > 0. To have a chance of a i n+b i being rime for an integer n, we would need that gcda i,b i =, so let us assume that this holds for i =,2,...,k. There is another condition that needs to hold. Consider the case of the two olynomials x and x+. They both have ositive leading coefficients and the constant terms are corime to the leading coefficients. But the only value of n where n and n+ are both rime is n = 2. What s haening here is that the roduct nn+ is always even, so one of the two factors must be divisible by 2. We thus ut on another necessary condition: there is no rime that divides a n+b a 2 n+b 2...a k n+b k 4

5 for all n. This criterion is easier to check than it might look at first. If the rime exceeds k then it automatically asses: the number of solutions to a n+b a 2 n+b 2...a k n+b k 0 mod is at most k a nonzero olynomial of degree at most k has at most k roots in a field, so if > k, there is some residue n mod where the above congruence is not satisfied. We say the k distinct linear olynomials a i x + b i for i =,2,...,k are admissible if each a i > 0, if each gcda i,b i =, and if for each rime k there is some integer n such that a i n+b i for i =,2,...,k. For examle, the olynomials x and x+2 are admissible, as are the olynomials x,x+2,x+6, or the olynials, x,2x+. The rime k-tules conjecture: If the olynomials a i x+b i for i =,2,...,k are admissible, then there are infinitely many integers n such that a i n+b i are simultaneously rime. This conjecture is originally due to L. E. Dickson. The only case where the rime k-tules conjecture has been roved is the case k =, this is Dirichlet s theorem. The heuristics resented above for twin rimes can be coied over to any admissible collection of linear olynomials; this was worked out by Hardy and Littlewood in the early twentieth century. So, we even have the ability with modern comuters to test the conjecture numerically, and it never has disaointed us. Surely it is true, but we are still awaiting a roof. 3 The Bonferroni inequalities Recall the exact formula πx π x+ = d P x µd, d that we obtained from viewing the sieve of Eratosthenes as a counting roblem. Here P is the roduct of the rimes in [, x]. This can be viewed as the inclusion exclusion formula from combinatorics. That is, first, with d =, we include all of the numbers to x. Then with d running over rime divisors of P, we exclude the x/ multiles of. Then for d = q, we include the x/q multiles of q, where < q x are rimes. And so on. Let ωn denote the number of distinct rime divisors of n. The error caused by removing the floor symbols in is O2 ωp = O2 π x, which we have seen to be way too large. Somehow, we need to reduce the number of terms in. One way to do this is not sieve with all of the rimes u to x. Say, we take y with 2 y x and we let Py = y. The integers in [,x] which are corime to Py include and all of the rimes in x,x] as before, but now also other numbers may be included. The exact same argument that gave us 5

6 also gives us that S := gcdn,py= = x µd. 2 d Removing the floor symbols now creates an error of O2 πy. For this to be suitably small, we would need to choose y small. For examle, y = logx suffices. Then 2 πy 2 y < x 0.7, since log2 < 0.7. Thus, the number of integers n x with least rime factor exceeding logx is x/e γ loglogx. In articular πx = Ox/loglogx. The latter result while true is very weak. So we need another idea. What if we sto early and do not comlete the inclusion exclusion argument? That is, say we only consider divisors d of Py with at most h rime factors. Let S h = x µd. 3 d ωd h ωd h What is the relationshi of S h to the full inclusion exclusion S that aears in 2? Intuitively, if h is even, we are stoing with an inclusion, so maybe S h should be larger than S, and if we take h odd, we are stoing at an exclusion, and S h should be smaller. We can actually rove this, it is not hard. Note that unwinding our revious calculation that gave us, we have S h = x µd = d µd. d gcdn,py ωd h The following lemma will hel us with the inner sum above. Lemma. For an integer m > with ωm = k and for a nonnegative integer h, we have that k µd = h. h d m ωd h+ d m ωd h Proof. This is easily done by induction on h. If h = 0, then both sides of the equation are. Assume it is true for h 0. Then by the induction hyothesis, k k k µd = h µd = h + h+, h h h+ + d m ωd=h+ since m has exactly k h+ squarefree divisors d with ωd = h +, and for each such d, we have µd = h+. The non-squarefree divisors d have µd = 0, so they do not contribute. Note too that if h + > k, then there are no d to consider, but the binomial coefficient is 6

7 by definition 0 in this case. We now use Pascal s identity the rule of formation of Pascal s triangle, so that k k k k k h + h+ = h + h+ + h+ h h+ h h h+ k = h+. h+ This is the h+ case of the lemma, so the roof is comlete. S h = Using Lemma in the revious calculation, we have that gcdn,py= + In articular, 4 shows that d gcdn,py gcdn,py> µd = S+ h gcdn,py> ωgcdn,py. h S S h if h is even, S S h if h is odd. 5 These assertions are sometimes referred to in combinatorics and robability theory as the Bonferroni inequalities. 4 Return to the roblem Wehave defined S h asthetruncatedinclusion exclusion givenby 3, where Pyis theroduct of the rimes in [,y]. By the Bonferroni inequalities 5, we have S sandwiched between two consecutive values of S h, with an odd subscrit S h as the lower bound and an even subscrit S h as the uer bound. This will be good if we can come u with a good estimate for S h. We have S h = ωd h = x Py x µd = x d +O ωd h x µd d +O ωd>h +O d ωd h ωd h 4. 6 So we have two O-terms to consider. The sum in the second O-term is h ωpy h h ωpy j y j y h. 7 j j=0 j=0 7 j=0

8 Note that here we have used the notation. In analytic number theory and some other fields, this is used as a synonym for O; it is useful in chains of inequalities. For those who like to use latex, it is tyed as \ll. Basically, the hilosohy is that the larger we choose h, then the smaller the first O-term will be in 6, but the larger the second O-term will be, so somewhere in between is a hay medium which gives both O-terms smaller than the main term. The first O-term in 6 can be handled with a neat trick. Let u > be some number that we shall choose shortly. Since u > we have that ωd>h d u ωd h d = u h u ωd d. This works because if ωd > h, then d gets counted with weight at least u/d >, and if ωd h, then d gets counted with some sort of ositive weight, so the inequality is correct. This idea is known as Rankin s trick, and it is entirely analogous to Chebyshev s inequality or is it Markoff s inequality? in elementary robability theory. The effect of the trick is to remove the condition on ωd under the summation, so the exression can be more easily handled, and we have a new arameter u at our disosal for otimization. The final sum above has an Euler roduct, and so ωd>h d u h y + u u h ex y using that +θ < e θ for all θ > 0. We conclude that d ex hlogu+u y d P ωd>h. u, We are now ready to choose the arameter u. We do this in such a way as to minimize this last exression, and by calculus, this minimum will occur when we choose / u = h. But we also wish to have u >, so this will influence our choice of h. Let h be an integer at least as big as 5 y. Then u 5, which is a legal choice, and utting in the above choice for u, our inequality reads d ex hlogh+hlog + ex hlog5+, y y d P ωd>h y 8 y

9 using h 5 y. And using this one more time, we get that d ex 5log5 y d P ωd>h ωd h logy 5log5 < logy 3, where we have used Mertens in the weak form that y = loglogy +O. Let us see now what we have. If h is an integer that is at least 5 y, then S h = x µd = x d x +O +O y h. logy 3 y Further, if then S = gcdn,py= S S h if h is odd, S S h if h is even. We now settle on choices for the arameters y,h. We would like to be able to choose y as large as ossible so that the error terms are under control. Say y x /0loglogx, h [5loglogx 2, 5loglogx] Z. Then y h x /2 andfor largeenoughvalues ofxwe have h 5 y, since h 5loglogy as x. We have roved the following result. Theorem. For 2 y x /0loglogx, the number of integers n [,x] not divisible by any rime in [,y] is x x +O. logy 3 y That is, if y is not too large, the naive estimation of the number of integers not divisible by any rime to y is aroximately correct. As a corollary, we have with y at the to end of the allowed range that, πx xloglogx. logx Though this estimate is inferior to the Chebyshev Erdős estimate of πx x/ log x, the method here can be easily generalized, which we shall now do. 9