MONOTONE NONDECREASING SEQUENCES OF THE EULER TOTIENT FUNCTION

A BSTRACT . Let 𝑀 ( 𝑥 ) denote the largest cardinality of a subset of { 𝑛 ∈ ℕ ∶ 𝑛 ≤ 𝑥 } on which the Euler totient function 𝜑 ( 𝑛 ) is nondecreasing. We show that 𝑀 ( 𝑥 ) = (1+ 𝑂 ( (loglog 𝑥 ) 5 log 𝑥 )) 𝜋 ( 𝑥 ) for all 𝑥 ≥ 10 , answering questions of Erdős and Pollack–Pomerance–Treviño. A similar result is also obtained for the sum of divisors function 𝜎 ( 𝑛 ) .

2020 Mathematics Subject Classification.11A25. 1 See Section 1.1 for our asymptotic notation conventions.
The main result of this paper is to improve the upper bound on ( ) to asymptotically match the lower bound, with only a slight degradation in the error term.
Proof.We may assume is an integer.By summation by parts (or telescoping series), we have By Theorem 1.1, we have ∑ ∈∶ ≤ 1 ≤ 1 + log 5   2 log log for all large .The claim then follows from the absolute convergence of ∑ ≥10 log 5   2 log 2 , and the standard asymptotic ∑ 10≤ ≤ 1 log = log 2 + (1) for ≥ 100.
In [22] the much stronger claim was posed as an open question, with numerics in fact suggesting the even more precise conjecture ( ) = ( ) + 64 (1.4) for ≥ 31957 (see [22, §9], where the conjecture was verified 2 for = 10 for ≤ 7).Unfortunately, there appear to be several obstacles to establishing such a strong result, which we discuss in Section 4. Nevertheless, some improvements in the error term in (1.2) may be possible, particularly if one is willing to establish results conditional on such conjectures as Cramér's conjecture [4] or quantitative forms of the Dickson-Hardy-Littlewood prime tuples conjecture [6], [15]; again, see Section 4.
The arguments in [22] were largely based on an analysis of equations of the form ( ) = ( + ).To get the more precise asymptotics of Theorem 1.1, we proceed differently, as we shall now explain.Perhaps surprisingly, we do not use deep results from analytic number theory; aside from the prime number theorem with classical error term (Lemma 1.6), all of our tools are elementary.
It is first necessary to establish the following sharp inequality regarding the (weighted) multiplicity of the function ↦ ( )∕ .

Proposition 1.4 (Preliminary inequality). For any positive rational number , one has
The reason we need to establish this inequality is as follows.Suppose for contradiction that the inequality failed, then one could find a rational number and a finite set  of natural numbers with ( ) = such that Observe that if = for ∈  and a prime larger than the largest element max  of , then ( ) = ( ) ( ) = ( − 1) = ( − ).
(1.5) To exploit this, let be the set of numbers in [ ] of the form , where ∈  and > max  is a prime, and let ′ be the set with any consecutive elements of with difference less than max  deleted.If 1 1 < 2 2 are consecutive elements of ′ , then by (1.5) we have In particular, is increasing on ′ , and one can check using the prime number theorem (as well as standard upper bound sieves to bound the number of elements deleted) that as → ∞, which would contradict Theorem 1.1.Thus we must prove Proposition 1.4 first.Fortunately, this is easy to accomplish thanks to a minor miracle: the sum appearing in Proposition 1.4 can be computed almost exactly whenever it is non-zero!See Lemma 2.1 below.
We now informally outline the proof of Theorem 1.1.Standard results from the anatomy of integers will tell us that all numbers in [ ] outside of a small exceptional set will be of one of the following two forms: ( 1 ) = , where is a very smooth number (all prime factors small), and is a large prime.
, where is a medium-sized prime, has all prime factors less than , and is an almost prime (a prime or product of two primes), with all prime factors of much larger than .
The precise definitions of 1 , 2 and are given in Section 3, with the decomposition justified by Lemma 3.1 and Proposition 3.2.
Suppose first that = is of type 2 .From multiplicativity we have the identity since is almost prime and has large prime factors we are led to the approximation The upshot of this approximation is that if we restrict to a fixed value, and restrict to a somewhat short interval, then ( ) can only be nondecreasing if the prime is also essentially nondecreasing.The presence of the unspecified factor allows one to decouple the magnitude of from the magnitude of , even when is fixed; as a consequence, this additional monotonicity constraint on turns out to significantly cut down the possible length of a sequence of numbers of this form in which is increasing, to the point where the contribution of this case is significantly less than log .
If instead = is of type 1 , then we similarly have the approximation and so the quantity ( ) must (heuristically, at least) be locally nondecreasing in order for ( ) to be nondecreasing.Upon applying the prime number theorem, one can basically conclude that the size of such a sequence is at most and one can use Lemma 1.4 to conclude (after performing a more careful accounting of error terms).
The argument used to establish Theorem 1.1 can also be adapted to treat the analogous problem for the sum of divisors function ( ) as well as the Dedekind totient function ( ): see Section 4.4.
The author is supported by NSF grant DMS-1764034.We thank Wouter van Doorn, Kevin Ford, Paul Pollack and Carl Pomerance for helpful comments, suggestions, and references, and to Shengtong Zhang to providing the key ingredient needed to obtain the analogous results for the sum of divisors function.We also thank the anonymous referees for several useful comments and references, including [17] and Remark 4.7.
1.1.Notation and basic tools.We use the usual asymptotic notation ≪ , ≫ , or = ( ) to denote the claim | | ≤ for an absolute constant , ≍ to denote the claim ≪ ≪ , and = ( ) to denote the claim | | ≤ ( ) for some quantity ( ) that goes to zero as → ∞.
The symbol will be understood to range over primes unless otherwise indicated.
For any ≥ 2, we let ℕ ≤ denote the set of -smooth numbers, that is the natural numbers whose prime factors are all less than or equal to .We similarly let ℕ < denotes the natural numbers whose prime factors are all strictly less than .
We record a useful (if somewhat crude) estimate: Lemma 1.5 (Rankin trick).If ≥ 10, and is a non-negative real number for every ∈ ℕ ≤ , then ∑ One could obtain sharper bounds here by using standard bounds on smooth numbers here, such as those in [5], but the above crude estimate suffices for our application.In this lemma we allow the quantities on either side to be infinite, though in applications we will only use the lemma when the right-hand side is finite, as the claim is trivial otherwise.
Proof.By Euler product factorization and Mertens' theorems, we have ∑ The claim now follows by bounding We also need some well known bounds for primes in intervals: Lemma 1.6 (Primes in intervals).Let ≥ 10, and let be an interval in [2, ].Let | | denote the length of .Then the number of primes in is equal to for some absolute constant > 0. In particular, we have the crude upper bound of ( ∕ log ).
Proof.Immediate from the prime number theorem with classical error term.
As a corollary of the prime number theorem, we obtain a Mertens-type estimate.
Proof.By Mertens' second theorem we have which is acceptable if > 2 (in which case we may discard the exp(− √ log ) term).In the remaining case ≤ ≤ 2 ,one can bound 1 ≪ 1 and apply Lemma 1.6.

PROOF OF PRELIMINARY INEQUALITY
We now prove Proposition 1.4.It turns out that (somewhat remarkably) one has a reasonably exact formula for the sum of interest, which implies Proposition 1.4 as an immediate corollary: Lemma 2.1 (Exact formula).Let = ∕ be a positive rational number in lowest terms.Then the expression ∑ for some finite set of primes  whose largest element is equal to the largest prime factor of (with  empty when = 1).In particular, Proposition 1.4 holds.
Proof.When = 1, the only possible value of that is of the form ( ) is = 1, which is attained precisely when = 1, so the identity holds in this case.
To handle the > 1 case we induct on the largest prime factor 1 of , assuming that the claim has already been proven for smaller values of 1 (with the convention that 1 = 1 when = 1).
We may assume that is of the form ( ) for at least one .For such a , we have From this identity we see that the largest prime factor 1 of must also be the largest prime factor of .Thus we can write = 1 ′ where ≥ 1, ′ ∈ ℕ < 1 and ( ′ ) ′ = ′ , where ′ = ′ ′ is equal to 1 1 −1 in lowest terms.By the same computation, the largest prime factor ′ 1 of ′ is equal to that of ′ , and thus , then the largest prime factor of ′ must equal ′ 1 , so that ′ is coprime to 1 , and hence any of the form = 1 ′ with ≥ 1 is such that Thus we have ∑ The claim now follows from the induction hypothesis.
Remark 2.2.From Lemma 2.1 we see that equality in Proposition 1.4 only holds when = 1 or = 1 2 , with the improved inequality holding in all other cases.The equality when = 1  2 produces an almost viable counterexample to conjectures such as (1.3); see Section 4.2 below.Remark 2.3.Lemma 2.1 is closely related to the well known fact (see e.g., [13]) that if , are natural numbers, then ( )∕ and ( )∕ are equal if and only if , have exactly the same set of prime factors.

MAIN ARGUMENT
We now prove Theorem 1.1.We may of course take to be larger than any given absolute constant.We will need some intermediate scales There is some flexibility in how to select these scales; we will make the choice Note that these scales are widely separated, in the sense that any fixed power of one of the expressions in (3.1) is still less than the next expression in (3.1), if is large enough.For instance, the savings exp(− √ log ) arising from Lemma 1.6 can easily absorb any polynomial losses in and .In a similar spirit, we have log 2 ≍ log ≍ log , log ≍ log ≍ log , log( ) ≍ log , and so forth.
We observe the triangle inequality3 as well as the trivial bound We shall cover the set [ ] by three (slightly overlapping) sets which by (3.5), (3.6) implies that We will control each of the three terms on the right-hand side separately.
The exceptional set is defined to be the set of all ∈ [ ] having an unusual prime factorization in at least one of the following senses: (iii) (Large square factor) is divisible by 2 for some > .
(iv) (Large smooth factor) is divisible by some ∈ ℕ ≤ with > .(v) (Smooth times two nearby primes) = 2 1 where ∈ ℕ ≤ and 1 , 2 are primes with ≤ 2 ≤ 1 ≤ 2 .(vi) (Integer times three nearby primes) = 3 2 1 where 1 , 2 , 3 are primes with 2 ≤ The primary set 1 is defined to be the set of all ∈ [ ] of the form = where ∈ ℕ ≤ with ≤ , is a prime, and > .The secondary set 2 is defined to be the set of all ∈ [ ] of the form = where is a prime greater than , ∈ ℕ < , and is an almost prime (either a prime or product of two primes) whose prime factors all exceed .Proof.We may assume that avoids all the cases (i)-(vi) of .Let 1 ≥ 2 ≥ 3 ≥ 4 be the four largest prime factors of (with the convention that = 1 if has fewer than prime factors).Since avoids the cases (i), (ii) of we have > and 1 > .We divide into three cases.
• If < 2 ≤ 1 , then we must have 3 ≠ 2 (since avoids case (iii) of ).One then checks that one is in the set 2 (with = 2 , = 1 , and = ). • Finally, if 2 ≤ , then the quantity ∶= 1 is at most (since avoids case (iv) of ), and then one checks that one is in the set 1 (with = 1 ).Now we verify that the exceptional set is small: Proposition 3.2 ( small).We have # ≪ The set of obeying (i) clearly has size ( ∕ ), which is acceptable (with plenty of room to spare).By Lemma 1.5, the set of obeying (ii) has size at most 4∑ which is also acceptable (with room to spare) from (3.4).The set of obeying (iii) has size at most which is again acceptable (with plenty of room to spare).
The set of obeying (iv) has size at most Applying Lemma 1.5, we can bound this by ≪ log 1∕ log , which is acceptable (with substantial room to spare) from (3.2), (3.3).
By the prime number theorem (Lemma 1.6), we see that for any choice of , 2 , the number of possible values of 1 is at most We conclude that this contribution to the size is at most From the prime number theorem (Lemma 1.6) we have Finally, if obeys (vi), then we may factor where 1 , 2 , 3 are primes in [ ] with 5 Alternatively, we can use the Euler product If we fix 1 , 2 , 3 , there are at most choices of , thus this contribution to the size is at most .
By two applications of Lemma 1.7, this is , which is acceptable thanks to Mertens' theorem and (3.4), (3.2).
We remark that one could obtain much sharper asymptotics on the size of if desired using more sophisticated results from the anatomy of integers, such as [12].Now we control 2 .
Proof.Let ′ be a subset of 2 on which is nondecreasing.Our task is to show that This will be the first time we will use the monotonicity of in our arguments; on the other hand, there is a lot of room here, and we can afford to lose several powers of log in what follows.If ∈ ′ , then we may factor where is a prime in the range < ≤ 1∕2 , is an element of ℕ < with ≤ 2 , and is an almost prime, with all prime factors larger than .In particular and thus by multiplicativity We exploit this relation as follows.First we perform dyadic decomposition to estimate where ( , ) ∶= ∑ and the inner sum is restricted to those for which = obeys the estimate (3.9).We will establish the estimate ( , ) ≪ log 4  (3.10) for each with < 2 ≪ 1∕2 and all ∈ ℕ <2 ; the claim will then follow by summing (3.10) over all , and using crude estimates (or Lemma 1.5).
Putting these estimates together (and recalling that is assumed to be large), we conclude that and hence by monotonicity of on ′ we have ′ > .This gives the claim.
As a corollary of this monotonicity, we see that the intervals , have bounded overlap in [1, ] as , vary (in fact any can belong to at most two of the , ).In particular, ∑ ∑ | , | ≪ . (3.12) We have the 6 trivial bound and hence by (3.11) we may bound Since there are only (2 log 6 ) values of , and the cover an interval { ∶ ≍ 2 } with bounded overlap, we can crudely estimate Next, by discarding the requirement that is prime, we have the trivial bound By (3.12), we conclude that Combining these estimates, we obtain and (3.10) follows from (3.2).Finally, we treat the primary set 1 .Proposition 3.4 ( 1 contribution).We have ( 1 ) ≤ 1 + log 2 log log . 6One could save an additional factor of log or so here by using a Brun-Titchmarsh inequality for almost primes, but this turns out not to significantly improve the bounds in our main result.
Proof.Let ′ be a subset of 1 on which is nondecreasing.Our task is to show that In particular (from (3.3)) we certainly have Thus is coprime to , and hence If we define the set then  is a finite set of rationals in (0, 1], of some cardinality = # with ≤ . (3.18) We arrange the elements of  in increasing order as For each = 1, … , , we let  denote the set of all ∈ ℕ ≤ with ≤ and We also partition [ , ] into ( 3 log ) disjoint intervals of the form ( , (1 + ( 1 3 )) ].For = 1, … , , and any , let , denote the smallest interval containing all = in ′ ∩ with ∈  and a prime obeying (3.15).Then we have The key observation is that the intervals , are all disjoint.To see this, first observe that since , ⊂ and the are disjoint, we clearly have then , ′ lies strictly to the right of , (so in particular the two sets are disjoint).Indeed, if = ∈ , and ′ = ′ ′ ∈ , ′ then from (3.17), (3.16), (3.19) one has Similarly we have Also, as , ′ both lie in ′ , we have while since ′ > , we have ′ > .From (3.14), the rational numbers ′ , have denominator at most , hence and thus (since ≤ 1) Combining all these statements (and recalling that and hence is large), we conclude that and hence ′ > by monotonicity, giving the claim.
From the prime number theorem (Lemma 1.6) and (3.15) we have for an absolute constant > 0. From Lemma 1.4 we have the crucial inequality Inserting these bounds back into (3.20),we conclude that From disjointness of the , we have Since there are ( 3 log ) values of , we conclude

OBSTACLES TO IMPROVEMENT, VARIANTS, AND OPEN QUESTIONS
It seems likely that small improvements to Theorem 1.1, such as the lowering of the exponent 5 in the power of log 2 , could be achieved with minor modifications of the method.However, we now present two obstacles that seem to prevent one from unconditionally establishing significantly stronger results, such as (1.3), and pose some open problems that might still be achievable despite these obstacles.
4.1.Legendre conjecture obstruction.Legendre's conjecture asserts that for every ≥ 2, there is a prime between ( − 1) 2 and 2 .This conjecture remains open even for sufficiently large , and even if one assumes the Riemann hypothesis (RH); it is roughly equivalent to showing that the gap +1 − between consecutive primes is significantly smaller than 1∕2 , whereas the best known bound is ( 1∕2 log ) on RH [4], or ( 0.525 ) unconditionally [2].
Legendre's conjecture is also open if one restricts to be prime.This special case of the conjecture will need to be resolved if one wishes to establish (1.3), due to the following implication: Proposition 4.1.Suppose that Legendre's conjecture is false for infinitely many primes; that is to say, there exist infinitely many primes such that there is no prime between ( − 1) 2 and 2 .Then ( ) − ( ) goes to infinity as → ∞.
Stated contrapositively: if (1.3) holds, then Legendre's conjecture is true for all sufficiently large primes.
Proof.If is a prime counterexample to Legendre's conjecture, then for primes ′ larger than 2 we have and for primes ′ less than 2 , we have ′ < ( − 1) 2 and hence Thus one can insert 2 into the sequence of primes without disrupting the monotonicity of .If there are infinitely many counterexamples to Legendre's conjecture at the primes, then one can then insert an infinite number of additional elements into the primes to create a larger sequence with nondecreasing, giving the claim.Remark 4.2.One can replace Legendre's conjecture here with the very slightly stronger conjecture of Oppermann [21] that there is always a prime between ( − 1) and 2 for ≥ 3. (Oppermann also made the analogous conjecture concerning 2 and ( + 1), but that is of less relevance here.)Remark 4.3.With a little more computation, one can show that the conjecture (1.4) implies that Legendre's conjecture holds at all primes (not just sufficiently large ones).Remark 4.4.In [26], the estimate is proven for large assuming the Riemann hypothesis, where denotes the th prime.One consequence of this is that the counterexamples to Legendre's conjecture are quite rare on RH, since each such counterexample contributes ≫ 1 to this sum.Thus this obstruction only worsens the gap between ( ) and ( ) by at most (log 3 ), on RH.
4.2.Dickson-Hardy-Littlewood conjecture obstruction.The Dickson-Hardy-Littlewood conjecture [6], [15] asserts, among other things, that if , are coprime integers with > 0, that the number of primes ∈ [ ] with + also prime is equal to as → ∞ for some positive quantity , (known as the singular series) which has an explicit form which we will not give here.Among other things, this conjecture (together with standard bounds for averages of singular series) would imply for any fixed natural number that the number of primes ∈ [ ] such that the quantity ⌈ 2 ⌉ (the first integer greater than or equal to ∕2 ) is also prime is ≍ log 2 , if is sufficiently large depending on .Such results are out of reach of current technology; note for instance that the Dickson-Hardy-Littlewood conjecture implies the notoriously open twin prime conjecture.
The following result shows that a severe breakdown of the Dickson-Hardy-Littlewood conjecture would imply that the bounds in Theorem 1.1 cannot be improved (except perhaps for removing the factors of log 2 ).This proposition suggests that significant progress on the Dickson-Hardy-Littlewood conjecture would need to be made in order to improve the error term in Theorem 1.1 to be better than ( 1 log ); in particular, this would need to be accomplished in order to have any hope of establishing (1.3).Note that standard probabilistic heuristics (cf.[14]) predict that for 2 ≤ 2 log , a positive fraction of the intervals [ , + 2 ] in [ ] will be devoid of primes, and so the set {⌈ 2 ⌉ ∶ ∈ [ ]} should avoid a positive density subset of [ ∕2 ].Thus, in the absence of the Dickson-Hardy-Littlewood conjecture it is a priori conceivable that the primes in [ ∕2 ] mostly avoid such a set; we do not know how to rule out such a scenario with current technology.We remark that the Maynard sieve [19, Theorem 3.1] (when combined with the calculations in [23]; see also [13,Lemma 2]) applied to the linear functions ↦ 2 + 1 for = 1, … , 50 implies that there exists 1 ≤ ≤ 49 such that the number of primes in [ ] with ⌈ 2 ⌉ also prime is ≫ log 50 for infinitely many , but this bound is not strong enough to contradict the hypothesis of the above proposition.
Proof.The construction here is motivated by the observation (see Remark 2.2) that Proposition 1.4 holds with equality not only at = 1, but also at = 1 2 , as well as the construction discussed after Proposition 1.4 in the introduction.We thank Kevin Ford for pointing out an improvement to the argument that gave superior quantitative bounds.
Let be sufficiently large, and let 0 be the unique natural number with log < 2 0 ≤ 2 log .
Let be the set of all numbers ∈ [ ] of the form = 2 where 1 ≤ ≤ 0 and is an odd prime.Observe from the prime number theorem (Lemma 1.6) and the logarithmic integral asymptotic We have 0 = (log 2 ), 0 Note for comparison that ( ) = log + log 2 + log 3  .
Since log 4 > 1, we see that is slightly denser than the primes.
Unfortunately, is not quite nondecreasing on ; however, the exceptions to monotonicity are quite rare and can be described explicitly as follows.Suppose that 2 , 2 ′ ′ are elements of with 2 < 2 ′ ′ and (2 ) > (2 ′ ′ ).Evaluating the totient functions, we obtain which we can rearrange as Since the left-hand side is positive, we must then have ′ > .Dividing by 2 ′ , we conclude in particular that By hypothesis, this scenario can only occur for ( log 2 log 3

2
) primes for each choice of , ′ .Since there are at most (log 2 2 ) choices for , ′ , we thus see that by deleting at most ( log 2 log 2 ) elements from , one can obtain a slightly smaller set ′ that still obeys the asymptotic (4.1), but on which is now nondecreasing.Since # ′ − ( ) ≫ log 2 for sufficiently large , we obtain the claim.

4.3.
A conditional result?Despite these two obstructions, it is possible that a more refined version of the analysis in this paper can be used to show that a significant improvement to the bounds in Theorem 1.1 can be achieved unless one of the above two scenarios occurs (breakdown of Legendre-type conjectures, or breakdown of Dickson-Hardy-Littlewood type conjectures).Indeed, the fact (see Remark 2.2) that Lemma 1.4 gains a factor of two when ≠ 1, 1 2 implies (by a routine modification of the arguments in this paper) that, if one sets to be the elements of [ ] that are not a prime, or a power of two times a prime, then ( ) ≤ ( 1 2 + ( log 5 2 log )) ( ).Thus largest subsets of [ ] on which are increasing must have almost half or more of their elements consist of primes, or powers of two times a prime.With a suitable form of the Dickson-Hardy-Littlewood conjectures, one might be able to show that the sets which contain a large number of powers of two times a prime lead to a set with fewer than ( ) elements, and so one would be left with studying sets that largely consist of primes.Here, one could take advantage of the phenomenon (already observed in [22, §9]) that for any prime , the first composite number > with ( ) ≥ ( ) must in fact exceed + √ − 1, which suggests that (in the absence of extremely large prime gaps, as in Proposition 4.1), it could rule out competitors to the set of primes in which one or more primes in the "interior" of the set are replaced with composite numbers.In view of known results on the anatomy of integers in short intervals, one might hope to obtain a bound such as ( ) = ( ) + ( ) for some 0 < < 1 assuming additional conjectures, such as a sufficiently quantitative version of the Dickson-Hardy-Littlewood conjecture.In a closely related spirit, one could also seek to obtain a "short interval" version of the main theorem of the form for some 0 < < 1.

4.4.
Analogues for the sum of divisors function.It turns out (as already conjectured in [10]) that the Euler totient function ( ) may be replaced 7 Theorem 3] (see also [7, p. 212-213] for an alternate proof).In [9, p. 63] it was also claimed that (4.2) could be shown "with more trouble", but without a full proof provided.Such problems are also related to those of counting solutions to ( ) = ( + ) for fixed ; see [13].Since the initial release of this preprint, a full proof of (4.2) using the methods in [8] has been kindly provided to us (via the author's personal blog) by Shengtong Zhang, which we reproduce below.Given this inequality, the arguments in Section 3 can be modified as follows in order to obtain the analogue of Theorem 1.1 for the sum of divisors function: • All occurrences of should now be replaced with (including in the definition of ()).• In the proof of Proposition 3.3, all displayed expressions of the form 1 − … should be replaced with 1 + … (and similarly with replaced by ′ ), and " , ′ lies strictly to the right of , " should be replaced with " , ′ lies strictly to the left of , ". • In the proof of Proposition 3.4, all displayed expressions of the form 1 − 1 or 1 − ( 13 ) should be replaced with 1 + 1 or 1 + ( 13 ) respectively, and Proposition 1.4 must be replaced with (4.2).
(4.3)If = 1 then ′ = 1 and hence = ′ , contradiction, and similarly if ′ = 1.Thus we may assume , ′ > 1.At least one of , ′ is not divisible by 3, so without loss of generality we may assume that is not divisible by 3. Then the largest prime factor of divides but not ′ or ( ), contradicting (4.3), and the key observation follows.
We may thus write ∑ where ranges over powerful numbers, and = , is equal to the (necessarily unique) square-free number coprime to with ( ) = , or ∶= +∞ if no such number exists.Since every powerful number is the product of a square and a cube, we have (in fact the sum is about 1.94359 … ).This already establishes (4.2) but with an upper bound of 2 instead of 1.To refine the upper bound to 1, it will suffice to establish the lower bound Since the summand is non-negative, it will suffice to establish this bound for the powers of 2, thus we have reduced to showing that This can be verified by the following case analysis.
• The case 1 = 1 cannot occur since is assumed to not equal 1.
It suffices to show that the set of ∈ [ ] obeying each of the six cases (i)-(vi) of has size ( we can bound the previous contribution to # 3 by This is acceptable thanks to (3.2), (3.3),(3.4).

Proposition 4 . 5 . 2 )
Suppose that for all sufficiently large , and all natural numbers with 2 ≤ 2 log , that the number of primes ∈ [ ] with ⌈ 2 ⌉ also prime is bounded by ( log 2 log 3 (in contradiction with what the Dickson-Hardy-Littlewood conjecture would predict).Then one has ( ) − ( ) ≫ log 2 for sufficiently large .

≥ 2
and coprime to , and so the left-hand side of (4.4) is 1 If 1 = ∞, then the left-hand side of (4.4) is clearly at least 1 denominator, which implies for any ≥ 2 and coprime to 2 that(2