aa r X i v : . [ m a t h . N T ] F e b THE NATURAL DENSITY OF SOME SETS OFSQUARE-FREE NUMBERS
RON BROWN
Abstract.
Let P and T be disjoint sets of prime numbers with T finite. A simple formula is given for the natural density of the setof square-free numbers which are divisible by all of the primes in T and by none of the primes in P . If P is the set of primes congruentto r modulo m (where m and r are relatively prime numbers), thenthis natural density is shown to be 0. Main results
In 1885 Gegenbauer proved that the natural density of the set ofsquare-free integers, i.e., the proportion of natural numbers which aresquare-free, is 6 /π [1, Theorem 333; reference on page 272]. In 2008J. A. Scott conjectured that the proportion of natural numbers whichare odd square-free numbers is 4 /π or, equivalently, the proportionof natural numbers which are square-free and divisible by 2 is 2 /π [3]. The conjecture was proven in 2010 by G. J. O. Jameson, in anargument adapted from one computing the natural density of the setof all square-free numbers [4]. In this note we use the classical resultfor all square-free numbers to reprove Jameson’s result and indeed togeneralize it:1.1. Theorem.
Let P and T be disjoint sets of prime numbers with T finite. Then the proportion of all numbers which are square-free anddivisible by all the primes in T and by none of the primes in P is π Y p ∈ T
11 + p Y p ∈ P p p . As in the above theorem, throughout this paper P and T will bedisjoint sets of prime numbers with T finite. The letter p will alwaysdenote a prime number. The term numbers will always refer to positiveintegers. Empty products, such as occurs in the first product abovewhen T is empty, are understood to equal 1. If P is infinite, we willargue below that the second product above is well-defined. Date : February 12, 2021.
Examples.
1. Setting P = { } and T equal to the empty set in thetheorem we see that the natural density of the set of odd square-freenumbers is π = π ; taking T = { } and P equal to the empty setwe see that the natural density of the set of even square-free numbersis π = π . Thus one third of the square-free numbers are even andtwo thirds are odd. (These are Jameson’s results of course.)2. Set T = { , , } and P = { } in the theorem. Then the theoremsays that the natural density of the set of square-free numbers divisibleby 30 but not by 7 is π , so the proportion of square-freenumbers which are divisible by 30 but not by 7 is = .Our interest in the case that P is infinite arose in part from a questionposed by Ed Bertram: what is the natural density of the set of square-free numbers none of which is divisible by a prime congruent to 1modulo 4? The answer is zero; more generally we have the:1.2. Theorem.
Let r and m be relatively prime numbers. Then thenatural density of the set of square-free numbers divisible by no primecongruent to r modulo m is zero. This theorem is a corollary of the previous theorem since, as we shallsee in Section 5, for any r and m as above, Y p ≡ r mod m p p = 0 . A basic lemma
For any real number x and set B of numbers, we let B [ x ] denote thenumber of elements t of B with t ≤ x . Recall that if lim x →∞ B [ x ] /x exits, then it is by definition the natural density of B [2, Definition11.1].Let A denote the set of square-free numbers. Then we let A ( T, P )denote the set of elements of A which are divisble by all elements of T and by no element of P (so, for example, A = A ( ∅ , ∅ )). The set ofsquare-free numbers analyzed in Theorem 1.1 is A ( T, P ).The next lemma shows how the calculation of the natural densityof the sets A ( T, P ) reduces to the calculation of the natural density ofsets of the form A ( ∅ , S ) and, when P is finite, also reduces to the thecalculation of the natural density of sets of the form A ( S, ∅ ).2.1. Lemma.
For any finite set of primes S disjoint from T and from P and for any real number x , we have A ( T, S ∪ P )[ x ] = A ( T ∪ S, P )[ xs ] where s = Q p ∈ S p . Moreover, the set A ( T ∪ S, P ) has a natural density HE NATURAL DENSITY OF SOME SETS OF SQUARE-FREE NUMBERS 3 if and only if A ( T, P ∪ S ) has a natural density, and if D is the naturaldensity of A ( T ∪ S, P ) , then the natural density of A ( T, P ∪ S ) is sD .Proof. The first assertion is immediate from the fact that multiplicationby s gives a bijection from the set of elements of A ( T, S ∪ P ) less thanor equal to x to the set of elements of A ( T ∪ S, P ) less than or equalto xs . This implies that A ( T, P ∪ S )[ x ] x = s A ( T ∪ S, P )[ xs ] xs . The lemma follows by taking the limit as x (and hence xs ) goes toinfinity. (cid:3) Remarks.
1. We might note that if we assume that for all T the sets A ( T, ∅ ) have natural densities, then it is easy to compute these naturaldensities. After all, for any T the set A is the disjoint union over allsubsets S of T of the sets A ( T \ S, S ), so by Lemma 2.1 for any realnumber x A [ x ] = X S ⊆ T A ( T \ S, S )[ x ] = X S ⊆ T A ( T, ∅ )[ d S x ]where for any S ⊆ T we set d S = Q p ∈ S p . Hence A [ x ] x = X S ⊆ T d S A ( T, ∅ )[ d S x ] d S x . Taking the limit as x goes to ∞ we have 6 /π = ( P S ⊆ T d S ) G where G denotes the natural density of A ( T, ∅ ). But X S ⊆ T d S = X d | d T d = Y p ∈ T (1 + p )[2, Theorem 4.5], so indeed G = 6 π Y p ∈ P
11 + p .
2. In the language of probability theory, Theorem 1.1 says that theprobability that a number in A is divisible by a prime p is 1 / ( p + 1) (sothe probability that it is not is p/ ( p + 1)) and, moreover, for any finiteset S of primes not equal to p , being divisible by p is independent ofbeing divisible by all of the elements of S . RON BROWN Proof of Theorem 1.1 when P is finite First suppose that P = ∅ . We will prove Theorem 1.1 in this caseby induction on the number of elements of T . The next lemma givesthe induction step.3.1. Lemma.
Let p be a prime number not in T . If the set A ( T, ∅ ) hasnatural density D , then the set A ( { p } ∪ T, ∅ ) has natural density p +1 D .Proof. For any real number x we set E ( x ) = A ( { p } ∪ T, ∅ )[ x ]. Let ǫ >
0. The theorem says thatlim x →∞ E ( x ) x = 1 p + 1 D. Therefore it suffices to show for all choices of ǫ above that, for allsufficiently large x (depending on ǫ ), (cid:12)(cid:12)(cid:12)(cid:12) E ( x ) x − p + 1 D (cid:12)(cid:12)(cid:12)(cid:12) < ǫ. Note that A ( T, ∅ ) is the disjoint union of A ( { p }∪ T, ∅ ) and A ( T, { p } ).Hence by Lemma 2.1 (applied to A ( T, { p } )) for any real number x , A ( T, ∅ )[ x/p ] = E ( x/p ) + E ( x )and so by the choice of D there exists a number M such that if x > M then (cid:12)(cid:12)(cid:12)(cid:12) E ( x ) x/p + E ( x/p ) x/p − D (cid:12)(cid:12)(cid:12)(cid:12) < ǫ/ . We next pick an even integer k such that p k < ǫ . Then(1) (cid:12)(cid:12) E ( x/p k ) (cid:12)(cid:12) ≤ x/p k < ǫ x and also (using the usual formula for summing a geometric series)(2) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − Dx k X i =1 ( − p ) i − Dx p + 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = Dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( − p ) − ( − p ) k +1 − ( − p ) + 1 p + 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = Dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − p k p + 1 + 1 p + 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < p k Dx < ǫ π x < ǫ x. Now suppose that x > p k M . Then for all i ≤ k we have x/p i > M and hence (applying the choice of M above),(3) (cid:12)(cid:12)(cid:12)(cid:12) E ( x ) + E ( x/p ) − D xp (cid:12)(cid:12)(cid:12)(cid:12) < ǫ xp HE NATURAL DENSITY OF SOME SETS OF SQUARE-FREE NUMBERS 5 and similarly (cid:12)(cid:12)(cid:12)(cid:12) − E ( x/p ) − E ( x/p ) + D xp (cid:12)(cid:12)(cid:12)(cid:12) < ǫ xp and (cid:12)(cid:12)(cid:12)(cid:12) E ( x/p ) + E ( x/p ) − D xp (cid:12)(cid:12)(cid:12)(cid:12) < ǫ xp and (cid:12)(cid:12)(cid:12)(cid:12) − E ( x/p ) − E ( x/p ) + D xp (cid:12)(cid:12)(cid:12)(cid:12) < ǫ xp ... (4) (cid:12)(cid:12)(cid:12)(cid:12) − E ( x/p k − ) − E ( x/p k ) + D xp k (cid:12)(cid:12)(cid:12)(cid:12) < ǫ xp k . Using the triangle inequality to combine the inequalities (1) and (2)together with all those between (3) and (4) (inclusive) and dividingthrough by x , we can conclude that (cid:12)(cid:12)(cid:12)(cid:12) E ( x ) x − p + 1 D (cid:12)(cid:12)(cid:12)(cid:12) < ǫ k X i =1 p i ! + ǫ ǫ < ǫ. (cid:3) Theorem 1.1 now follows in the case that P is empty from the abovelemma by induction on the number of elements of T . That it is truewhen P is finite but not necessarily empty follows from Lemma 2.1:in the statement of the lemma replace P by ∅ and S by P ; then thenatural density of A ( T, P ) is6 π Y p ∈ P p Y p ∈ T ∪ P
11 + p = 6 π Y p ∈ T
11 + p Y p ∈ P p p . Proof of Theorem 1.1 when P is infinite We begin by proving the theorem in the case that T is empty. Let p , p , p , · · · be the strictly increasing sequence of elements of P . Sinceall the quotients p i / (1 + p i ) are less than 1, the partial products ofthe infinite product Q i p i / (1 + p i ) form a strictly decreasing sequencebounded below by 0; thus Q p ∈ P p/ (1 + p ) converges and its limit, say α , is independent of the order of the factors.First suppose that α = 0. Then P p ∈ P /p < ∞ . After all, we have − log α = − X p ∈ P log p p = X p ∈ P log(1+ p ) − log p > X p ∈ P
11 + p > X p ∈ P /p. RON BROWN
Now observe that
A \ A ( ∅ , P ) is the disjoint union A \ A ( ∅ , P ) = ∪ k ≥ A ( { p k } , { p , · · · , p k − } )since for all b ∈ A \ A ( ∅ , P ) there exists a least k with p k | b , so that b ∈ A ( { p k } , { p , · · · , p k − } ).For all n and k we have A ( { p k } , { p , · · · , p k − } )[ n ] n ≤ |{ j : 1 ≤ j ≤ n, p k | j }| n ≤ p k . Hence by Tannery’s theorem (see [6, p. 292] or [7, p. 199]) the naturaldensity of
A \ A ( ∅ , P ) islim n →∞ ( A \ A ( ∅ , P ))[ n ] n = lim n →∞ ∞ X k =1 A ( { p k } , { p , · · · , p k − } )[ n ] n = ∞ X k =1 lim n →∞ A ( { p k } , { p , · · · , p k − } )[ n ] n = ∞ X k =1 π
11 + p k Y i
0. By hypothesis there exists a number M with π Q i ≤ M p i p i < ǫ/
2. Then by our proof of the theorem in thecase that P is finite there exists a number L such that if n > L then( A ( ∅ , { p , p , · · · , p M } )[ n ] n < ǫ π Y i ≤ M p i p i . Then if n > L we have0 < A ( ∅ , P )[ n ] n ≤ A ( ∅ , { p , p , · · · , p M } )[ n ] n < ǫ ǫ ǫ. HE NATURAL DENSITY OF SOME SETS OF SQUARE-FREE NUMBERS 7
Therefore, lim n →∞ A ( ∅ , P )[ n ] n = 0 = Y p ∈ P p p . This completes the proof of the theorem in the case that T = ∅ .The general case where T is arbitrary then follows from Lemma 2.1,applied with T and S replaced respectively by ∅ and T : then if we set s = Q p ∈ T p we have that the natural density of A ( T, P ) equals 1 /s times the natural density of A ( ∅ , P ∪ T ), i.e., equals6 π Y p ∈ T p Y p ∈ T p p Y p ∈ P p p = 6 π Y p ∈ T
11 + p Y p ∈ P p p , which completes the proof of Theorem 1.1.5. Proof of Theorem 1.2
It suffices by Theorem 1.1 to prove that Q p ∈ P p/ (1 + p ) = 0. Let M be any number. By a lemma of K. K. Norton [5, Lemma 6.3] thereexists a constant B independent of the choices of r, m, and M suchthat (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X M>p ∈ P p − log log Mφ ( m ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < B log(3 m ) φ ( m ) . As we observed earlier, for any p we have1 / (2 p ) < / (1 + p ) < log(1 + p ) − log p so log Y M>p ∈ P p p = X M>p ∈ P (log p − log(1 + p )) < − X M>p ∈ P p < (cid:18) − log log Mφ ( m ) + B log(3 m ) φ ( m ) (cid:19) which has limit −∞ as M → ∞ . Hence Y p ∈ P p p = lim M →∞ Y M>p ∈ P p p = 0 , completing the proof of Theorem 1.2. RON BROWN
References [1] G.H. Hardy and E.M. Wright,
An Introduction to the Theory of Numbers (5thed.) (Oxford Univ. Press, 1979).[2] I. Niven and H. S. Zuckerman,
An Introduction to the Theory of Numbers (4thed.) (Wiley, 1980).[3] J.A. Scott, Square-free integers once again,
Math. Gazette (2008), 70 – 71.[4] G. J. O. Jameson, Even and odd square-free numbers, Math. Gazette (2010),123–127.[5] K. K. Norton, On the number of restricted prime factors of an integer, IllinoisJ. Math. (1976), 681 – 705.[6] J. Tannery, Introduction `a la Th´eorie des Fonctions d’une Variable, 2 ed., Tome1, Libraire Scientifique A (Hermann, Paris, 1904).[7] J. Hofbauer, A simple proof of 1 + 1 /
22 + 1 /
32 + · · · = π / American Math. Monthly (2002), 196 – 200.
Department of Mathematics, University of Hawaii, 2565 McCarthyMall, Honolulu, Hawaii 96822
Email address ::