THE RECIPROCAL SUM OF THE AMICABLE NUMBERS

Transcription

1 THE RECIPROCAL SUM OF THE AMICABLE NUMBERS HANH MY NGUYEN AND CARL POMERANCE ABSTRACT. In this paper, we improve on several earlier attempts to show that the reciprocal sum of the amicable numbers is small, showing this sum is < INTRODUCTION Let σ(n denote the sum-of-divisors function; that is, σ(n = d n d. A pair of distinct numbers n, n are said to form an amicable pair if σ(n = σ(n = n + n, and we call an integer amicable if it is a member of such a pair. This concept was first noted by Pythagoras who used the function s(n = σ(n n. Thus, n is amicable if and only if s(s(n = n and s(n n. There are about 2 million amicable pairs nown, but we do not now if there are infinitely many of them. Though studied by many since antiquity, the amicable numbers were not nown to comprise a set of asymptotic density 0 until 955, when this was shown by Erdős [6]. And it was not nown until 98 that the amicable numbers have a finite reciprocal sum, see [2]. Roughly using the approach of [2], Bayless and Klyve [2] were able to show the reciprocal sum of the amicable numbers is less than This is in contrast to the lower bound of computed from the nown amicable numbers, so there is clearly a huge gap between this upper bound and the lower bound! The paper [2] on the distribution of the amicable numbers was improved in the recent paper [3], and using some ideas from this paper, the first-named author [7] was able to about halve the gap (on a logarithmic scale, showing the reciprocal sum of the amicable numbers is less than Here we mae further progress. Theorem.. The reciprocal sum of the amicable numbers is smaller than 222. One of the ideas from [7], namely using an averaging argument to show there are few abundant numbers (s(n > n among the odd numbers, is taen further here, to include numbers that are 2 (mod 4 and not divisible by 5. In addition, we establish some new estimates on the reciprocal sum of numbers without large prime factors. These estimates may prove to be useful in other problems, such as in []. We carve out various subsets of the amicable numbers, such as the odd amicables and the even pairs which do not agree (mod 4. In particular, these two subsets have a considerably smaller reciprocal sum than what we are able to prove for the complementary set. 2. LEMMAS Lemma 2.. With γ as Euler s constant, we have for x > 0 that n (log x + γ < x. n x Date: August 4, Mathematics Subject Classification. A25, N25, N64.

2 2 HANH MY NGUYEN AND CARL POMERANCE Proof. The result holds trivially when 0 < x <, so assume x. By partial summation n = x x x + t x t 2 dt = log x + x + t t t t t 2 dt + t 2 dt. n x The next-to-last integral is γ so that x x = log x + γ + n x n x x t t t 2 Since this last integral is positive and smaller than /x, the result follows. Let ϕ denote Euler s function, let µ denote the Möbius function, and let ω denote the function which counts the number of distinct prime divisors. Corollary 2.. For x > 0 and u a positive integer, n ϕ(u u (log x + γ + d u n x gcd(n,u= µ(d log d d dt. x < 2ω(u x. Proof. This result follows immediately from Lemma 2. and the identity n = µ(d n = µ(d d n. n x d u n x d u n x/d gcd(n,u= d n Lemma 2.2. For any z > 0 we have z<n ez Let S be a set of positive integers. We have n < z<n ex z<n ez s S, s n n < + z. s S, s ez s + z s S, s ez Proof. The first estimate is trivial if z <, so assume that z. Then n z + n < ez z + dt t + z. s S, s ez z<n ez s n z + n ez For the second estimate, we have that the sum in question is at most (2. n = s m < s S, s ez using the first estimate, and the result follows. z<n ez s S, s n gcd(n,u= z/s<m ez/s z. s S, s ez ( + s, s z Lemma 2.3. Let S be a set of positive integers coprime to the positive integer u. We have n < ϕ(u u s + 2ω(u ( + /e. z s S, s ez s S, s ez

3 THE RECIPROCAL SUM OF THE AMICABLE NUMBERS 3 Proof. We use Corollary 2. for the sum on m in (2.. Lemma 2.4. For a real number x e, we have log log x < < log log x + n log(x/n log x. n x/e Further, for x 6, n log(x/n < log log x log log(x/2 + 2 log x < 7/5 log log x + 2 log x, n x/e n odd n x/e 2 n, 3 n n log(x/n < 2 log log(x/2 6 log log(x/6 + 2 log(x/2 < 3/4 log log x + 3 log x. Proof. The function /(t log(x/t is decreasing in t on the interval [, x/e]. Since it has antiderivative log log(x/t, we have x/e n x/e For the lower bound, we use n log(x/n < log x + n x/e dt t log(x/t = + log log x. log x x/e n log(x/n > dt t log(x/t. The last two assertions follow from the first displayed result and some simple calculations. Lemma 2.5. For positive integers j, n, let τ j (n denote the number of ordered factorizations of n into j positive factors. We have for any x > 0 that τ j (n n j! (j + log xj. n x This result is [8, (4.9]. We always use the letters p, q, r to represent prime numbers. Lemma 2.6. Let H(x = p. p x With B = the Mertens constant and x 286, we have H(x (log log x + B < 2(log x 2. Further, x<p ex p < log x + 2(log x 2. Proof. The first assertion is [4, Theorem 5], and the second assertion follows from this and also the inequality log log(ex log log x + 2(log x 2 + 2(log(ex 2 < log x + 2(log x 2.

4 4 HANH MY NGUYEN AND CARL POMERANCE Lemma 2.7. For x >, we have p>x p 2 < x log x. Proof. We easily verify that the lemma holds when x 0 4 (in fact, the sum is smaller than 0.92/(x log x in this range, so assume that x > 0 4. Let θ(t denote the Chebyshev function log t. It follows from [4] and [5] that p t (2.2 t 2 t < θ(t < t (423 t 0 9, θ(t t < We have p>x p 2 = p>x log p p 2 log p = θ(x x 2 log x + x t (log t 3 (t ( 2 θ(t t 3 log t + t 3 (log t 2 dt, via partial summation. Assume that x 0 9, so that (2.2 implies that p 2 < x 2 x ( x 2 log x + 2 t 2 log t + t 2 (log t 2 dt = x 2 x x 2 log x + 2 x log x p>x In addition, = x log x + 2 x 3/2 log x x x x dt t 2 (log t 2. dt t 2 (log t 2 > ex dt ( (log ex 2 x t 2 = e x(log ex 2. x dt t 2 (log t 2 Using this estimate in the prior one, we have the lemma in the range 0 4 x 0 9. The range x > 0 9 follows in the same way by using the second inequality in (2.2 If a, m are coprime integers with m > 0, let π(x; m, a = Lemma 2.8. For a, m coprime as above and x > m, π(x; m, a < Moreover, if A, B are numbers with m < A < B, then A<p B p a (mod m p x p a (mod m. 2x ϕ(m log(x/m. p < 2 ϕ(m log(b/m + 2 (log log(b/m log log(a/m. ϕ(m Proof. The first assertion is the version of the Brun Titchmarsh inequality in Montgomery Vaughan []. The second assertion follows directly by partial summation. Lemma 2.9. For x > y 2 and 0 < s <, let S(x, y =, ζ(s, y = n n s = P (n y p y n>x, P (n y ( + p s.

5 THE RECIPROCAL SUM OF THE AMICABLE NUMBERS 5 Then S(x, y x s ζ( s, y. Further, if 2 y 0 < y, then ( y S(x, y x s s 0 exp y0 s y 0 <p y p s p y 0 ( + p s. Proof. The first inequality is clear since if n > x we have /n < x s /n s. The second inequality follows from + α < e α for α > 0 and the fact that z s /(z s is decreasing in z for z 2. Lemma 2.0. Let x > y 2, u = log x/ log y, and assume that u 3 and log(u log u/ log y /3. With S(x, y as in Lemma 2.9, we have where ε = S(x, y < 25e (+εu (u log u u (2 log(u log u/ log y. Proof. Let s = log(u log u/ log y and apply Lemma 2.9. Then x s = (u log u u and we have ( (2.3 S(x, y (u log u u exp log ( + /(p s. We have (2.4 ( log + p s < p s + p y p y p p y ( ( log + p s p s < p y +0.83, p s using s /3. Let f(t = /(t s log t. Note that from [4], [3] (also see [0, Proposition 2.], we have (2.5 θ(t < ( + εt (t > 0, where ε = By partial summation and (2.2, (2.5, we have p s = y f(p log p = θ(yf(y θ(tf (t dt < ( + εyf(y ( + ε p y p y 2 y 2 tf (t dt, using that f (t < 0 for t 2. Integrating by parts, we have y (2.6 < ( + ε2f(2 + ( + ε f(t dt = ( + ε(li(y s Li(2 s + 2 s / log 2, p s p y where Li(t = t 2 Li(2 s = dt/ log t. Note that 2 2 s dt log t < 2 2 dt 2 s (t 2 (t = log(2s + log(3 2 s. 2 Using this in (2.6 and noting that y s = u log u, we have p s < ( + ε(li(u log u log(2s + log(3 2 s + 2 s / log 2. p y Finally, using this in (2.4 and (2.3, noting that Li(u log u < u and log(3 2 s + 2 s / log < log 25, we have the lemma. Remar 2.. We can use some of the techniques in the proof of Lemma 2.0 to help numerically with the estimate in Lemma 2.9. In particular, we have y 0 <p y ( p s < ( + ε Li(y s Li(y0 s + ys 0 log y 0 θ(y 0 ys 0 log y 0.

6 6 HANH MY NGUYEN AND CARL POMERANCE We find that in the ranges we are using Lemma 2.9, it is helpful to tae s = log(e γ u log u/ log y. Let S odd (x, y = n, S even(x, y = n, S even, no 3(x, y = n. n>x, P (n y n odd n>x, P (n y n even n>x, P (n y 3 n, n even In Lemma 2.9, if we now our summand n is odd, as in S odd (x, y, we may remove the factor ( + /(2 s from the product. And if we now our summand is even, as in S even, we may replace the factor ( +/(2 s with /(2 s. In the latter case, if we also now our summand is coprime to 3, as in S even, no 3, we may also remove the factor ( + /(3 s. 3.. Parity and number of primes. 3. AMICABLE NUMBERS OF MODERATE SIZE Proposition 3.. Let A 0 denote the set of amicable numbers n such that either ( n < 0 4, (2 n belongs to a pair of opposite parity, or (3 0 4 < n < e 300 and 4 σ(n. The reciprocal sum of the members of A 0 is < Proof. The amicable numbers to 0 4 have been completely enumerated, and their reciprocal sum is < 0.02, as reported in [2]. If n belongs to an amicable pair of opposite parity, then σ(n is odd. This implies that n is either a square or the double of a square. There are no examples up to 0 4. Further, as is easy to see, (3. n 2 >0 4 n 2 + 2n 2 < n 2 >0 4 If n is even and 2 σ(n, then n = pm, where p (mod 4 and m is either an even square or the double of one. So, the reciprocal sum of such n in (0 4, e 300, when p > 0 4, is at most 2 pm = 3 4 ζ(2(h(e300 H(0 4 < 2.753, 0 4 <p<e 300 m or 2m= using Lemma 2.6. For the case p < 0 4, we use that for x > 0, j 2 < x + x t 2 dt = + x x. We have 4 p<0 4 Similarly, we have m>0 4 /(4p m= 8 p<0 4 j 2 >x pm < 4 ( 4p 4p + p = p π( p<0 4 p<0 4 m>0 4 /(8p m= pm < p<0 4 p + π( We now that π(t < Li(t for t < 0 9, see [4]. Using this we compute that p < p<0 4

7 THE RECIPROCAL SUM OF THE AMICABLE NUMBERS 7 We also now the exact value of π(0 4, it is Adding these estimates to our prior one when p > 0 4 and to (3., we have less than 2.84 for the reciprocal sum of the members of A 0. Remar 3.. In the sequel we will only consider amicable pairs of the same parity. We shall also assume a simple, but useful result of Lee [9] that no amicable number in an even-even pair is divisible by 3. We would lie to extend the third property in Proposition 3. to all even amicable numbers, but this will require some tools, which will be of use later as well. Proposition 3.2. Let A denote the set of amicable numbers n not in A 0 with ω(n > 4 log log n. The sum of reciprocals of those amicable numbers with at least one of the pair > e 00 and at least one of the pair in A is less than Proof. Note that τ 4 (n 4 ω(n, using the notation in 2. For any integer K 0, we have n n < 4 4 log( τ 4 (n n n<e n>e K ω(n>4 log log n K+ e <n<e ω(n>4 log( < 24 K+ (4 + 4 ( 4 log 4, K+ by Lemma 2.5 We can use this inequality to capture the reciprocal sum of those amicable numbers n > e K with ω(n > 4 log log n. We must also sum /n for such numbers n. If n > n, n + n < 2 n. Suppose n < n and ω(n 4 log log n. If n is even, then we may assume that n is even as well, so that n > n/2, and (3.2 n + n < 3 n. Now assume that n, n are odd. Let µ be the product of p/(p over the first 4 log odd primes. Since ω(n 4 log log n < 4 log log n < 4 log, We have n + n = σ(n < µ n, so that (3.3 n + n < µ n. Since µ > 3 for 0, we have in all cases that (3.3 holds. It follows from [4, Theorem5] that if s(n > e 00, then n > e 97. We compute that 24 K µ (4 + 4 < ( 4 log 4 for K = 97. For larger values of, we use some estimates in [4], in particular, (3. and (3.30. From these we deduce that (3.4 µ <.3 log( + 4 log. We compute that log( + 4 log (4 + 4 ( 4 log 4 <

8 8 HANH MY NGUYEN AND CARL POMERANCE This completes the proof Multipliers. We have seen in the proof of Proposition 3.2 that if n, n form an odd amicable pair with n > n and e < n < e, then (3.3 holds, while if n > n form an even amicable pair, then (3.2 holds. Here µ is the product of p/(p as p runs over the first 4 log odd primes, and that 3 < µ <.3 log( + 4 log. We can do better in certain cases. For example, suppose that n > n and h(n 2.5. Then n/n.5 and /n + /n 2.5/n. We shall see shortly that there are very few odd amicables where one of the pair is so abundant, so in moderate ranges we can tae the odd multiplier as 2.5. The multiplier for even amicable numbers can be improved from the 3 in (3.2 when we now that 2 j n, n. It can be taen as (2 j+ /(2 j. Indeed, if n > n, then s(n/n > s(2 j /2 j = 2 j. Thus, n > ( 2 j n, and so /n + /n < ( + ( 2 j /n Proper prime powers. Let L(x = exp( log x /5 and let L = L(e = e /5. We have L(x = L for all x (e, e ]. Proposition 3.3. Let A 2 denote the set of amicable numbers n not in A 0 nor A such that either ( n > e 750, n is even, and n is divisible by a proper prime power > 5L(n, (2 n > e 500, n is odd, s(n/n.5 when n < e 5000, and n is divisible by a proper prime power > 5L(n, (3 n > e 300 and P (n 2 n. The reciprocal sum of those amicable numbers n with n or n in A 2 is < Proof. Let S be the reciprocal sum of all odd proper prime powers, so that S = p a = p(p. p 3 a 2 p 3 We compute that ( < S < By a fairly trivial argument, for B 2 we have, (3.6 p a = p(p + p a >B, a 2 We also have that for x 200, (3.7 e <n<e s S, 2s n gcd(n,3= p> B p a x, a 2 = j 2 p B, p a >B p a < + π( B < 2. B B B π(x /j < x /2. Let S = {p a : p 5, a 2}, S = S (5L, e. We have, by Lemma 2.2, Lemma 2.3, and (3.7, that for any positive integer, n < 3 s + 3e #S < (S 3 s s S s S, s L and e <n<e s S, s n n odd n < 2 + 3e /2 s + 3e #S < (S + 3e /2 2 s s S s S, s L

9 THE RECIPROCAL SUM OF THE AMICABLE NUMBERS 9 Using that even amicable numbers are not divisible by 3 (Remar 3., if e < n < e is an even amicable number divisible by a proper prime power > 5L, then either n coprime to 3 is divisible by a power of 2 that is > 5L or n coprime to 3 is divisible by the double of a member of S. We have =750 e <n<e n amicable n even s S, s n ( n + n =750 e <n<e s S, 2s n gcd(n,3= n < Since S leaves out powers of 2, in the even case we should also be summing 2/(5L. (The factor 2 reflects the multiplier 3 and the fact that n is not divisible by 3. This adds on < summing to infinity. For the remaining even amicables > e we use (3.7 and (3.6 with the above method to find the reciprocal sum is < In total, the contribution to the reciprocal sum in case ( is < For odd amicable numbers, using multiplier 2.5 below e 5000, we have 5000 =500 e <n<e n amicable n odd s S, s n ( n + n =500 e <n<e s S, s n n, odd n < Beyond 5000 we use multiplier.3 log( + 4 log from (3.4 for the odd amicables and find their contribution to e is < Using (3.6 beyond e the contribution is < Finally, since S leaves out powers of 3, we add on the sum from = 500 to 5000 of.25/(5l and the sum beyond = 5000 of (/2.3 log( + 4 log /(5L, which is < In all, the contribution to the reciprocal sum in case (2 is <.306. If n is an amicable number > e 300 and n, n A, then n > e 298. Since ω(n 4 log log n it follows that the largest prime power p a (proper or not that divides n is > n /(4 log log n. If a = then p = P (n and n is not in case (3. If a >, then (3.6 and (3.7 imply that the reciprocal sum in question at most 299 (.3 log( + 4 log 2 e ( /(8 log( + e /2 < For an integer n >, the largest prime power that divides n is at least n /ω(n. If ω(n 4 log log n and n is not divisible by a proper prime power > 2L(n, then for n 20, we have P (n n/4 log log n and P (n 2 n. We apply this to the numbers n, n in an amicable pair with n, n not in A j, j < 3. It follows that we may write n = pm where p = P (n m, and similarly, n = p m where p = P (n m. We now complete the argument for 4 σ(n, showing that this may be assumed for even amicable numbers, since those that do not satisfy this property having a fairly small reciprocal sum. Proposition 3.4. Let A 3 denote the set of amicable numbers n with n, n A j for j < 3, with 4 σ(n. The reciprocal sum of those amicable numbers with at least one of the pair > e 300 and with n, n A 3 is < Proof. We have just seen that we have n = pm, n = p m where p, p are the largest primes in n, n, and they are indeed large. Thus σ(n = σ(n are both even. If they are not divisible by 4, then both

10 0 HANH MY NGUYEN AND CARL POMERANCE m, m are either squares or doubles of squares. It is shown in [3] that m, m uniquely determine n, n. We have mm = nn pp < n /4 log log n n /4 log log n. Suppose that e < n < e. Then n < (µ n, so that (3.8 mm < (µ e 2.5/ log log((µ e = x, say. Let S denote the set of numbers that are either squares or the doubles of squares, with counting function S(x. Then S(x < 2 x for x. The number of pairs m, m in S satisfying (3.8 is at most x < 2 m < (4 + 2 log x x, m<x, m S m <x /m, m S m<x, m S where we have used partial summation for the last estimate. Thus, the number of n is upper-bounded by this last estimate, so the reciprocal sum is at most (4 + 2 log x x e. Summing this expression for 299 we get a contribution of at most Corollary 3.. If n > 0 4 is an amicable number with n, n A j, j 3, then 2 n if and only if 2 n Odd amicables of moderate size. For the rest of this section we have K 50 an integer. Proposition 3.5. We have n<e K n odd, amicable < K , n n<e K n odd, abundant n<e K n odd, amicable h(n or h(n >2.5 n < K Proof. Let h(n = σ(n/n. Then n is abundant if and only if h(n > 2. For any positive integer j we have n < h(n j 2 j n. n<e K n odd Let f j (n be the multiplicative function with f j (p a = h(p a j h(p a j for prime powers p a, so that (3.9 h(n j = d n f j (d. Thus, n<e K n odd, abundant By Corollary 2. with u = 2, we have m e K m odd n < 2 j d<e K d odd f j (d d m<e K /d m odd m. m < 2 K + 2 γ + 2 log e K < 2 K

11 THE RECIPROCAL SUM OF THE AMICABLE NUMBERS using K 50. Thus, d<e K d odd f j (d d m<e K /d m odd m < f j (d (K d d odd and so n<e K n odd, amicable n < 2 n<e K n odd, abundant n < 2 j (K +.28 d odd f j (d d. Note the Euler product (3.0 d odd f j (d d = p>2 ( + f j(p + f j(p 2 p p , which allows us, for any particular value of j, to compute this sum to high accuracy. We find that the optimal value of j is 8, and 2 j d odd f j (d d < This completes the proof of the first assertion. The second assertion follows by exactly the same method, where the factor 2 j is replaced with 2.5 j. The minimum value of , which occurs at j = 44. We shall use K = 500 in the first inequality of Proposition 3.5 and K = 5000 in the second. We have (3. n<e 500 n odd, amicable n + e 500 <n<e 5000 n odd, amicable h(n or h(n >2.5 n < Even amicables of moderate size. We now turn to even amicable numbers < e K, where as before, K 50 is an integer. Proposition 3.6. We have n<e K, 2 n 5 nn n amicable < K n Proof. Using that 6 n from Remar 3., the sum in question is at most 2 n<e K, 2 n gcd(n,5= n abundant n.

12 2 HANH MY NGUYEN AND CARL POMERANCE If 2 n and gcd(n, 5 =, then n = 2l where gcd(l, 30 =. Since h(2 = 4/3, we have h(n > 2 if and only if h(l > 4/3. Thus, for any positive integer j, we have n<e K h(n>2 2 n, gcd(n,5= n = 2 = 2 l<e K /2 h(l>4/3 gcd(l,30= ( 3 4 j l < 2 d<e K /2 gcd(d,30= using (3.9. By Corollary 2., the inner sum here is at most ( 3 j h(l j 4 l l<e K /2 gcd(l,30= f j (d d m<e K /2d gcd(m,30= 4 8 (K log 2 + γ e K /2 < 4 5 K + 0.4, using K 50. Further, using the Euler product in (3.0 starting at p = 7, we find that when j = 35, ( 3 j f j (d < d gcd(d,30= m, Thus, n<e K, 2 n gcd(n,5= n amicable n < 2 This completes the proof. n<e K, 2 n gcd(n,5= n abundant n < 2 ( K < K For the remaining amicables with 2 n we have two remaining (possibly overlapping cases: ( 5 n, n deficient, 5 n, (2 5 n. Note that in case ( we have /n < /n, so the reciprocal sum in case ( is less than the reciprocal sum in case (2. Thus, ( <n<e K 2 n, 5 n or n n amicable n < <n<e K 2 n, gcd(n,5=5 n < 5 K 2.49 for K 50, using Corollary 2. and Remar 3.. When 2 2 n, we can use that 7 n. Indeed, if 7 n, since σ(2 2 = 7, we would have 7 n. But by Corollary 3., this implies that 28 n, n, so that both n, n are abundant, a contradiction. Using too that 3 n from Remar 3., we have that 0 4 <n<e K 2 2 n n amicable n 0 4 <n<e K 2 2 n gcd(n,2= n < 4 K

13 THE RECIPROCAL SUM OF THE AMICABLE NUMBERS 3 If 2 3 n, since 5 σ(2 3 and 20 is abundant, we have that not only 3 n, but 5 n. Thus, n < 30 K <n<e K 2 3 n n amicable We finally consider 2 4 n. We consider two cases: 5 n and 5 n. In the first case, if n/80 is divisible by any of the 59 primes to 277, then h(n > 7/3, and so n cannot belong to an amicable pair with both members divisible by 4. Thus, 0 4 <n<e K 80 n n amicable 0 4 <n<e K 6 n, gcd(n,5= n < K, again using K 50. The remaining even amicables to e K have reciprocal sum at most n < 30 K.074. Adding together all of the contributions in this subsection, we have < K n In particular, taing K = 750, ( <n<e K n even, amicable 0 4 <n<e 750 n even, amicable n < LARGE AMICABLE NUMBERS We consider odd amicable numbers in (e 500, e 5000 with h(n, h(n 2.5, odd amicable numbers > e 5000, and even amicable numbers > e 750. Proposition 4.. Let A 4 denote the set of amicable numbers n such that n, n A j for j < 4 and gcd(n, s(n is divisible by a prime > 3L(n. The reciprocal sum of those even amicable numbers with at least one of the pair > e 750 and at least one of the pair in A 4 plus the reciprocal sum of those odd amicable numbers with at least one of the pair > e 500 and at least one in the pair in A 4 is at most Proof. Let n be an amicable number in the interval (e, e. Let n = min{n, n }. If n is even, then n > n/2, if n < e 5000 is odd, then n > n/.5, and if n > e 5000 is odd, then n > n/(µ. In all cases, if e < n < e, then we have n > n/(µ. Let L = exp(( log(µ /5. If n or n is in A 4, since n = s(n and n = s(n, then gcd(n, n is divisible by a prime q > 3L. Thus, it suffices to sum the reciprocals of such numbers n without the need for a multiplier. Suppose that e < n < e, q gcd(n, n, and q > 3L. Since q σ(n, there is a prime power r a n with q σ(r a. We have r a > 2 σ(ra > 2 q, so that ra > 5.5L > 5L for 750. Thus, since we are assuming that n A 2, we have a = and so r (mod q. In particular, r 2q. It simplifies matters a little if we dispose of the case r = 2q. In this case, n is divisible by q(2q. Using Lemma 2.7, we have that the sum of /(q(2q for q > 3L is less than /(3L log(3l, while the number of integers q(2q < e is at most e /2. It thus follows

14 4 HANH MY NGUYEN AND CARL POMERANCE from Lemma 2.2 and a calculation that the reciprocal sum of such n which are even and > e 750 plus the reciprocal sum for such n which are odd and > e 500 is less than So, we now assume that n is divisible by qr where q > 3L, r (mod q, and r 4q. Using Lemma 2.8, the reciprocal sum of such numbers qr < e is at most q>3l 2 log( log(3l q(q < 2 log( log(3l (3L log(3l, using Lemma 2.7. Summing one-half of this for 750 we get < , using Lemma 2.7, and this contributes to the reciprocal sum of even n A 4. The parallel contribution for odd n > e 500 is < We also must count the number of integers qr < e. We could use Lemma 2.8 again, but it s simpler to not use that r is prime. For a given q, the number of integers r with q < r < e /q and r (mod q is at most e /q 2. Using Lemma 2.2 and summing e times this estimate for 750 (using Lemma 2.7 adds on < to the reciprocal sum for even, and the parallel contribution for odd n > e 500 is < Now, totalling the various contributions, we have that the sum in the proposition is at most Proposition 4.2. Let A 5 denote the set of amicable numbers n such that n, n are not in A j for j < 5 and mm n/(0l(n. Then the reciprocal sum of those amicable numbers such that at least one of the pair is > e 500 in the odd case and > e 750 in the even case, and at least one of the pair is in A 5 is at most Proof. By Proposition 3.4, we may assume that we re in one of the 3 cases m, m odd, m m 2 (mod 4, m m 0 (mod 4. As in the proof of Proposition 3.4, the pair m, m determines the pair n, n. Suppose we are in the odd-odd case. We distinguish two ranges for n: e 500 < n < e 5000 and n > e In the first range we have multiplier 2.5, since by (3. we are assuming that h(n, h(n 2.5. In the second range, we have multiplier µ, where = log n. Say n, n are an amicable pair and n/(0l(n > mm. If n > n, then n /(0L(n > mm. Suppose that n < n. Then n > n/.5 in the first range, so if.5n/(0l(n > mm, then n /(0L(n > mm. In the second range, if n < n, we have n > n/(µ, so, if (µ n/(0l(n > mm, then n /(0L(n > mm. For n or n > e 500, p = P (n > n /(4 log log n > So, if n is abundant, then h(m = p p + h(n > 2p p + > Also note that if n > e 500, then n > e 999 and if n > e 5000, then n > e Let ν be the appropriate multiplier, so that ν = 2.5 in the small odd range and ν =.3 log( + 4 log for large odd cases. Let N 0 (t be the number of odd amicable numbers n t with mm < (ν n/(0l(n. By partial summation, (4. n or n A 5 nn odd n or n >e 500 ( n + n 500 ( N0 (e e N 0(e e N 0 (t N 0 (t e + e t 2 dt e 499 t 2 dt.

15 THE RECIPROCAL SUM OF THE AMICABLE NUMBERS 5 Let t = (ν t/(0l(t. If {m, m } = {m, m 2 } where h(m < h(m 2, then N 0 (t 2 t. m 2 m 2 <t, m 2 odd h(m 2 > m 2 <t m 2 odd h(m 2 > <m t /m 2 m odd m 2 <t m 2 odd h(m 2 > (Note that m, since all amicable numbers are composite. We now follow the argument in the proof of Proposition 3.5. We have for any positive integer j that < ( j h(m 2 j m 2 Taing j = 8, we get n or n A 5 nn odd n or n >e 500 m 2 <t, m 2 odd h(m 2 > m 2 <t, m 2 odd m 2 < 2 (log(t +.28( j d odd m 2 < 2 (log t , f j (d d. so that N 0 (t < (t + (log t Let ν = ν = 2.5 when 5000 and ν = µ when > We conclude from (4. that ( n + n < e 499 t 2 (t + (log t +.28 dt < = e e (ν t t 2 (log t + log(ν log(0l +.29 dt 0L (ν ( /2 + log(ν log(0l L < We now turn to the 2 (mod 4 case, which has multiplier ν = 3. First suppose that 5 nn. By Remar 3. we have 3 mm. Let N (t denote the number of amicable numbers n t with n 2 (mod 4, 3 mm, 5 mm, and mm < 2n/(0L(n. As in the odd-odd case, we wish to upper bound e N 749 (t/t 2 dt. Say {m, m } = {m, m 2 } where h(m < h(m 2. Similarly as in the odd-odd case, since n, n > e 749, we have h(m 2 > Let t = 2t/(0L(t = t/(5l(t and let N,0 (t be the contribution to N (t when m 2 t /00 and N, (t be the contribution when m 2 > t /00. Note that N,0 (t 2 ( t m 2 m 2 t /00, 2 m 2 gcd(m 2,5= h(m 2 >2 0 4 m t /m 2, 2 m gcd(m,5= 2(.04 5 m 2 t /00, 2 m 2 gcd(m 2,5= h(m 2 >2 0 4 t m 2 t /00, 2 m 2 gcd(m 2,5= h(m 2 >2 0 4 m 2.

16 6 HANH MY NGUYEN AND CARL POMERANCE For any positive integer j, the inner sum is < (2 0 4 j h(m 2 j < 2 m 2 t /00, 2 m 2 gcd(m 2,5= m 2 = 2( 3 2 ( j( 4 5 (log(t /200 + γ Taing j = 35, this last expression is Thus, j (2 0 4 j gcd(d,30= m t /200 gcd(m,30= f j (d d. < 2 ( (log t < 2 5 ( (log t N,0 (t < ( t (log t < t log t t For N, (t we have N, (t m 00, 2 m m 2 t /m, 2 m 2 gcd(m,5= gcd(m 2,5= h(m 2 >2 0 4 The inner sum is ( 3 < j Taing j = 35 again, we have m t /2m gcd(m,30= N, (t < t 2 ( m 00, 2 m gcd(m,5= With the prior estimate for N,0 (t, we have m 00, 2 m m t /2m gcd(m,5= gcd(m,30= h(m>(2/3(2 0 4 h(m j t 2m ( j gcd(d,30= f j (d d. h(m j m < t 2 ( (0.825 < t. N (t < t log t t. As in the odd-odd case, we deduce that the contribution when 2 m, m and 5 mm is < ( /2 log(5l < L 750 We now bound the contribution when 2 m, m and 5 mm. If N 2 (t denotes the number of pairs, we have N 2 (t + m t /2 m 2 t /4m gcd(m,30=5 gcd(m 2,6= m t /2 gcd(m,30=5 3 ( t m m t /2 m 2 t /4m gcd(m,30= gcd(m 2,30=5 m t /2 gcd(m,30= < 60 t ( 3 (log( t /0 + γ ( t m + 60 t ( 4 5 (log( t /2 + γ m + t.

17 Thus, for t > e 999, THE RECIPROCAL SUM OF THE AMICABLE NUMBERS 7 N 2 (t < 200 t log t t. As before, we have the contribution to our sum being < 0.005( /2 log(5l < L 750 We now consider the case when m, m are both multiples of 4. We divide this into a few subcases: ( v 2 (m = 2, v 2 (m = 2, (2 {v 2 (m, v 2 (m } = {2, 3}, (3 {v 2 (m, v 2 (m } = {2, 4}, (4 v 2 (m = 2, v 2 (m 5 or v 2 (m 5, v 2 (m = 2, (5 v 2 (m 3, v 2 (m 3, In all of these cases we have 3 mm. In cases (-(4, since 7 σ(n = σ(n, we have 7 mm. Similarly, in case (2, we have 5 mm since 5 σ(n = σ(n. We also have 5 mm in cases (4 and (5 since s(20 20 s(32 s(4 >, 32 4 s(60 s(40 >, s(8 >. 8 In cases (-(4, we have multiplier 7/3 and in case (5, we have multiplier 5/7. All cases are symmetric in m, m, so we may assume that m t. Using the same method as above, we find that ( n + n < n or n A 5 n,n 0 (mod 4 n or n >e 750 Totalling the contributions in the various cases completes the proof. Proposition 4.3. Let A 6 denote the set of amicable numbers n such that n, n are not in any A j for j < 6 and p > n 3/4 L(n. The reciprocal sum of those even amicable numbers with at least one of the pair in A 6 and at least one > e 750 plus the corresponding reciprocal sum of odd amicable pairs with at least one of the pair > e 500 is < Proof. Assume that t > e 750 and let N(t denote the number of n A 6 with n t. For n A 6, we have m < n /4 /L(n, so since n A 5, we have m > 0 n3/4. This then implies that p < 0n /n 3/4. Let ν be less than the appropriate multiplier, so that ν =.5 in the smaller odd case, ν = 2 in the even case, and ν =.3 log( + 4 log in the larger odd case. In particular, n < νn, so we have p < 0νn /4. Write n = q q 2... q l, where the q i s are pairwise coprime prime powers (possibly first powers of primes and q > q 2 > > q l. We have q = p, so all of the q i s are < 0νn /4 0νt /4. Assume that n > t 0.84, and choose i minimally so that D := q q 2... q i > t/l(t. Then D < 0νt 3/4 / L(t. If D is divisible by a prime < 3L(t, then in fact D is smaller, it is < 3L(t t/l(t < t 0.5. Further, (3L(t 4 log log t < t Thus, if n > t 0.84 and n is counted by N(t, then the fact that n is not in A nor A 2 implies that all of the prime factors of D are greater than 3L(n. Since n A 4, we have gcd(d, σ(d =. Write n = DM. It is shown in [3] that σ(mdm mσ(m (mod σ(d.

18 8 HANH MY NGUYEN AND CARL POMERANCE Thus, N(t is at most t 0.84 plus the number of solutions M to these congruences with M < νt/d, as m runs to t /4 /L(t and D runs over the interval ( t/l(t, 0νt 3/4 / L(t. For a given choice of m, D, the number of solutions is at most + using gcd(d, σ(d =. We have νt/d νtσ(m + σ(d/ gcd(σ(md, σ(d D 2, m<t /4 /L(t D<0νt 3/4 / L(t < 5νt/L(t 3/2 +, both in the case m even and in the case m odd. Further, using the inequality m<b σ(m < B2, νt n or n A 6 e <n<e m<t /4 /L(t D> t/l(t σ(m D 2 < νt 3/2 L(t 2 D> t/l(t where we also used that D>B D 2 < /B + /B 2. We have ( n + e n < (ν + N(t e t 2 dt < D 2 < νt/l(t 3/2 + νt /2 /L(t, e ν + 6(ν + ν + e t.6 L 3/2 t + ν + t 2 < e (ν + ν/l 3/2 + (ν + /( 2. For evens starting at = 750, we have ν = 2, and the contribution is < For odds from = 500 to 5000, we have ν =.5 and the contribution is < 0.058, and the contribution for odds with > 5000 is < In all, the total contribuion is < Proposition 4.4. Let A 7 denote the set of amicable numbers n such that neither n nor n is in A j for j < 7, and such that P (σ(m 00L(n. Then the reciprocal sum of the amicable numbers n with either n or n > e 750 in the even case and > e 500 in the odd case, and either n or n A 7 is at most.399. Proof. Let M = e ( /4 /L. Since n A 6, if n (e, e, then m > M. Let u = /4 and let q = P (m. We consider three cases: ( q 0 7 M /u and m < e /2, (2 q 0 7 M /u and m > e /2, (3 q > 0 7 M /u and P (q + 00L. If n is not in any of these cases, then q > 0 7 M /u > 5L, so from n A 2, we have q m. Also P (σ(m P (q + > 00L, so that n A 7, so it suffices to bound the reciprocal sums for the three cases above. For a given value of and e < n < e, the reciprocal sum in case ( is at most m p. M <m<e /2 P (m 0 7 M /u e /m<p<e /m dt

19 THE RECIPROCAL SUM OF THE AMICABLE NUMBERS 9 Since e /m > e /2, Lemma 2.6 implies that the inner sum over p is smaller than 2/( 2 + 2/( 2 2. Thus, the reciprocal sum in case ( in the odd and even cases, respectively, is at most ( ( 2 2 S odd (M, 0 7 M /u, ( ( 2 2 S even, no 3 (M, 0 7 M /u, using the notation of Remar 2.. Summing the first expression using Lemma 2.9 and Remar 2. with y 0 = e 0 for and using multiplier 2.5, we get an estimate of < Summing the second expression for with multiplier 3, we get an estimate of < Summing for > 5000 and using a multiplier of.3 log( + 4 log, using Lemma 2.0, we get < The second case is done in almost the same way. Now we must estimate m>e /2 P (m 0 7 M /u m e /m<p<e /m We now that p > n /4 log log n > e ( /(4 log(, so the inner sum here is 0 unless m is such that e /m e ( /4 log(. With a( := ( /(4 log(, Lemma 2.6 then implies the inner sum above is at most /a( + /(2a( 2. Thus, the reciprocal sum in case ( in the odd and even cases, respectively, is at most ( a( + 2a( 2 S odd (e /2, 0 7 M /u, p. ( a( + 2a( 2 S even, no 3 (e /2, 0 7 M /u. Summing the first expression using Lemma 2.9 and Remar 2. with y 0 = e 0 for and using multiplier 2.5, we get an estimate of < Summing the second expression for with multiplier 3, we get an estimate of < Summing for > 5000 and using a multiplier of.3 log( + 4 log, using Lemma 2.0, we get < We now turn to case (3. Write l = n/q. Here the reciprocal sum for e < n < e is at most q>0 7 M /u P (q+ 00L q e /q<l<e /q where l is odd in the odd case, and in the even case, l is even and not divisibly by 3. Using Corollary 2. for the inner sum, we have a quantity at most ( M /u e q>0 7 M /u P (q+ 00L l, ( q < M /u 0 7 M /u + e 0 7 M /u S even (0 7 M /u, 00L in the odd case, with the same estimate but with 3 in place of 2 in the even case. Here we have relaxed the condition that q is prime, eeping only that it is odd, so that q + is even. Summing this using Lemma 2.9 from = 750 to = 5000, using x = 0 7 M /u, y = 00L, s = log(2u log u/ log y, and multiplier 3, we get < in the even case. For the odd case we sum from = 500 to 5000 using multiplier 2.5, getting an estimate of < We sum for 500 using Lemma 2.0 and multiplier.3 log( + 4 log getting < Thus, the total contribution to the reciprocal sum from A 7 is <.3988.

20 20 HANH MY NGUYEN AND CARL POMERANCE 5. CONCLUSION We are now faced with summing the reciprocals of those amicable numbers n such that both n, n are > e 750 in the even case, and > e 500 in the odd case, and neither is in any set A j. As before, we have n = pm, n = p m, where p = P (n m, p = P (n m, and p p. We shall assume that p > p and sum /n, using an appropriate multiplier to tae into account the numbers /n. Let r = P (σ(m, so since n A 7, we have r > 00L(n. Since r σ(m σ(n = σ(n, there are prime powers q α m, q α n with r σ(q α and r σ(q α. Then q α, q α > 2r > 50L(n, so since n, n A 2, we have α = α =. In particular, q q (mod r. Since q > r > 00L(n and since n A 4, we have q n. Since q n = s(n = ps(m + σ(m, we have ps(m + σ(m 0 (mod q. This implies that if q σ(m, then q s(m, which implies that q m, a contradiction. So, we have q σ(m and the above congruence places p in a residue class a(m, q (mod q for a given choice of m and q. Also note that p > p q. Write m = qm. For a given value of 750, we have S := n in this case e <n<e n < r>00l q<e /2 q (mod r q m q <e + m <e /q q (mod r e /qm <p<e /qm p a(qm,q (mod q p>q We begin with the inner sum. Fix q, m, q and let a be in the residue class a(qm, q (mod q with 0 < a < q. First suppose that q is large. If q > e /qm, then the sum on p is 0. (In particular, we may assume that q < e /q. Suppose that q > e 2 /qm. Using only that q is odd, that p is an odd number in the interval (e /qm, e /qm, and that p a (mod q with p > q, we have that the sum on p is at most /q < qm /e 2. Let w = e /qq and assume that q e 2 /qm ; that is, m w/e. Let z = e /qm. By Lemma 2.8, we have that z<p<ez p a (mod q p>q m <ew p < 2 (q log(z/q + 2 ( + log(z/q q log log(z/q < 4 (q log(z/q = 4 (q log(w/m. We now sum on m. Since q < ez = e /qm, we have m < e /qq = ew, so that we have qm m e m (q log(w/m. m w/e We distinguish the even and odd cases. Using Lemma 2.4 and w = e /qq, we have the sum on m is < e2 2q + 2 e2 q log (odd case, < 3q + 4/3 q log (even case. What we have at this point is S < c r>00l q<e /2 q (mod r q q <e /q q (mod r ( 4 e2 q log + q, p.

21 THE RECIPROCAL SUM OF THE AMICABLE NUMBERS 2 where c = /2 in the odd case and c = /3 in the even case. Let ι = /(00L. By Lemma 2.8, using the fact that the least prime in the residue class (mod r is at least 2r and log log((2r /r < 0.37, the sum on q is at most ( + ι 2 2(4 log + e2 (log r Similarly, the sum on q is at most 2(log(/ ( + ι, r so we are left with S < c( + ι 3 4(4 log + e 2 (log (log(/ r 2. r>00l We use Lemma 2.7 for the sum over r. In the odd case we sum our bound for S from = 500 to 5000 with multiplier 2.5, getting <.525. The remainder of the odds, using multiplier.3 log( + 4 log adds on < For the even case, using multiplier 3 and summing for 750, we get < In total, the contribution is < ACKNOWLEDGMENT We are grateful to Paul Kinlaw for pointing out to us that our original version of Lemma 2.7 could be improved. REFERENCES [] J. Bayless and P. Kinlaw, Consecutive coincidences of Euler s function, Int. J. Number Theory 2 (206, [2] J. Bayless and D. Klyve, On the sum of reciprocals of amicable numbers, Integers A (20, Article 5. [3] J. Büthe, Estimating π(x and related functions under partial RH assumptions, Math. Comp. 85 (206, [4], An analytic method for bounding ψ(x, Math. Comp., to appear, Also see arxiv.org [math.NT]. [5] P. Dusart, Explicit estimates of some functions over primes, Ramanujan J. (206, doi:0.007/s [6] P. Erdős, On amicable numbers, Publ. Math. Debrecen 4 (955, 08. [7] H. M. Nguyen, The reciprocal sum of the amicable numbers, Senior honors thesis, Mathematics, Dartmouth College, 204. [8] S. H. Kim and C. Pomerance, The probability that a random probable prime is composite, Math. Comp. 53 (989, [9] E. Lee, On divisibility by nine of the sums of even amicable pairs, Math. Comp. 23 (969, [0] J. D. Lichtman and C. Pomerance, Explicit estimates for the distribution of numbers free of large prime factors, submitted for publication, see arxiv: [math.nt]. [] H. L. Montgomery and R. C. Vaughan, The large sieve, Mathematia 20 (973, [2] C. Pomerance, On the distribution of amicable numbers, II, J. reine angew. Math. 325 (98, [3] C. Pomerance, On amicable numbers, in Analytic number theory (in honor of Helmut Maier s 60th birthday, M. Rassias and C. Pomerance, eds., Springer, Cham, Switzerland, 205, pp [4] J. B. Rosser and L. Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois J. Math. 6 (962, HMN: DARTMOUTH COLLEGE, CLASS OF 204 address: hanhmn9@gmail.com CP: MATHEMATICS DEPARTMENT, DARTMOUTH COLLEGE, HANOVER, NH 03755, USA address: carl.pomerance@dartmouth.edu