COMPUTING π(x): THE MEISSEL, LEHMER, LAGARIAS, MILLER, ODLYZKO METHOD

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1 MATHEMATICS OF COMPUTATION Volume 65, Number 213 Januar 1996, Pages COMPUTING π(): THE MEISSEL, LEHMER, LAGARIAS, MILLER, ODLYZKO METHOD M. DELEGLISE AND J. RIVAT Abstract. Let π() denote the number of primes. Our aim in this paper is to present some refinements of a combinatorial method for computing single values of π(), initiated b the German astronomer Meissel in 1870, etended and simplified b Lehmer in 1959, and improved in 1985 b Lagarias, Miller and Odlzko. We show that it is possible to compute π() ino( 2/3 log 2 )time and O( 1/3 log 3 log log ) space. The algorithm has been implemented and used to compute π(10 18 ). 1. Introduction One of the oldest problems in mathematics is to compute π(), the eact number of primes. The most obvious method for computing π() is to find and count all primes p, for instance b the sieve of Eratosthenes. According to the Prime Number Theorem (1) π() log,. Therefore, such a method cannot compute π() with less than about log operations. Despite its time compleit, the sieve of Eratosthenes has been for a ver long time the practical wa to compute π(). In the second half of the 19th centur, the astronomer Meissel discovered a practicable combinatorial method that is faster than finding all primes. He used his algorithm to compute b hand π(10 8 )and π(10 9 ) (which turned out to be too small b 56) [4, 5, 6, 7]. In 1959, Lehmer etended and simplified Meissel s method. He used an IBM 701 computer to obtain the value of π(10 10 ) (his value was shown [1] to be too large b 1). In 1985, Lagarias, Miller and Odlzko [2] adapted the Meissel-Lehmer method and proved that it is possible to compute π() with O( 2/3 log ) operations using O( 1/3 log 2 log log ) space. The used their algorithm to compute several values of π() upto= The also corrected the value of π(10 13 ) given in [1], which was too small b 941. In 1987, Lagarias and Odlzko [3] described a completel different method, based on numerical integration of certain integral transforms of the Riemann ζ-function, for computing π(), using O( 1/2+ε )timeando( 1/4+ε ) space for each ε>0. Received b the editor Januar 12, 1994 and, in revised form, December 1, Mathematics Subject Classification. Primar 11N05, 11Y c 1996 American Mathematical Societ

2 236 M. DELEGLISE AND J. RIVAT Despite its asmptotic superiorit, this algorithm has never been implemented. Its authors noted [2] that the implied constants are probabl large, and therefore that it would not be competitive with their version of the Meissel-Lehmer method for In this paper we describe a modified form of the algorithm presented in [2] which computes π() usingo( 2/3 log 2 )timeando(1/3 log 3 log log ) space. 2. Outline of the method For clarit we will describe the whole method given in [2], in order to introduce the quantities needed for the analsis. For the convenience of the reader we adopt the notations used in [2], and follow as long as possible the same approach. In particular, 3, 4, 5 are close to [2]. The idea that man special leaves (see below, 6) could be computed at the same time, saving much computation (and a log factor in the compleit) was also present in [2]. We develop this idea further, and show that it is possible to compute more special leaves at the same time, saving another log factor in the compleit (see 6.2 and below). 3. The Meissel-Lehmer method Let p 1,p 2,p 3,... denote the primes 2, 3, 5,... numbered in increasing order. Let (, a) denote the partial sieve function, which counts numbers with all prime factors greater than p a : (2) (, a) =#{n ; p n p>p a } and let (3) P k (, a) =#{n ; n=q 1 q 2 q k, q 1,...,q k >p a }, which counts numbers with eactl k prime factors, all larger than p a.weset P 0 (, a) =1. If we sort the numbers b the number of their prime factors greater than p a, we obtain the following identit: (, a) =P 0 (, a)+p 1 (, a)+ +P k (, a)+, where the sum on the right has onl finitel man nonzero terms, because P k (, a) = 0forp k a >. Let denote an integer such that 1/3 1/2,andleta=π(). From P 1 (, a) =π() aand P k (, a) =0fork 3 we deduce (4) π() =(, a)+a 1 P 2 (, a). Hence, for computing π() it remains to compute (, a) andp 2 (, a). 4. Computing P 2 (, a) B (3) we have to count all pairs (p, q) of prime numbers such that <p q and. We first remark that p [ +1, ]. Furthermore, for each p, wehaveq [p, /p]. Thus, P 2 (, a) = ( ) (5) π π(p)+1. p <p

3 COMPUTING π() 237 When p [ +1, ], we have p [1, ]. Hence, P 2(, a) can be computed b completel sieving the interval [1, ] and then adding up π( p ) π(p) + 1 for all primes p [ +1, ]. In order to reduce the space compleit of the above method, we can work with blocks of length L. ForL=, we can compute P 2 (, a) in O( log log ) timeando() space. 5. The sieve machiner for computing (, a) For b a the set of all integers whose prime factors are >p b 1 is composed of two classes: 1. those that are multiples of p b, 2. those not divisible b p b. The first class has ( p b,b 1) elements, while the second has (, b) elements. Hence we conclude: Lemma 5.1. The function satisfies the following identities: (6) (7) (u, 0) = [u], (, b) =(, b 1),b 1. p b A straightforward method for computing (, a) can be deduced from this lemma: it suffices to appl repeatedl the recurrence (7) until we get terms of the form (u, 0), which are eas to compute using (6). One ma think of this process as creating a rooted binar tree starting with the root node (, a);seefig.1.using this method, we obtain the following formula: (, a) = 1 n P + (n) [ µ(n), n] where µ(n) denotes the Möbius function and P + (n) denotes the greatest prime factor of n. (, a) (, a 1) ( p, a 1) a (, a 2) ( p, a 2) a 1 ( p, a 2) (, a 2) a p a p a 1 Figure 1. A binar tree for computing (, a): the sum of the terminal nodes is (, a)

4 238 M. DELEGLISE AND J. RIVAT Unfortunatel, this sum has too man terms for our purpose: as 1/3,ifwe onl count the n s which are the product of three primes, we get at least about such terms. log 3 In order to limit the growth of the tree, we must replace the trivial truncation rule, Truncation Rule 1. Do not split a node µ(n)( n,b) if b =0 with the more powerful: Truncation Rule 2. Do not split a node µ(n)( n,b) if either of the following holds: 1. b =0and n, 2. n>. We are now able to define two clases of leaves: 1. ordinar leaves are those of the form µ(n)( n, 0) satisfing n, 2. special leaves are those of the form µ(n)( n,b 1) satisfing n>with n = mp b and m. We conclude that: Lemma 5.2. We have (8) (, a) =S 0 +S, where S 0 is the contribution of the ordinar leaves, S 0 = [ (9) µ(n), n] n and S is the contribution of the special leaves, S = ( ) (10) µ(n) n n,π(δ(n)) 1, δ(n) <n where δ(n) denotes the smallest prime factor of n. The computation of S 0 can be achieved in O( log log ) time, which is negligible. It remains to compute S. (11) We have S = p 6. Computing S δ(m)>p m <mp µ(m) mp,π(p) 1.

5 COMPUTING π() 239 We now write with S 1 = S 2 = S 3 = 1 3 <p δ(m)>p m <mp 1 4 <p 1 3 p 1 4 S = S 1 + S 2 + S 3 δ(m)>p m <mp δ(m)>p m <mp µ(m) mp,π(p) 1, µ(m) µ(m) mp,π(p) 1, mp,π(p) 1. First observe that the m s involved in S 1 and S 2 are all prime: otherwise, since δ(m) >p> 1/4,wewouldhavem>p 2 >, a contradiction with m. Moreover, the condition pm is true when mp > 1/2. Hence we have S 1 =,π(p) 1, S 2 = 1 3 <p 1 4<p 1 3,π(p) Computing S 1. Since <1/3 <p, we have,π(p) 1 =1. Hence all terms involved in S 1 are equal to 1. So we have to count all pairs (p, q) such that Thus, 1/3 <. S 1 = (π() π(1/3 ))(π() π( 1/3 ) 1). 2 This takes constant time to compute S Computing S 2. We have S 2 = 1/4 <p 1/3,π(p) 1.

6 240 M. DELEGLISE AND J. RIVAT We split S 2 in two parts, depending on q>/p 2 or q /p 2 : S 2 = U + V with and U = V = 1/4 <p 1/3 1/4 <p 1/3 q> p 2 q p 2,π(p) 1,π(p) Computing U. The condition q>/p 2 implies p 2 >/q / and p> /. Thus, U= <p 1/3 q> p 2,π(p) 1. From /p 2 <qwe deduce / < p and (/, π(p) 1) = 1. Each term in the sum U equals 1. Hence, U = # {q; p } 2 <q. <p 1/3 Thus, U = <p 1/3 ( ) π() π p 2. Since /p 2 <,thesumucan be calculated in O() operationsoncewehave tabulated π(t) for t Computing V. For each term involved in V we have p <1/2 <p 2. Hence,,π(p) 1 =1+π (π(p) 1) Thus, with =2 π(p)+π V = V 1 + V 2 ( ).

7 V 1 = V 2 = 1/4 <p 1/3 COMPUTING π() 241 p<q min( p 2,) (2 π(p)), π 1/4 <p 1/3 p<q min( p 2,). Computing V 1 can be achieved in O( 1/3 ) time once we have tabulated π(t) for t. In order to speed up the computation of V 2, we observe that for each p we can split the summation over q into sums over q on intervals where the function q π( ) is constant. Thus, we onl need the length of these intervals, and the set of values of q where q π( ) is changing. More precisel, we first split V 2 in two parts in order to simplif the condition q min(/p 2,): V 2 = We now write 1/4 <p π + <p 1/3 V 2 = W 1 + W 2 + W 3 + W 4 + W 5 p<q p 2 π. with W 1 = W 2 = W 3 = W 4 = W 5 = 1/4 <p 2 <p 2 <p 2 <p 1/3 <p 1/3 p<q p p <q π p<q p π, p <q p 2 π π,, π,. Computing W 1 and W 2. These two quantities need values of π(/) with< / < 1/2. The are computed simultaneousl with a sieve of the interval [1, ]. The sieving is done b blocks, and for each block we sum π(/) for the pairs (p, q) subjected to the conditions of the sums W 1 or W 2 andsuchthat/ lies in the block. Computing W 3. For each p we speed up the computation of the sum over q b computing in O(1) operations the sums of the π(/) for the values of q for which π(/) is constant. When we obtain a new value of q, we compute π(/) with

8 242 M. DELEGLISE AND J. RIVAT the table of values of π(t) fort. Then a table of all primes gives t such that π(t) <π(t+1)=π( ). We then deduce the net value of q for which π(/) is changed. Computing W 4. We simpl sum over (p, q). There would be no advantage to proceed as for W 3 since most of the values π(/) are distinct. Computing W 5. We proceed as for W Computing S 3 We sieve the interval [1, ] successivel b all primes less than 1/4. Assoonaswe have sieved b p k 1,wesumall µ(m)( mp k,k 1) for all squarefree m [/p, ] such that δ(m) > p k. This computation can be done b blocks, see [2]. The main idea is that we maintain a binar tree (as eplained in [2, pp. 545, 546]) in connection with the interval we are sieving, to keep track of the intermediate results after sieving b all primes up to a given prime. It is then possible to know the number of unsieved elements in the interval less than a given value, using onl O(log ) operations. 8. Time and space compleit The time and space significant computations are: 1. The computation of P 2 (, a), 2. The computation of W 1,W 2,W 3,W 4,W 5, 3. The computation of S Cost of computing P 2 (, ). We have alread seen that it costs O( log log ) time and O() space Cost of computing W 1,W 2,W 3,W 4,W 5. For W 1 and W 2 the sieve costs O( log log ) timeando() space, working b blocks of length. ThetimenecessartocomputethesumW 1 is about π 2 π() =O log 2. ThetimenecessartocomputethesumW 2 is about O <p 2 π ( ) 3/4 = O p 1/4 log 2. In W 3,foreachpwe have /p. Hence, π(/) takes at most π( /p) values. For each such value it costs constant time to determine the number of q s concerned. Hence, the time necessar to compute the sum W 3 is about O <p 2 π ( ) 3/4 = O p 1/4 log 2.

9 COMPUTING π() 243 ThetimenecessartocomputethesumW 4 is about ( ) 2/3 O π = O p log 2. <p 1/3 We proceed for W 5 as for W 3, and the time necessar to compute the sum W 5 is about ( ) 2/3 O π = O log 2. <p 1/ Cost of computing S 3. The sieve. Owing to the necessit of quickl retrieving the values (u, b), we have to maintain a data structure such that each access costs O(log ) instead of O(1) in a normal sieve. Hence the cost is O( log log log ). The sum. For each term in the sum we have to access the data structure mentioned above, doing O(log ) operations. It remains to count the terms in the sum. All leaves are of the form ±(/mp b,b 1) with m and b<π( 1/4 ). Hence, the number of these leaves is O(π( 1/4 )). The total cost of S 3 is O log log log + 1/ Total cost. We have described an algorithm taking O() space and ( O log log + ) log log log + 1/4 + 2/3 log 2 time. If we choose = 1/3 log 3 log log, the time compleit is O( 2/3 log 2 ). 9. Practical considerations We describe here some modifications which improve the time of computation without changing the asmptotic compleit. In the truncation rule 2, we ma replace b some z>. It is possible to prove that the time compleit for computing S 3 then becomes 1/4 O log log log + z log + z3/2. This also gives a good wa for checking the computations b changing the value of z. For clarit we chose to split the sum S at 1/4, but in fact we onl need to have p <p2. One can take advantage of this, but the asmptotic compleit remains the same. Precomputing the sieving b the first primes 2, 3, 5 saves some more time.

10 244 M. DELEGLISE AND J. RIVAT 10. Results The algorithm has been implemented in C + +. All the computations were done using a HP 730 workstation (SPEC92INT=47.8). The 64-bit integers were emulated b the long long tpe of GNU C/C++ Compiler. For comparison we tried our program for some specific values on a DEC Alpha 3000 Model 600 at 175 Mhz (which has 64-bit integers, SPEC92INT=114). The latter turned out to be more than three times faster. The difference could be greater because our program was optimized for a 32-bit computer, which is a drawback on a 64-bit computer. We confirmed all the values alread computed in [2]. Table 1 gives the new values compared with the corresponding values of Li() = 0 dt log t Table 1. Results and times of computation on HP-730 π() Li() π() R() π() Time (s)

11 COMPUTING π() 245 and R() = n=1 µ(n) n Li(1/n ). References 1. J. Bohman, On the number of primes less than a given limit, BIT12 (1972), MR 48 # J. C. Lagarias, V. S. Miller, and A. M. Odlzko, Computing π(): The Meissel-Lehmer method, Math. Comp. 44 (1985), MR 86h: J. C. Lagarias and A. M. Odlzko, Computing π(): An analtic method, J. Algorithms 8 (1987), MR 88k: E. D. F. Meissel, Über die Bestimmung der Primzahlenmenge innerhalb gegebener Grenzen, Math. Ann. 2 (1870), , Berechnung der Menge von Primzahlen, welche innerhalb der ersten hundert Millionen natürlicher Zahlen vorkommen, Math. Ann. 3 (1871), , Über Primzahlenmengen, Math. Ann. 21 (1883), , Berechnung der Menge von Primzahlen, welche innerhalb der ersten Milliarde natürlicher Zahlen vorkommen, Math. Ann. 25 (1885), Département de Mathématiques, Université Lon 1, 43 Blvd. du 11 Novembre 1918, Villeurbanne Cede, France address: deleglis@lmdi.univ-lon1.fr address: rivat@caissa.univ-lon1.fr