The aliquot constant, after Bosma and Kane

The aliquot constant, after Bosma and Kane Abstract Let s(n) be the sum of those positive divisors of the natural number n other than n itself. A conjecture of Catalan–Dickson is that the ‘aliquot’ sequence of iterating s starting at any n terminates at 0 or enters a cycle. There is a ‘counter’ conjecture of Guy–Selfridge that while Catalan–Dickson may be correct for most odd numbers n, for most even seeds, the aliquot sequence is unbounded. Lending some support for Catalan–Dickson, Bosma and Kane recently showed that the geometric mean of the numbers s(n)/n for n even tends to a constant smaller than 1. In this paper, we reprove their result with a stronger error term and with a finer calculation of the asymptotic geometric mean. In addition, we solve the analogous problems for certain subsets of the even numbers, such as the even squarefrees and the multiples of 4. 1. Introduction Let s(n)=σ(n)−n denote the sum of the ‘proper’ divisors of the natural number n; that is, the sum of the positive divisors of n that are smaller than n. The Catalan–Dickson conjecture asserts that every ‘aliquot’ sequence n,s(n),s(s(n)),… either terminates at 0 or enters into a cycle, so it is always bounded. The first n in doubt is 276. The Guy–Selfridge counter conjecture, see [7], is that Catalan–Dickson is correct for asymptotically all odd numbers and false for asymptotically all even numbers. On average, the ratios s(n)/n are ζ(2)−1=0.6449⋯<1, perhaps lending some credence to the Catalan–Dickson conjecture. However, restricted to odd numbers, the average ratio s(n)/n is 34ζ(2)−1=0.2337… and restricted to even numbers it is 54ζ(2)−1=1.0561>1, perhaps lending some credence to the Guy–Selfridge counter conjecture. (Since the function s(n) usually, but not always, satisfies s(n)≡n(mod2), it seems appropriate to separate the problem by parity.) In a recent paper [1], Bosma and Kane take the view that if one is to look at averages, the geometric mean is more appropriate than the arithmetic mean. Further they found that restricted to odd numbers (larger than 1) the geometric mean of the numbers s(n)/n is o(1) and restricted to even numbers it is <1. Specifically, they prove the following result. Theorem 1.1 We have 2x∑1<n≤xnoddlog(s(n)/n)=−2e−γloglogx+O(logloglogx),where γ is Euler’s constant, and there is a constant λ<−0.03 such that 2x∑n≤xnevenlog(s(n)/n)=λ+O(1/logx). (1.1) They call the negative number λ the ‘aliquot constant’ and they suggest that their theorem lends some support to the Catalan–Dickson conjecture. In this note, we strengthen Theorem 1.1 by computing λ to higher precision and we provide a power-saving error estimate in (1.1). Theorem 1.2 To 13 decimal places we have λ=−0.03325 94808 010…, and 2x∑n≤xnevenlog(s(n)/n)=λ+O(x−0.08317). In addition, we discuss the average geometric mean over certain subsets of the even numbers, such as the even squarefrees and those that are 0 (mod 4). The function s(n) not only usually preserves parity, it also usually preserves the property of being squarefree, the property of being not squarefree, the property of being 0 (mod 4) and the property of being 2 (mod 4). So, investigating the average geometric mean over these subsets has some relevance. Finally, we remark that our high-order precision calculation of λ is not accomplished through heroic computation, but rather through some standard ideas for accelerating the convergence of certain series. It appears after numerical computations in [1] and [2] of the sum in (1.1) that the convergence to λ is fairly rapid, perhaps of order x−1+o(1) as x→∞. It is clear that it cannot be O(x−1) as individual terms can be of a slightly larger order. In [2], some numerical experiments are done for higher iterates of s. Let sk(n) denotes the kth iterate, when it exists. It is conjectured in [11] that averaged over those even numbers for which sk(n)>0, we have log(sk(n)/sk−1(n)) the same for every fixed k, namely the aliquot constant λ. This is proved for k=2 in [11]. In the numerical experiments in [2] convergence to λ seems possible for higher values of k, but if so, it is quite slow. 2. Average value of a multiplicative function The following result corrects an oversight and broadens [1, Lemma 3.12]. Proposition 2.1 Let fbe a multiplicative function and suppose that κis a positive integer with ∣f(pm)−1∣≤κ/pfor every prime pand positive integer m. For x≥max{eκ,20}, ∑n≤xf(n)=x∏p((1−1p)∑m≥0f(pm)pm)+O(1(2κ)!(logx+2κ)2κx1.4/loglogx),where the O-constant is absolute. (In [1, Lemma 3.12], it is assumed that each f(pm) is in [0,1], and the error term is asserted to be O((logx)C). Even if C is allowed to depend on κ, the proof there does not seem to support such a small error estimate.) Proof Let g be the multiplicative function which satisfies g(pm)=f(pm)−1 for each prime power pm. Writing u∣∣v if u∣v and gcd(u, v/u) = 1, we have ∑n≤xf(n)=∑n≤x∏pm∥n(1+g(pm))=∑n≤x∑d∥ng(d)=∑d≤xg(d)∑k≤x/dgcd(k,d)=11. The inner sum here, via an inclusion–exclusion over the squarefree divisors of d, is xφ(d)/d2+O(2ω(d)), so that ∑n≤xf(n)=x∑d≤xg(d)φ(d)d2+O(∑d≤x∣g(d)∣2ω(d)), (2.1) where ω(d) is the number of different primes that divide d. Letting rad(d) denote the largest squarefree divisor of d, note that ∑d≤x∣g(d)∣2ω(d)≤∑d≤x(2κ)ω(d)rad(d)=∑n≤x(2κ)ω(n)n∑d≤xrad(d)=n1. (2.2) The inner sum here is O(x1.4/loglogx) for x>e, uniformly for any n. (A stronger result is implicit in the proof of [6, Theorem 11] and explicit in [10, Lemma 4.2].) Further, as is easy to show by an induction argument ∑n≤xkω(n)n≤1k!(logx+k)k (2.3) for all positive integers k and for all x≥1. Thus, (2.2) implies that ∑d≤x∣g(d)∣2ω(d)=O(1(2κ)!(logx+2κ)2κx1.4/loglogx). (2.4) It remains to consider the main term in (2.1). We will show that ∑d>x∣g(d)∣φ(d)d2=O(1κ!(logx+κ)κx−1+1.4/loglogx), (2.5) which implies that ∑d∣g(d)∣φ(d)/d2 converges. Assume this for now. Note that 1+(1−1p)∑m≥1g(pm)pm=1+(1−1p)∑m≥1f(pm)−1pm=(1−1p)∑m≥0f(pm)pm, so that ∑dg(d)φ(d)d2=∏p(1+(1−1p)∑m≥1g(pm)pm)=∏p((1−1p)∑m≥0f(pm)pm). With (2.5), we thus have x∑d≤xg(d)φ(d)d2=x∏p(1−1p)∑m≥0f(pm)pm+O(1κ!(logx+κ)κx1.4/loglogx). (2.6) To show (2.5) first note that as in (2.2), for t>e, ∑d≤t∣g(d)∣≤∑n≤tκω(n)n∑d≤trad(d)=n1≪1κ!(logt+κ)κt1.4/loglogt, using (2.3). Since the derivative of −1.6t−1+1.4/loglogt is ≥t−2+1.4/loglogt for t≥20, ∑d>x∣g(d)∣d≪∫x∞1κ!(logt+κ)κt−2+1.4/loglogtdt≤∫x∞1κ!(logt+κ)κd(−1.6t−1+1.4/loglogt)=1.6κ!(logx+κ)κx−1+1.4/loglogx+∫x∞1.6(κ−1)!(logt+κ)κ−1t−2+1.4/loglogtdt. Under the assumption that t≥x≥max{eκ,20}, 1.6(κ−1)!(logt+κ)κ−1≤0.8κ!(logt+κ)κ, so that the prior calculation implies that ∑d>x∣g(d)∣d≪1κ!(logx+κ)κx−1+1.4/loglogx, which implies (2.5). Using (2.6) with (2.4) and (2.1) completes the proof.□ Corollary 2.2 Suppose that f,κare as in Proposition2.1and denote by E(x)the expression in the O-term. For any positive integer Aand for all xwith x/A≥max{eκ,20}we have ∑n≤xgcd(n,A)=1f(n)=φ(A)Ax∏p∤A((1−1p)∑m≥0f(pm)pm)+O(E(x)),∑n≤xA∥nf(n)=φ(A)f(A)A2x∏p∤A((1−1p)∑m≥0f(pm)pm)+O(∣f(A)∣E(x/A)). Proof Let fA(n) be the same as f(n) except that it is 0 when gcd(n,A)>1. The first assertion follows directly from Proposition 2.1 applied to fA. The second assertion follows directly from the first assertion.□ Remark. The estimate (2.3) in the proof of Proposition 2.1 is proved, as mentioned, by induction. However, the proof in mind is for the larger function τk(n) which is the number of ordered factor-izations of n into k positive integral factors. Some difficulties are entailed working directly with kω(n), but if one is willing to push through them, as in Exercises 55 and 56 in [12], one can do better. This in turn would lead to a better error estimate in Theorem 1.2. 3. A geometric mean In this section, we compute the geometric mean of the numbers σ(n)/n. Let α=∑pmlog(1+1/(σ(pm)−1))pm, (3.1) where the summation is over all prime powers pm with m≥1. Proposition 3.1 For x≥3, we have ∑n≤xlog(σ(n)/n)=αx+O(loglogx). Proof Let h(n)=σ(n)/n. For prime powers pm with m≥1 let Λσ(pm)=log(h(pm)/h(pm−1)), and for all other positive integers, let Λσ(n)=0. Note that Λσ(pm)=log(1+1σ(pm)−1)∈(0,1pm)and∑d∣nΛσ(d)=log(h(n)). We thus have ∑n≤xlog(σ(n)/n)=∑n≤x∑d∣nΛσ(d)=∑d≤xΛσ(d)⌊xd⌋. Using this identity, we immediately have the upper bound ∑n≤xlog(σ(n)/n)≤x∑dΛσ(d)d=αx. (3.2) For a lower bound, we have ∑d≤xΛσ(d)⌊xd⌋≥(α−∑d>xΛσ(d)d)x−∑d≤xΛσ(d). The proposition follows upon noting that ∑d>xΛσ(d)/d≪1/x and ∑d≤xΛσ(d)≪loglogx.□ As with Corollary 2.2, we now have the following result. Corollary 3.2 Let Abe a positive integer. For x≥3, we have ∑n≤xgcd(n,A)=1log(σ(n)/n)=α1(A)x+O(2ω(A)loglogx),∑n≤xA∥nlog(σ(n)/n)=α2(A)x+O(2ω(A)(loglogx+log(σ(A)/A))),where α1(A)=φ(A)A(α−∑gcd(d,A)>1Λσ(d)d),α2(A)=α1(A)A+φ(A)A2log(σ(A)A). 4. The geometric mean of s(n)/n To compute the geometric mean of s(n)/n, we use the identity log(s(n)/n)=log(σ(n)/n−1)=log(σ(n)/n)−∑j≥11j(n/σ(n))j, (4.1) which holds for all n>1. Let fj(n)=(n/σ(n))j. We have 1>fj(pm)>(1−1p)j≥1−jp for all primes p and positive integers m,j. We thus may apply Proposition 2.1 and Corollary 2.2 to fj with κ=j. Let βp,j=(1−1p)∑m≥0fj(pm)pm,βp,j′=(1−1p)∑m≥1fj(pm)pm, (4.2) and let Mj=∏pβp,j−12∏p>2βp,j=β2,j′∏p>2βp,j. (4.3) It follows from Proposition 2.1 and Corollary 2.2 that ∑n≤x2∣nfj(n)=Mjx+O(1(2j)!(logx+2j)2jx1.4/loglogx) (4.4) for x≥max{40,2ej}. In addition to (4.4), we have the trivial estimate ∑n≤x2∣nfj(n)≤12(23)jx. (4.5) Indeed, if n is even then n/σ(n)≤23 and so fj(n)≤(23)j. Thus, (4.5) follows since the sum there has at most 12x terms. Dividing (4.4) by x and letting x→∞, the estimate (4.5) implies that Mj≤12(23)j, (4.6) so that β:=∑j1jMj (4.7) converges. Let λ=2α−2α1(2)−2β, where α is defined in (3.1) and α1(2) is defined in Corollary 3.2. Theorem 4.1 We have 2x∑n≤x2∣nlog(s(n)/n)=λ+O(x−0.08317). Proof Using (4.1), we have ∑n≤x2∣nlog(s(n)/n)=A−B, where A=∑n≤x2∣nlog(σ(n)/n)andB=∑j≥11j∑n≤x2∣nfj(n). By Corollary 3.2, we have A=(α−α1(2))x+O(loglogx). (4.8) Let J be a large number to be determined shortly. By (4.4), we have ∑j≤J1j∑n≤x2∣nfj(n)=x∑j≤JMjj+O(∑j≤J1(2j)!(logx+2j)2jx1.4/loglogx). (4.9) Further, from (4.5) and (4.6), we have ∑j>J1j∑n≤x2∣nfj(n)≪(23)Jx,∑j>JMjj≪(23)J. Using this with (4.9), we have B=βx+O((23)Jx+∑j≤J1(2j)!(logx+2j)2jx1.4/loglogx). (4.10) It remains to choose the optimal value of J, which is close to 0.205131logx. With this choice, a calculation shows the error term in (4.10) is O(x0.91683), which together with (4.8), proves the theorem.□ With a few simple changes, we have the following results. Corollary 4.2 With δ=0.08317, we have 4x∑n≤x2∥nlog(s(n)n)=λ2+O(x−δ)and4x∑n≤x4∣nlog(s(n)n)=λ4+O(x−δ),where λ2=4α2(2)−∑j≥11j(23)j∏p>2βp,j,λ4=2λ−λ2. 4.1. The even squarefree case We can also obtain an estimation for ∑n≤x2∣nμ(n)2=1log(s(n)n). To this end, it is straightforward by following the proof of Proposition 3.1 to obtain the following result: ∑n≤xμ(n)2=1log(σ(n)n)=6α0π2x+O(x1/2), where α0=∑plog(1+1/p)p+1. Further, as in Corollary 3.2, ∑n≤xμ(n)2=12∤nlog(σ(n)n)=4π2(α0−log(3/2)3)x+O(x1/2), so that ∑n≤xμ(n)2=12∣nlog(σ(n)n)=(2α0π2+4log(3/2)3π2)x+O(x1/2). (4.11) For a multiplicative function f satisfying the hypothesis of Proposition 2.1, it is possible to prove an analog of that result for μ(n)2f(n) (with an extra factor of x1/2 in the error term), but we leave this for another time. For now, we content ourselves with a non-uniform result which follows directly from Tenenbaum [12, Theorem I.3.12]: ∑n≤xμ(n)2=12∣nfj(n)=x4fj(2)∏p>2((1−1p)(1+fj(p)p))+o(x),(x→∞). We now use (4.1). Since the number of even squarefree numbers to x is 2π2x+O(x1/2), we have the average value of log(s(n)/n) for even squarefree numbers to x is ∼λ0x as x→∞, where λ0=α0+23log(3/2)−π28∑j≥11j(23)j∏p>2((1−1p)(1+fj(p)p)). 5. Evaluation of constants In this section, we discuss the numerical evaluation of the various constants introduced, and in particular α, defined in (3.1), and β, defined in (4.7). Calculations were done with Mathematica. 5.1. The calculation of α Let N0=15 485 863 denote the one-millionth prime. Let Ap=1plog(1+1p)+log(1−1p2). A minor calculation shows that −1/p3<Ap<−1/(2p3) for all p. Note, as well, that p−mlog(1+1σ(pm)−1)<p−2m. Let mp=⌊logN0/logp⌋. We have α−logζ(2)=∑pAp+∑p,m≥21pmlog(1+1σ(pm)−1)=∑p≤N0Ap+∑p≤N0∑2≤m≤2mp1pmlog(1+1σ(pm)−1)+E, (5.1) where −∑p>N01p3<E<−∑p>N012p3+∑p>N01p2(p2−1)+∑p≤N01p2mp(p2−1). It is not difficult to get a reasonable estimate for the sum on the left. One can use what is known about the evaluation of the ‘prime zeta-function’ (for example, see Glaisher [8]), or merely compute logζ(3)+∑p≤N0log(1−p−3). We find that 1.22×10−16<∑p>N0p−3<1.23×10−16. (5.2) We have ∑p>N01p2(p2−1)<1N02∑p>N01p2−1<1N03<2.7×10−22,∑p≤N01p2mp(p2−1)<2.4×10−20, so that −1.23×10−16<E<−6.0×10−17. With (5.1), we have α=0.44570 89175 47339 15±4×10−17. (5.3) 5.2. The calculation of β Lemma 5.1 Using the notation of (4.2), we have for each prime pand positive integer j, 1−jp2<βp,j<1−jp2+j2(p−1)3. Proof For the lower bound note that βp,j>(1−1p)(1+1p−1(1−1p)j)>(1−1p)(1+1p−1(1−jp))=1−jp2. The upper bound trivially holds for p≤j since βp,j<1. So assume that p≥j+1. Then βp,j<(1−1p)(1+1p−1(1−1p+1)j)<(1−1p)(1+1p−1(1−jp+1+(j2)1(p+1)2))=1−jp(p+1)+(j2)1p(p+1)2<1−jp2+j2(p−1)3, completing the proof.□ A simple calculation now verifies the following consequence. Corollary 5.2 For p>jwe have ∣log((1−1p2)−jβp,j)∣<2j2p3. We use this for p large. For p somewhat smaller we use the following result. Recall that fj(n)=(n/σ(n))j. Lemma 5.3 For every prime pand positive integers j,m1, we have (1−1/p)jpm1(p−1)+∑m=0m1fj(pm)pm<βp,j1−1/p<fj(pm1+1)pm1(p−1)+∑m=0m1fj(pm)pm. Recalling the definition of Mj in (4.3), we also have the following simple result. Lemma 5.4 For each positive integer j, we have Mj+1/Mj<2/3. Proof For every p,j,m we have fj+1(pm)fj(pm)=pmσ(pm)≤1, with the inequality strict for m≥1. Thus βp,j+1<βp,j. In the case p=2 and m≥1, the ratio above is maximal when m=1 and is 23 there. Thus, β2,j+1′≤23β2,j′. This completes the proof.□ As before, N0 is the one-millionth prime and mp=⌊logN0/logp⌋. For each prime p≤N0, let βp,j−=(1−1p)((1−1/p)jpmp(p−1)+∑m=0mpfj(pm)pm),βp,j+=(1−1p)(fj(pmp+1)pmp(p−1)+∑m=0mpfj(pm)pm), and similarly let βp,j′± be the corresponding quantities where the sums start at m=1. Let Mj,N0±=(34ζ(2))−jβ2,j′±∏2<p≤N0βp,j±(1−p−2)−j. We have from Lemma 5.3 that Mj,N0−∏p>N0βp,j(1−p−2)−j<Mj<Mj,N0+∏p>N0βp,j(1−p−2)−j. With Corollary 5.2, we thus have for j<N0 that Mj,N0−exp(−2j2∑p>N0p−3)<Mj<Mj,N0+exp(2j2∑p>N0p−3). (5.4) Using (5.2), Mj,N0−exp(−2.46×10−16j2)<Mj<Mj,N0+exp(2.46×10−16j2). (5.5) We have computed that 2.14×10−13<M65<2.141×10−13, so by Lemma 5.4, we have that 3.29×10−15<∑j≥651jMj<9.65×10−15. (5.6) We have also computed that ∑j=1641jMj,N0−exp(−2.46×10−16j2)>0.36578 82599 73748 99,∑j=1641jMj,N0+exp(2.46×10−16j2)<0.36578 82599 73751 53. Thus, with (5.5) and (5.6), we have β=∑j1jMj=0.36578 82599 73757 23±5×10−15. (5.7) 5.3. The calculation of λ We have 2α−2α1(2)=α+∑m≥1log(1+1/(2m+1−2))2m. The infinite sum is easily computed to be 0.25260 81215 99091 81384…, so that with (5.3) and (5.7), λ=2α−2α1(2)−2β=−0.03325 94808 01084±1.1×10−14. (5.8) We have done similar calculations for λ0,λ2,λ4. In particular, to 13 decimal places, we have λ0=−0.33843 54384 114…,λ2=−0.24129 50555 350…,λ4=+0.17477 60939 329…. 6. Discussion According to Bosma and Kane [1], the fact that the average value of log(s(n)/n) for n even is negative lends support to the Catalan–Dickson conjecture. The reasoning is as follows. One should think of the sequence n,s(n),s(s(n)),… as usually approximately geometric. Let sj(n) denote the jth iterate of s at n. In fact, it is shown in Erdős [4] that for each fixed k and ϵ>0, the ratio sk+1(n)/sk(n) is at least s(n)/n−ϵ on a set of n of asymptotic density 1. He also claims the analogous result that sk+1(n)/sk(n) is at most s(n)/n+ϵ almost always, but this was later retracted in [5]. It is still thought though that this is the case, just the claim of proof is retracted. In [5], the assertion for k=1 is proved. But why do we restrict to even numbers? It is easy to see that s(n)≡n(mod2) except if n is of the form 2ab2, and of course such numbers are very sparsely distributed. So, as a simplifying assumption, perhaps it is reasonable to assume that at high levels, an aliquot sequence tends to maintain its parity. It is less obvious that it maintains other properties. For example, if n is even, then s(n)≡n(mod4) except if the odd part of n is of the form pb2, where p is a prime with p≡1(mod4). So, s(n) maintains parity with ‘probability’ about 1/n, but for even n it maintains the residue mod 4 with ‘probability’ about 1/logn. Persistence of divisibility or lack thereof by other small primes is even weaker. For example, for an odd prime p, the ‘probability’ that p∤s(n) given that p∣n is about 1/(logn)1/(p−1), and the same holds for p∣s(n) given that p∤n. However, as noted in Guy–Selfridge [7], these events are not independent. For example, if n≡2(mod4), then 3∣σ(n), so 3∣n if and only if 3∣s(n). This gives rise to the idea of a ‘driver’. If it should turn out for example that n≡6(mod12) (and n>6), then not only do we have s(n)>n, but it is highly likely that s(n)≡6(mod12), and so on. What it would take to break this behavior is to arrive at a point where we pick up an extra factor of 2, which may occur with a chance about 1/logn, as discussed. So, the persistence of 3 is not a 1/(logn)1/2 chance of breaking, but rather 1/logn. Note that if n≡±6(mod36), then it is certain that s(n)≡2mod4 and so it is certain that s(n)≡6(mod12), and so it is certain that 6∣s2(n). However, the chance of breaking being ±6 (mod 36) is about 1/(logn)1/2. One complication of this kind of thinking is that we do not know much about what is ‘normal’ for sk(n), even for a fixed number k≥2. We do know that a positive proportion of even integers are not values of s(n) (see Erdős [3]) and that a positive proportion of even integers are values of s(n) (see Luca–Pomerance [9]), but we know very little about even numbers that are of the form s2(n). While proofs can be, and often are, complicated, if they are to be useful and believable, heuristics should be simple. So perhaps the Bosma–Kane point of view is helpful. But consider the situation for those n divisible by 4. If the chance for this to break is 1/logn and the aliquot sequence is about geometric with ratio eλ4=1.1909…, by the time we move logn steps further, the numbers we are dealing with are larger by a factor of about nλ4, and so now the chance of breaking divisibility by 4 is somewhat less. It is perhaps not so convincing, but it may be most aliquot sequences that start at a multiple of 4 will diverge to infinity. Acknowledgements I thank Anton Mosunov for his interest in this paper. In addition, he suggested that it was of interest to compute the average of log(s(n)/n) over even squarefree numbers. I am grateful to Mits Kobayashi for helping me with a Mathematica issue. I also thank the referee for a careful reading. References 1 W. Bosma and B. Kane , The aliquot constant , Quart. J. Math. 63 ( 2012 ), 309 – 323 . Google Scholar Crossref Search ADS 2 K. Chum , R. K. Guy , M. J. Jacobson Jr. and A. S. Mosunov , Numerical and statistical analysis of aliquot sequences, preprint, 2017 . 3 P. Erdős , Über die Zahlen der Form σ(n)−n und n−φ(n) , Elem. Math. 28 ( 1973 ), 83 – 86 . 4 P. Erdős , On asymptotic properties of aliquot sequences , Math. Comp. 30 ( 1976 ), 641 – 645 . Google Scholar Crossref Search ADS 5 P. Erdős , A. Granville , C. Pomerance and C. Spiro , On the normal Behavior of the Iterates of Some Arithmetic Functions, Analytic number theory (Allerton Park, IL, 1989), Progr. Math. Vol. 85, Birkhäuser , Boston, Boston, MA , 1990 , 165 – 204 . 6 P. Erdős , F. Luca and C. Pomerance , On the Proportion of Integers Coprime to a Given Integer, Proceedings of the Anatomy of Integers Conference, Montreal, March 2006 (Eds. J.-M. De Koninck, A. Granville and F. Luca), CRM Proceedings and Lecture Notes, Vol. 46 ( 2008 ), 47–64. 7 R. K. Guy and J. L. Selfridge , What drives an aliquot sequence? , Math. Comp. 29 ( 1975 ), 101 – 107 . Google Scholar Crossref Search ADS 8 J. W. L. Glaisher , On the sums of inverse powers of the prime numbers , Quart. J. Pure Appl. Math. 25 ( 1891 ), 347 – 362 . 9 F. Luca and C. Pomerance , The range of the sum-of-proper-divisors function , Acta Arith. 168 ( 2015 ), 187 – 199 . Google Scholar Crossref Search ADS 10 P. Pollack , The greatest common divisor of a number and its sum of divisors , Michigan Math. J. 60 ( 2011 ), 199 – 214 . Google Scholar Crossref Search ADS 11 C. Pomerance , The First Function and Its Iterates, Connections in Discrete Mathematics: A Celebration of the Work of Ron Graham (Eds. S. Butler, J. Cooper and G. Hurlbert), Cambridge U. Press, to appear. 12 G. Tenenbaum , Introduction to Analytic and Probabilistic Number Theory , 3rd edn , Graduate Studies in Mathematics, Vol. 163. American Math. Soc , Providence , 2015 . © The Author(s) 2018. Published by Oxford University Press. All rights reserved. For permissions, please e-mail: journals.permissions@oup.com This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/open_access/funder_policies/chorus/standard_publication_model) http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png The Quarterly Journal of Mathematics Oxford University Press

The aliquot constant, after Bosma and Kane

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Abstract Let s(n) be the sum of those positive divisors of the natural number n other than n itself. A conjecture of Catalan–Dickson is that the ‘aliquot’ sequence of iterating s starting at any n terminates at 0 or enters a cycle. There is a ‘counter’ conjecture of Guy–Selfridge that while Catalan–Dickson may be correct for most odd numbers n, for most even seeds, the aliquot sequence is unbounded. Lending some support for Catalan–Dickson, Bosma and Kane recently showed that the geometric mean of the numbers s(n)/n for n even tends to a constant smaller than 1. In this paper, we reprove their result with a stronger error term and with a finer calculation of the asymptotic geometric mean. In addition, we solve the analogous problems for certain subsets of the even numbers, such as the even squarefrees and the multiples of 4. 1. Introduction Let s(n)=σ(n)−n denote the sum of the ‘proper’ divisors of the natural number n; that is, the sum of the positive divisors of n that are smaller than n. The Catalan–Dickson conjecture asserts that every ‘aliquot’ sequence n,s(n),s(s(n)),… either terminates at 0 or enters into a cycle, so it is always bounded. The first n in doubt is 276. The Guy–Selfridge counter conjecture, see [7], is that Catalan–Dickson is correct for asymptotically all odd numbers and false for asymptotically all even numbers. On average, the ratios s(n)/n are ζ(2)−1=0.6449⋯<1, perhaps lending some credence to the Catalan–Dickson conjecture. However, restricted to odd numbers, the average ratio s(n)/n is 34ζ(2)−1=0.2337… and restricted to even numbers it is 54ζ(2)−1=1.0561>1, perhaps lending some credence to the Guy–Selfridge counter conjecture. (Since the function s(n) usually, but not always, satisfies s(n)≡n(mod2), it seems appropriate to separate the problem by parity.) In a recent paper [1], Bosma and Kane take the view that if one is to look at averages, the geometric mean is more appropriate than the arithmetic mean. Further they found that restricted to odd numbers (larger than 1) the geometric mean of the numbers s(n)/n is o(1) and restricted to even numbers it is <1. Specifically, they prove the following result. Theorem 1.1 We have 2x∑1<n≤xnoddlog(s(n)/n)=−2e−γloglogx+O(logloglogx),where γ is Euler’s constant, and there is a constant λ<−0.03 such that 2x∑n≤xnevenlog(s(n)/n)=λ+O(1/logx). (1.1) They call the negative number λ the ‘aliquot constant’ and they suggest that their theorem lends some support to the Catalan–Dickson conjecture. In this note, we strengthen Theorem 1.1 by computing λ to higher precision and we provide a power-saving error estimate in (1.1). Theorem 1.2 To 13 decimal places we have λ=−0.03325 94808 010…, and 2x∑n≤xnevenlog(s(n)/n)=λ+O(x−0.08317). In addition, we discuss the average geometric mean over certain subsets of the even numbers, such as the even squarefrees and those that are 0 (mod 4). The function s(n) not only usually preserves parity, it also usually preserves the property of being squarefree, the property of being not squarefree, the property of being 0 (mod 4) and the property of being 2 (mod 4). So, investigating the average geometric mean over these subsets has some relevance. Finally, we remark that our high-order precision calculation of λ is not accomplished through heroic computation, but rather through some standard ideas for accelerating the convergence of certain series. It appears after numerical computations in [1] and [2] of the sum in (1.1) that the convergence to λ is fairly rapid, perhaps of order x−1+o(1) as x→∞. It is clear that it cannot be O(x−1) as individual terms can be of a slightly larger order. In [2], some numerical experiments are done for higher iterates of s. Let sk(n) denotes the kth iterate, when it exists. It is conjectured in [11] that averaged over those even numbers for which sk(n)>0, we have log(sk(n)/sk−1(n)) the same for every fixed k, namely the aliquot constant λ. This is proved for k=2 in [11]. In the numerical experiments in [2] convergence to λ seems possible for higher values of k, but if so, it is quite slow. 2. Average value of a multiplicative function The following result corrects an oversight and broadens [1, Lemma 3.12]. Proposition 2.1 Let fbe a multiplicative function and suppose that κis a positive integer with ∣f(pm)−1∣≤κ/pfor every prime pand positive integer m. For x≥max{eκ,20}, ∑n≤xf(n)=x∏p((1−1p)∑m≥0f(pm)pm)+O(1(2κ)!(logx+2κ)2κx1.4/loglogx),where the O-constant is absolute. (In [1, Lemma 3.12], it is assumed that each f(pm) is in [0,1], and the error term is asserted to be O((logx)C). Even if C is allowed to depend on κ, the proof there does not seem to support such a small error estimate.) Proof Let g be the multiplicative function which satisfies g(pm)=f(pm)−1 for each prime power pm. Writing u∣∣v if u∣v and gcd(u, v/u) = 1, we have ∑n≤xf(n)=∑n≤x∏pm∥n(1+g(pm))=∑n≤x∑d∥ng(d)=∑d≤xg(d)∑k≤x/dgcd(k,d)=11. The inner sum here, via an inclusion–exclusion over the squarefree divisors of d, is xφ(d)/d2+O(2ω(d)), so that ∑n≤xf(n)=x∑d≤xg(d)φ(d)d2+O(∑d≤x∣g(d)∣2ω(d)), (2.1) where ω(d) is the number of different primes that divide d. Letting rad(d) denote the largest squarefree divisor of d, note that ∑d≤x∣g(d)∣2ω(d)≤∑d≤x(2κ)ω(d)rad(d)=∑n≤x(2κ)ω(n)n∑d≤xrad(d)=n1. (2.2) The inner sum here is O(x1.4/loglogx) for x>e, uniformly for any n. (A stronger result is implicit in the proof of [6, Theorem 11] and explicit in [10, Lemma 4.2].) Further, as is easy to show by an induction argument ∑n≤xkω(n)n≤1k!(logx+k)k (2.3) for all positive integers k and for all x≥1. Thus, (2.2) implies that ∑d≤x∣g(d)∣2ω(d)=O(1(2κ)!(logx+2κ)2κx1.4/loglogx). (2.4) It remains to consider the main term in (2.1). We will show that ∑d>x∣g(d)∣φ(d)d2=O(1κ!(logx+κ)κx−1+1.4/loglogx), (2.5) which implies that ∑d∣g(d)∣φ(d)/d2 converges. Assume this for now. Note that 1+(1−1p)∑m≥1g(pm)pm=1+(1−1p)∑m≥1f(pm)−1pm=(1−1p)∑m≥0f(pm)pm, so that ∑dg(d)φ(d)d2=∏p(1+(1−1p)∑m≥1g(pm)pm)=∏p((1−1p)∑m≥0f(pm)pm). With (2.5), we thus have x∑d≤xg(d)φ(d)d2=x∏p(1−1p)∑m≥0f(pm)pm+O(1κ!(logx+κ)κx1.4/loglogx). (2.6) To show (2.5) first note that as in (2.2), for t>e, ∑d≤t∣g(d)∣≤∑n≤tκω(n)n∑d≤trad(d)=n1≪1κ!(logt+κ)κt1.4/loglogt, using (2.3). Since the derivative of −1.6t−1+1.4/loglogt is ≥t−2+1.4/loglogt for t≥20, ∑d>x∣g(d)∣d≪∫x∞1κ!(logt+κ)κt−2+1.4/loglogtdt≤∫x∞1κ!(logt+κ)κd(−1.6t−1+1.4/loglogt)=1.6κ!(logx+κ)κx−1+1.4/loglogx+∫x∞1.6(κ−1)!(logt+κ)κ−1t−2+1.4/loglogtdt. Under the assumption that t≥x≥max{eκ,20}, 1.6(κ−1)!(logt+κ)κ−1≤0.8κ!(logt+κ)κ, so that the prior calculation implies that ∑d>x∣g(d)∣d≪1κ!(logx+κ)κx−1+1.4/loglogx, which implies (2.5). Using (2.6) with (2.4) and (2.1) completes the proof.□ Corollary 2.2 Suppose that f,κare as in Proposition2.1and denote by E(x)the expression in the O-term. For any positive integer Aand for all xwith x/A≥max{eκ,20}we have ∑n≤xgcd(n,A)=1f(n)=φ(A)Ax∏p∤A((1−1p)∑m≥0f(pm)pm)+O(E(x)),∑n≤xA∥nf(n)=φ(A)f(A)A2x∏p∤A((1−1p)∑m≥0f(pm)pm)+O(∣f(A)∣E(x/A)). Proof Let fA(n) be the same as f(n) except that it is 0 when gcd(n,A)>1. The first assertion follows directly from Proposition 2.1 applied to fA. The second assertion follows directly from the first assertion.□ Remark. The estimate (2.3) in the proof of Proposition 2.1 is proved, as mentioned, by induction. However, the proof in mind is for the larger function τk(n) which is the number of ordered factor-izations of n into k positive integral factors. Some difficulties are entailed working directly with kω(n), but if one is willing to push through them, as in Exercises 55 and 56 in [12], one can do better. This in turn would lead to a better error estimate in Theorem 1.2. 3. A geometric mean In this section, we compute the geometric mean of the numbers σ(n)/n. Let α=∑pmlog(1+1/(σ(pm)−1))pm, (3.1) where the summation is over all prime powers pm with m≥1. Proposition 3.1 For x≥3, we have ∑n≤xlog(σ(n)/n)=αx+O(loglogx). Proof Let h(n)=σ(n)/n. For prime powers pm with m≥1 let Λσ(pm)=log(h(pm)/h(pm−1)), and for all other positive integers, let Λσ(n)=0. Note that Λσ(pm)=log(1+1σ(pm)−1)∈(0,1pm)and∑d∣nΛσ(d)=log(h(n)). We thus have ∑n≤xlog(σ(n)/n)=∑n≤x∑d∣nΛσ(d)=∑d≤xΛσ(d)⌊xd⌋. Using this identity, we immediately have the upper bound ∑n≤xlog(σ(n)/n)≤x∑dΛσ(d)d=αx. (3.2) For a lower bound, we have ∑d≤xΛσ(d)⌊xd⌋≥(α−∑d>xΛσ(d)d)x−∑d≤xΛσ(d). The proposition follows upon noting that ∑d>xΛσ(d)/d≪1/x and ∑d≤xΛσ(d)≪loglogx.□ As with Corollary 2.2, we now have the following result. Corollary 3.2 Let Abe a positive integer. For x≥3, we have ∑n≤xgcd(n,A)=1log(σ(n)/n)=α1(A)x+O(2ω(A)loglogx),∑n≤xA∥nlog(σ(n)/n)=α2(A)x+O(2ω(A)(loglogx+log(σ(A)/A))),where α1(A)=φ(A)A(α−∑gcd(d,A)>1Λσ(d)d),α2(A)=α1(A)A+φ(A)A2log(σ(A)A). 4. The geometric mean of s(n)/n To compute the geometric mean of s(n)/n, we use the identity log(s(n)/n)=log(σ(n)/n−1)=log(σ(n)/n)−∑j≥11j(n/σ(n))j, (4.1) which holds for all n>1. Let fj(n)=(n/σ(n))j. We have 1>fj(pm)>(1−1p)j≥1−jp for all primes p and positive integers m,j. We thus may apply Proposition 2.1 and Corollary 2.2 to fj with κ=j. Let βp,j=(1−1p)∑m≥0fj(pm)pm,βp,j′=(1−1p)∑m≥1fj(pm)pm, (4.2) and let Mj=∏pβp,j−12∏p>2βp,j=β2,j′∏p>2βp,j. (4.3) It follows from Proposition 2.1 and Corollary 2.2 that ∑n≤x2∣nfj(n)=Mjx+O(1(2j)!(logx+2j)2jx1.4/loglogx) (4.4) for x≥max{40,2ej}. In addition to (4.4), we have the trivial estimate ∑n≤x2∣nfj(n)≤12(23)jx. (4.5) Indeed, if n is even then n/σ(n)≤23 and so fj(n)≤(23)j. Thus, (4.5) follows since the sum there has at most 12x terms. Dividing (4.4) by x and letting x→∞, the estimate (4.5) implies that Mj≤12(23)j, (4.6) so that β:=∑j1jMj (4.7) converges. Let λ=2α−2α1(2)−2β, where α is defined in (3.1) and α1(2) is defined in Corollary 3.2. Theorem 4.1 We have 2x∑n≤x2∣nlog(s(n)/n)=λ+O(x−0.08317). Proof Using (4.1), we have ∑n≤x2∣nlog(s(n)/n)=A−B, where A=∑n≤x2∣nlog(σ(n)/n)andB=∑j≥11j∑n≤x2∣nfj(n). By Corollary 3.2, we have A=(α−α1(2))x+O(loglogx). (4.8) Let J be a large number to be determined shortly. By (4.4), we have ∑j≤J1j∑n≤x2∣nfj(n)=x∑j≤JMjj+O(∑j≤J1(2j)!(logx+2j)2jx1.4/loglogx). (4.9) Further, from (4.5) and (4.6), we have ∑j>J1j∑n≤x2∣nfj(n)≪(23)Jx,∑j>JMjj≪(23)J. Using this with (4.9), we have B=βx+O((23)Jx+∑j≤J1(2j)!(logx+2j)2jx1.4/loglogx). (4.10) It remains to choose the optimal value of J, which is close to 0.205131logx. With this choice, a calculation shows the error term in (4.10) is O(x0.91683), which together with (4.8), proves the theorem.□ With a few simple changes, we have the following results. Corollary 4.2 With δ=0.08317, we have 4x∑n≤x2∥nlog(s(n)n)=λ2+O(x−δ)and4x∑n≤x4∣nlog(s(n)n)=λ4+O(x−δ),where λ2=4α2(2)−∑j≥11j(23)j∏p>2βp,j,λ4=2λ−λ2. 4.1. The even squarefree case We can also obtain an estimation for ∑n≤x2∣nμ(n)2=1log(s(n)n). To this end, it is straightforward by following the proof of Proposition 3.1 to obtain the following result: ∑n≤xμ(n)2=1log(σ(n)n)=6α0π2x+O(x1/2), where α0=∑plog(1+1/p)p+1. Further, as in Corollary 3.2, ∑n≤xμ(n)2=12∤nlog(σ(n)n)=4π2(α0−log(3/2)3)x+O(x1/2), so that ∑n≤xμ(n)2=12∣nlog(σ(n)n)=(2α0π2+4log(3/2)3π2)x+O(x1/2). (4.11) For a multiplicative function f satisfying the hypothesis of Proposition 2.1, it is possible to prove an analog of that result for μ(n)2f(n) (with an extra factor of x1/2 in the error term), but we leave this for another time. For now, we content ourselves with a non-uniform result which follows directly from Tenenbaum [12, Theorem I.3.12]: ∑n≤xμ(n)2=12∣nfj(n)=x4fj(2)∏p>2((1−1p)(1+fj(p)p))+o(x),(x→∞). We now use (4.1). Since the number of even squarefree numbers to x is 2π2x+O(x1/2), we have the average value of log(s(n)/n) for even squarefree numbers to x is ∼λ0x as x→∞, where λ0=α0+23log(3/2)−π28∑j≥11j(23)j∏p>2((1−1p)(1+fj(p)p)). 5. Evaluation of constants In this section, we discuss the numerical evaluation of the various constants introduced, and in particular α, defined in (3.1), and β, defined in (4.7). Calculations were done with Mathematica. 5.1. The calculation of α Let N0=15 485 863 denote the one-millionth prime. Let Ap=1plog(1+1p)+log(1−1p2). A minor calculation shows that −1/p3<Ap<−1/(2p3) for all p. Note, as well, that p−mlog(1+1σ(pm)−1)<p−2m. Let mp=⌊logN0/logp⌋. We have α−logζ(2)=∑pAp+∑p,m≥21pmlog(1+1σ(pm)−1)=∑p≤N0Ap+∑p≤N0∑2≤m≤2mp1pmlog(1+1σ(pm)−1)+E, (5.1) where −∑p>N01p3<E<−∑p>N012p3+∑p>N01p2(p2−1)+∑p≤N01p2mp(p2−1). It is not difficult to get a reasonable estimate for the sum on the left. One can use what is known about the evaluation of the ‘prime zeta-function’ (for example, see Glaisher [8]), or merely compute logζ(3)+∑p≤N0log(1−p−3). We find that 1.22×10−16<∑p>N0p−3<1.23×10−16. (5.2) We have ∑p>N01p2(p2−1)<1N02∑p>N01p2−1<1N03<2.7×10−22,∑p≤N01p2mp(p2−1)<2.4×10−20, so that −1.23×10−16<E<−6.0×10−17. With (5.1), we have α=0.44570 89175 47339 15±4×10−17. (5.3) 5.2. The calculation of β Lemma 5.1 Using the notation of (4.2), we have for each prime pand positive integer j, 1−jp2<βp,j<1−jp2+j2(p−1)3. Proof For the lower bound note that βp,j>(1−1p)(1+1p−1(1−1p)j)>(1−1p)(1+1p−1(1−jp))=1−jp2. The upper bound trivially holds for p≤j since βp,j<1. So assume that p≥j+1. Then βp,j<(1−1p)(1+1p−1(1−1p+1)j)<(1−1p)(1+1p−1(1−jp+1+(j2)1(p+1)2))=1−jp(p+1)+(j2)1p(p+1)2<1−jp2+j2(p−1)3, completing the proof.□ A simple calculation now verifies the following consequence. Corollary 5.2 For p>jwe have ∣log((1−1p2)−jβp,j)∣<2j2p3. We use this for p large. For p somewhat smaller we use the following result. Recall that fj(n)=(n/σ(n))j. Lemma 5.3 For every prime pand positive integers j,m1, we have (1−1/p)jpm1(p−1)+∑m=0m1fj(pm)pm<βp,j1−1/p<fj(pm1+1)pm1(p−1)+∑m=0m1fj(pm)pm. Recalling the definition of Mj in (4.3), we also have the following simple result. Lemma 5.4 For each positive integer j, we have Mj+1/Mj<2/3. Proof For every p,j,m we have fj+1(pm)fj(pm)=pmσ(pm)≤1, with the inequality strict for m≥1. Thus βp,j+1<βp,j. In the case p=2 and m≥1, the ratio above is maximal when m=1 and is 23 there. Thus, β2,j+1′≤23β2,j′. This completes the proof.□ As before, N0 is the one-millionth prime and mp=⌊logN0/logp⌋. For each prime p≤N0, let βp,j−=(1−1p)((1−1/p)jpmp(p−1)+∑m=0mpfj(pm)pm),βp,j+=(1−1p)(fj(pmp+1)pmp(p−1)+∑m=0mpfj(pm)pm), and similarly let βp,j′± be the corresponding quantities where the sums start at m=1. Let Mj,N0±=(34ζ(2))−jβ2,j′±∏2<p≤N0βp,j±(1−p−2)−j. We have from Lemma 5.3 that Mj,N0−∏p>N0βp,j(1−p−2)−j<Mj<Mj,N0+∏p>N0βp,j(1−p−2)−j. With Corollary 5.2, we thus have for j<N0 that Mj,N0−exp(−2j2∑p>N0p−3)<Mj<Mj,N0+exp(2j2∑p>N0p−3). (5.4) Using (5.2), Mj,N0−exp(−2.46×10−16j2)<Mj<Mj,N0+exp(2.46×10−16j2). (5.5) We have computed that 2.14×10−13<M65<2.141×10−13, so by Lemma 5.4, we have that 3.29×10−15<∑j≥651jMj<9.65×10−15. (5.6) We have also computed that ∑j=1641jMj,N0−exp(−2.46×10−16j2)>0.36578 82599 73748 99,∑j=1641jMj,N0+exp(2.46×10−16j2)<0.36578 82599 73751 53. Thus, with (5.5) and (5.6), we have β=∑j1jMj=0.36578 82599 73757 23±5×10−15. (5.7) 5.3. The calculation of λ We have 2α−2α1(2)=α+∑m≥1log(1+1/(2m+1−2))2m. The infinite sum is easily computed to be 0.25260 81215 99091 81384…, so that with (5.3) and (5.7), λ=2α−2α1(2)−2β=−0.03325 94808 01084±1.1×10−14. (5.8) We have done similar calculations for λ0,λ2,λ4. In particular, to 13 decimal places, we have λ0=−0.33843 54384 114…,λ2=−0.24129 50555 350…,λ4=+0.17477 60939 329…. 6. Discussion According to Bosma and Kane [1], the fact that the average value of log(s(n)/n) for n even is negative lends support to the Catalan–Dickson conjecture. The reasoning is as follows. One should think of the sequence n,s(n),s(s(n)),… as usually approximately geometric. Let sj(n) denote the jth iterate of s at n. In fact, it is shown in Erdős [4] that for each fixed k and ϵ>0, the ratio sk+1(n)/sk(n) is at least s(n)/n−ϵ on a set of n of asymptotic density 1. He also claims the analogous result that sk+1(n)/sk(n) is at most s(n)/n+ϵ almost always, but this was later retracted in [5]. It is still thought though that this is the case, just the claim of proof is retracted. In [5], the assertion for k=1 is proved. But why do we restrict to even numbers? It is easy to see that s(n)≡n(mod2) except if n is of the form 2ab2, and of course such numbers are very sparsely distributed. So, as a simplifying assumption, perhaps it is reasonable to assume that at high levels, an aliquot sequence tends to maintain its parity. It is less obvious that it maintains other properties. For example, if n is even, then s(n)≡n(mod4) except if the odd part of n is of the form pb2, where p is a prime with p≡1(mod4). So, s(n) maintains parity with ‘probability’ about 1/n, but for even n it maintains the residue mod 4 with ‘probability’ about 1/logn. Persistence of divisibility or lack thereof by other small primes is even weaker. For example, for an odd prime p, the ‘probability’ that p∤s(n) given that p∣n is about 1/(logn)1/(p−1), and the same holds for p∣s(n) given that p∤n. However, as noted in Guy–Selfridge [7], these events are not independent. For example, if n≡2(mod4), then 3∣σ(n), so 3∣n if and only if 3∣s(n). This gives rise to the idea of a ‘driver’. If it should turn out for example that n≡6(mod12) (and n>6), then not only do we have s(n)>n, but it is highly likely that s(n)≡6(mod12), and so on. What it would take to break this behavior is to arrive at a point where we pick up an extra factor of 2, which may occur with a chance about 1/logn, as discussed. So, the persistence of 3 is not a 1/(logn)1/2 chance of breaking, but rather 1/logn. Note that if n≡±6(mod36), then it is certain that s(n)≡2mod4 and so it is certain that s(n)≡6(mod12), and so it is certain that 6∣s2(n). However, the chance of breaking being ±6 (mod 36) is about 1/(logn)1/2. One complication of this kind of thinking is that we do not know much about what is ‘normal’ for sk(n), even for a fixed number k≥2. We do know that a positive proportion of even integers are not values of s(n) (see Erdős [3]) and that a positive proportion of even integers are values of s(n) (see Luca–Pomerance [9]), but we know very little about even numbers that are of the form s2(n). While proofs can be, and often are, complicated, if they are to be useful and believable, heuristics should be simple. So perhaps the Bosma–Kane point of view is helpful. But consider the situation for those n divisible by 4. If the chance for this to break is 1/logn and the aliquot sequence is about geometric with ratio eλ4=1.1909…, by the time we move logn steps further, the numbers we are dealing with are larger by a factor of about nλ4, and so now the chance of breaking divisibility by 4 is somewhat less. It is perhaps not so convincing, but it may be most aliquot sequences that start at a multiple of 4 will diverge to infinity. Acknowledgements I thank Anton Mosunov for his interest in this paper. In addition, he suggested that it was of interest to compute the average of log(s(n)/n) over even squarefree numbers. I am grateful to Mits Kobayashi for helping me with a Mathematica issue. I also thank the referee for a careful reading. References 1 W. Bosma and B. Kane , The aliquot constant , Quart. J. Math. 63 ( 2012 ), 309 – 323 . Google Scholar Crossref Search ADS 2 K. Chum , R. K. Guy , M. J. Jacobson Jr. and A. S. Mosunov , Numerical and statistical analysis of aliquot sequences, preprint, 2017 . 3 P. Erdős , Über die Zahlen der Form σ(n)−n und n−φ(n) , Elem. Math. 28 ( 1973 ), 83 – 86 . 4 P. Erdős , On asymptotic properties of aliquot sequences , Math. Comp. 30 ( 1976 ), 641 – 645 . Google Scholar Crossref Search ADS 5 P. Erdős , A. Granville , C. Pomerance and C. Spiro , On the normal Behavior of the Iterates of Some Arithmetic Functions, Analytic number theory (Allerton Park, IL, 1989), Progr. Math. Vol. 85, Birkhäuser , Boston, Boston, MA , 1990 , 165 – 204 . 6 P. Erdős , F. Luca and C. Pomerance , On the Proportion of Integers Coprime to a Given Integer, Proceedings of the Anatomy of Integers Conference, Montreal, March 2006 (Eds. J.-M. De Koninck, A. Granville and F. Luca), CRM Proceedings and Lecture Notes, Vol. 46 ( 2008 ), 47–64. 7 R. K. Guy and J. L. Selfridge , What drives an aliquot sequence? , Math. Comp. 29 ( 1975 ), 101 – 107 . Google Scholar Crossref Search ADS 8 J. W. L. Glaisher , On the sums of inverse powers of the prime numbers , Quart. J. Pure Appl. Math. 25 ( 1891 ), 347 – 362 . 9 F. Luca and C. Pomerance , The range of the sum-of-proper-divisors function , Acta Arith. 168 ( 2015 ), 187 – 199 . Google Scholar Crossref Search ADS 10 P. Pollack , The greatest common divisor of a number and its sum of divisors , Michigan Math. J. 60 ( 2011 ), 199 – 214 . Google Scholar Crossref Search ADS 11 C. Pomerance , The First Function and Its Iterates, Connections in Discrete Mathematics: A Celebration of the Work of Ron Graham (Eds. S. Butler, J. Cooper and G. Hurlbert), Cambridge U. Press, to appear. 12 G. Tenenbaum , Introduction to Analytic and Probabilistic Number Theory , 3rd edn , Graduate Studies in Mathematics, Vol. 163. American Math. Soc , Providence , 2015 . © The Author(s) 2018. Published by Oxford University Press. All rights reserved. For permissions, please e-mail: journals.permissions@oup.com This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/open_access/funder_policies/chorus/standard_publication_model)

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The Quarterly Journal of MathematicsOxford University Press

Published: Sep 1, 2018

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