On the number of co-prime-free sets. Neil J. Calkin and Andrew Granville *

Transcription

1 On the number of co-prime-free sets. by Neil J. Calkin and Andrew Granville * Abstract: For a variety of arithmetic properties P such as the one in the title) we investigate the number of subsets of the positive integers, that have that property. In so doing we answer some questions posed by Cameron and Erdös. 1. Introduction. In [CE] Cameron and Erdös investigated subsets of the positive integers with certain given properties P ; in particular, how large such sets can be, and how many there are. The properties P that they were interested in are monotone decreasing, that is, if S has property P, and T is a subset of S, then T has property P. Thus if S is a maimal set of positive integers with property P then one knows that there are 2 S such sets. In this paper we improve various estimates in [CE] for the number of sets satisfying certain properties P : In Theorem 3.5 of [CE], Cameron and Erdös showed that the number of sets of positive integers, in which any two elements have a common factor, lies between 2 [/2] and 2 [/2]. Here we improve this to Theorem 1. The number of sets of integers, with any two elements having a common factor, is 1.1) 2 [/2] + 2 [/2] N + O 2 [/2] N ep C log 2 log log )), * The second author is supported, in part, by the National Science Foundation grant number DMS )

2 for some absolute constant C > 0, where N, which will be defined in the proof, satisfies 1.2) N = e γ + o1)) log log, and γ is the Euler-Mascheroni constant. In Theorem 3.3 of [CE], Cameron and Erdös showed that the number of sets of positive integers, in which any two elements are coprime, lies between 2 π) e 1/2+o1)) and 2 π) e 2+o1)). Here we improve this to Theorem 2. The number of sets of integers, with every pair of elements coprime, is 1.3) 2 π) e {1+Olog log /log )}. In Section 4.3 of [CE], Cameron and Erdös conjectured that there are cs) +o) sets of integers, with sum of reciprocals bounded by s, for some positive constant cs). We prove a quantitatative form of this conjecture here: Theorem 3. The number, ν), of sets {a 1, a 2,... a t } of positive integers, with i 1/a i s, is 1.4) cs) e O3/4), where cs) = and fs) is defined by 1.5) s = fs) du u1 + e u ). 1 + e fs)), Cameron and Erdo s observed that cs) 2 1 e s, for all s. We can provide some rather more accurate estimates based on an analysis of 1.4) and 1.5): As s 0, we have 1.6) cs) = 1 + s log 1/s) + log log 1/s) + O1)) ; as s, we have 1.7) cs) = 2 Ce 2s + Oe 4s ),

3 for some constant C On the number of co-prime-free sets 3 2. Sets where any two elements have a common factor. Proof of Theorem 1: Clearly there are 2 [/2] such sets that contain only even numbers. We now count such sets that contain at least one odd number: Let k be the largest integer for which q = p 1 p 2... p k where p j is the jth smallest odd prime); by the Prime Number Theorem, k log /log log. Define R = {n : n is divisible by 2p j for some j k}. Clearly any set of the form S {q}, where S is a subset of R, has the property that any two elements have a common factor. The number of such subsets S is 2 R, and R = [/2] N where N is the number of integers 2m that are coprime to q. From the combinatorial sieve we obtain N φq) q 2, and then Mertens Theoremplies 1.2). Our proof of the rest of Theorem 1 is based on the ideas of Pomerance given in Cameron and Erdös [CE]: We start by ordering the odd numbers as q = m 1, m 2,... so that φm 1 ) m 1 φm 2) m 2 φm 3) m Define f i ) to be the number of sets of integers, that contain but not +1, +2,..., and for which every pair of integers in the set have a common factor. If φ) 2/3 then #{n : n, ) > 1} p p p log ) p = log p 1 ) mi log 3/2), φ ) so that f i ) 2 log 3/2), which is part of the error tern 1.1).

4 4 Neil J. Calkin and Andrew Granville Define A m ) to be the number of even integers, that are coprime to m, and c i = A mi ). Clearly f i ) 2 i 1+[/2] c i, and so, in order to complete the proof of Theorem 1, we must show that 2.1) c i ) i N log 2 log log, for all i 2 for which φ) 2/3. Suppose that m = pr m = p r are odd, squarefree numbers, with p < 1/3. Noting that A r ) = A m ) + A r /p), we see that 2.2) A m ) A m ) = A r /p) \ A r /p ) 1 p 1 ) p log log p 2 log log. Thus, given any squarefree integer r, r q, we form a sequence r = r 0, r 1,... r j = q, as follows: If r i = p 1 p 2... p h for some h < k, then then let r i+1 = r i p h+1. Otherwise we construct r i+1 by dividing r i by its largest prime factor, and multiplying it by the smallest odd prime that does not yet divide it. Now as A q ) = N, and as the prime factors of q are all log, we find that, if i 2 then c i N /log 2 log log by 2.2); which implies 2.1) for i /log 2 log log. So we are left with those i for which φ) deal with those i 100/log log : 2/3 and i /log 2 log log. We first If A is any set of < i odd integers then φ ) min n, n odd n A φn) n. So we choose A to be the set of those [/log 3 ] odd integers, with the most distinct prime factors. Hardy and Ramanujan [HR] showed that there eists an absolute constant c such that the number of integers with eactly k distinct prime factors is log log log + c) k 1 ; k 1)!

5 On the number of co-prime-free sets 5 therefore the number of integers with at least 10log log distinct prime factors is /log 10. Thus, if n A then n has 10log log distinct prime factors, and so φn) n p log log ) ) p 1 log log log by Mertens Theorem. Therefore, by the combinatorial sieve, which implies 2.1) in this range of i. c i φ) log log log, Finally we come to those i in the range 100/log log i /2, with φ) 2/3: An immediate consequence of Proposition 4 of [PS] is that i < 3 / 20 φ) 1 φm ) i) 9 20 φ). On the other hand, by the combinatorial sieve, { } 1 c i 2 + o1) φ) > { } o1) i; and thus and so 2.1) is satisfied. c i i i log log, 3. Sets where any two elements are coprime. Proof of Theorem 2: Let t = π ). For the upper bound, note that the number of composite elements of each set is t, as these elements must all have distinct prime factors. All other elements are prime, and so the number of such sets is 2 π) t ) [] i i=0 π) t 2 t! 2 π) e ) t, t

6 6 Neil J. Calkin and Andrew Granville which gives 1.3) by the Prime Number Theorem. The proof here is the same as in Theorem 3.3 of [CE], ecept that they made a computational error in the final step.) and let To obtain the lower bound, we shall construct 1.3) such sets. Let ) log log k = t 1 log < q1 < q 2 <... < q k be the k smallest primes larger than. Note that, using the Prime Number Theoren the form π) = log 1 + O1/log )) we have 3.1) q j = 1/2 1 + j/t + O1/log )) and so q k < 2. We construct our sets as follows: Each prime in the interval 2, ] is in our set or not as desired, giving 2 π) π2 ) different options. We may put any number of the form p k q k in the set, where p k is a prime less than /q k giving π/q k ) choices). Then any p k 1 q k 1 where p k 1 is a prime /q k 1 giving π/q k 1 ) 1 choices). We continue in this fashion, taking, in general, any p k j q k j, where p k j is a prime /q k j, not already used as some p k i, giving us π/q k j ) j choices), for j = 0, 1,..., k 1. Thus the number of different sets constructed is k { ) π Now, by 3.1), ) π q i 2 π)+o /log ) i=0 1/2 k i) = π 1 + i/t = t { i/t + O q i } k i). ))) O k i) log ) 1 log log 1 + log log + i t },

7 On the number of co-prime-free sets 7 using the Prime Number Theorem again, )) log log = t log + i/t) i/t + O log Therefore, the number of sets is at least t log. which gives 1.3). 2 π)+o /log ) ) k t, log Remark: With some care it is possible to replace the log log in 1.3) with log log log ) 2, but we are currently unable to do better. 4. Sets whose sum of reciprocals is bounded. Proof of Theorem 3: Let y = [ 1/4 ], z = [ 1/2 ], and j = j/z for y j z. A given set of positive integers, whose sum of reciprocals is bounded by s, has, say, b j integers in the interval j, j+1 ] for y j < z with 0 b j /z + 1), and so satisfies 4.1) z 1 j=y b j j+1 s. So, if the b j are fied with these values then the number of sets of integers, with precisely b j integers from the interval j, j+1 ], is 4.2) 2 y/z y j<z ) [/z] + 1. So define α j := b j /[/z] + 1) for each j. Clearly 0 α j 1 for each j, and, by 4.1), they must satisfy 4.3) z 1 j=y α j j s 1 + O b j )) 1. y

8 8 Neil J. Calkin and Andrew Granville Moreover, by Stirling s formula, ) [/z] + 1 b j ep Olog ) ) z α jlog α j + 1 α j )log 1 α j )). Noting that there are no more than /z + 1) z choices for the b j, we thus see that ν) ep O 3/4 ) z min 0 α j 1,y j<z 4.3) holds y j<z α j log α j + 1 α j )log 1 α j )). By the method of Lagrange multipliers we find that the minimum occurs when each α j = 1/ 1 + e A/j) for some constant A > 0. Now z 1 j=y α j j = = z y A/z dt 1 t1 + e A/t ) + O ye + 1 A/y du u1 + e u ) + O ze A/z ) 1 ye A/y + 1 A/y)e A/y + 1 ze A/z ). To obtain equality in 4.3), we need to select A = 1/2 fs) + O 1/4 ). Thus y j<z α j log α j + 1 α j )log 1 α j ) = = z y 1 + e A/u) + By the substitution t = A/u, and the fact that d dt log 1 + e t ) t y j<z log A u 1 + e A/u) 1 + e A/j) + A j ) = 1 t 2 log 1 + e t ) + du + O1). t ) 1 + e t, e A/j ) this last line is zlog 1 + e A/z ) + O1), and so 1.4) is an upper bound for ν).

9 On the number of co-prime-free sets 9 To obtain a lower bound for ν) note that if we select integers b j such that 4.4) z 1 j=y b j j s, then the sum of the reciprocals of any set consisting of b j j, j+1 ), for each y j < z, is s. Clearly the number of such sets is 4.5) y j<z ) [/z]. b j integers from the interval Now the idea is to select each b j = /z 1 + e A/j) + O1), for some constant A > 0, so that 4.4) is satisfied; this can be done by the choice A = 1/2 fs) + O 1/4 ) the proof being almost identical to that for the upper bound). Now, when we estimate 4.5) with this value for A, we proceed as in the upper bound, and show that 4.5) is at least 1.4). Remarks: Cameron and Erdo s observed that any set of integers taken from [/e s, ] has sum of reciprocals s, and thus cs) 2 1 e s. We will derive 1.6) and 1.7): If is large then du u1 + e u ) = 1 { e 1 + O )} 1, which implies 1.6). On the other hand, 1.5) gives that, for fs) < 1 that is, s large ), 4.6) s = C log 1/fs)) fs) where C 1 = 1 du u1 + e u ) e u 1 ) du 2 u, e u 1 ) du 2 u. Therefore fs) e 2s, and so the last tern 4.6) is fs) e 2s. Thus fs) = Ce 2s +Oe 4s ), where C = e 2C 1 which can be computed eplicitly), which implies 1.7).

10 10 Neil J. Calkin and Andrew Granville References [CE] CAMERON, P.J. & ERDÖS, P., On the number of sets of integers with various properties, Number Theory ed. R.A. Mollin), de Gruyter, New York, 1990), [HR] [PS] HARDY, G.H. & RAMANUJAN, S,, The normal number of prime factors of a number n, Quarterly J. Math., ), POMERANCE, C. & SELFRIDGE, J.L., Proof of D.J. Newman s coprime mapping conjecture, Mathematika, ), Neil J. Calkin, School of Mathematics, Georgia Institute of Technology, Altanta, GA 30332, USA. Andrew J. Granville, Department of Mathematics, University of Georgia, GA 30602, USA.