Dartmouth College. The Reciprocal Sum of the Amicable Numbers

Transcription

1 Dartmouth College Hanover, New Hampshire The Reciprocal Sum of the Amicable Numbers A thesis submitted in partial fulfillment of the requirements for the degree Bachelor of Arts in Mathematics by Hanh My Nguyen Advisor: Carl Pomerance 204

2 Abstract Two different positive integers a and b are said to form an amicable pair if sa = b and sb = a, where sn denotes the sum of proper divisors of n In 98, Pomerance proved that the count of amicable numbers up to x is less than x/ exp log x /3 for x large It follows immediately that the sum P of the reciprocals of numbers belonging to an amicable pair is a constant In 20, Bayless & Klyve showed that P < 656,000,000 In this paper, we improve this upper bound by proving that P < 4084, based on recent work by Pomerance that shows a stricter bound for the count of amicable numbers plus some other ideas that are new to this thesis i

3 Acknowledgements First and foremost, I would like to thank my thesis advisor, Prof Carl Pomerance, for his continued patience and guidance throughout this project I also want to express my deepest gratitude to my professors at the Department of Mathematics for their support and mentorship during the past four years ii

4 Contents Introduction 2 Framework Description 2 2 Small amicable numbers 2 22 Large amicable numbers 4 3 Lemmas 6 4 Main argument 4 For n A k, ωn 4 log k 42 For n A k, the largest squarefull divisor of n is at most / For n A k, the largest squarefree divisor d of n with P d has d e k/ For n A k, P gcdn, sn / For n A k, mm > ek 5 46 For n A k, p e 3k/ For n A k, q For n A k, P σm 9 49 Remaining amicable numbers 23

5 Introduction Let σ denote the sum-of-divisors function, and let sn = σn n denote the sum-of-properdivisors function Two different positive integers n and n are said to form an amicable pair if and only if sn = n and sn = n We say n is amicable if it belongs to an amicable pair The smallest pair of amicable numbers, 220 and 284, was first known to Pythagoras in the 5 th century BC, and was ascribed with many mystical properties In the 9 th century, Thâbit ibn Qurra found a formula for amicable pairs, which states that if p = 3 2 n, q = 3 2 n, r = 9 2 2n are primes, then 2 n pq, 2 n r form an amicable pair For example, n = 2 yields the primes p =, q = 5, r = 7, which generates the pair 220, 284 It was not until 636 that a second pair of amicable numbers 7296, 846, which corresponds to n = 4, was found by Fermat Later, Descartes gave a third pair of amicable numbers , , which corresponds to n = 7 In the 8 th century, Euler generalized Thâbit s rule: if p = 2 n m + 2 n, q = 2 n m + 2 n, r = 2 n m n are primes, then 2 n pq, 2 n r form an amicable pair He also compiled a list of 64 amicable pairs, of which two were later shown to be not amicable Interestingly, Euler completely overlooked the second smallest amicable pair 84, 20, which was discovered by Paganini, a 6-year-old Italian, in 866 We now know exhaustively all amicable pairs with the smallest member up to 0 4 and nearly 2 million other pairs [6] However, the infinitude of the set of amicable numbers has yet to be proved Let A denote the set of amicable numbers, and let Ax = A [, x] In 954, Kanold initiated the quest to explore the asymptotic density of A, showing that #Ax < 0204x for x sufficiently large [4] In 955, Erdős was the first to prove that the set of amicable numbers has zero asymptotic density, showing that #Ax = ox [2] From 973 to 98, Rieger, Erdős, and Pomerance improved upon this result, proving that #Ax is less than x/log log log log x /2, O x/ log log log x, x/ exp log log log x /2, and x/ exp log x /3 for x sufficiently large [0][3][7][8] Most recently in 204, Pomerance showed that #Ax x/ explog x /2 as x [9] It follows immediately from the last two results that the reciprocal sum of the amicable numbers is finite In 20, Bayless & Klyve showed a numerical bound for the reciprocal sum of amicable numbers, which they denote by P : P < 656,000,000 In this paper, we improve Bayless & Klyve s result in [] by showing a smaller upper bound for the reciprocal sum of amicable numbers Theorem We have P = n A n < 4084 We will introduce some standard notations and describe the framework in part 2, prove some useful lemmas in part 3, and give detailed proof and numerical estimates in part 4

6 2 Framework Description Let n A be an amicable number and n = sn denote the corresponding friend For some K > log 0 4, we can write P = n A n = n A min{n,n }0 4 n + n A 0 4 <min{n,n }expk n + n A min{n,n }>expk From the exhaustive list of amicable pairs with the smaller member less than 0 4, we can compute P s = n = n A min{n,n }0 4 The list of amicable numbers is obtained from [6], and the reciprocal sum is computed in MatLab with standard machine precision This result differs from the lower bound shown in [] by approximately 0 9 An amicable pair n, n can fall into one of the following cases: n, n are both odd, n, n are both even, or n, n are of different parity If n and n are of different parity, then σn = σn = n + n is odd Since σp a = + p + p p a is odd if and only if p = 2 or 2 a, the odd member is a square, and the even member is either a square or twice one No odd-even amicable pair has been found, so the reciprocal sum of such amicable numbers, if they exist at all, is small and bounded by ε < 3 2 a>0 4 a 2 < From this point forward, we consider only amicable pairs whose members are either both odd or both even 2 Small amicable numbers For amicable pairs n, n such that 0 4 < min{n, n } expk, we will grossly overestimate their contributions to P since the asymptotic behavior outlined by the proof has yet to kick in In particular, we will consider even amicable pairs and the odd ones separately If n and n are both even, then n n = d > 2, d n d> and similarly n/n > /2, so 0 4 < n, n 2 expk If n and n are odd, then min{n, n } is an odd abundant number Recall that an integer is abundant if the sum of its proper divisors is greater than the number itself n 2

7 We have P K = n A 0 4 <min{n,n }expk /2<nexpK n = + 2 n n A n even 0 4 <n2 expk n + n odd abundant 0 4 <nexpk n A n odd 0 4 <min{n,n }expk n, n where and 0 4 /2<nexpK expk n < dt 0 4 /2 t = K log04 /2, n odd abundant 0 4 <nexpk is smaller Let s find out how small it is Let hn = σn/n, then h is a multiplicative function Let f j be the multiplicative function defined as follow { f = f j p a = hp a j hp a j For x 3, we have so that Here, and nx n odd mx m odd hn j n = n nx n odd x/2 m + k= dx d odd f j d d d n nx n odd hn j = nx n odd f j d = dx d odd f j d n f j d, d n mx/d m odd md dx d odd 2k + < + x/2 2 k < + + logx/2 < 2 k= d odd f j d d = is finite, as we shall see in Lemma 37 p odd f j d d mx m odd m log x , 2 + f jp + f jp 2 p p 2 + = π j, say, 3

8 Since hn > 2 for all n abundant, we have nx n odd abundant n 2 j nx n odd hn j n π jlog x j+ After investigating j [, 40], we pick j = 8, which yields π j < 62387, so that Thus, we have nx n odd abundant 22 Large amicable numbers 0089 log x n P K K For the remaining pairs, ie, n, n such that min{n, n } > expk, we will consider them in small intervals and build an ordered list of properties Let A k = {n A : e k < n < e k }, and let ωn denote the number of distinct primes dividing n Let µ k = Property : For n A k, ωn 4 log k 4 log k i=2 p i p i, where p i is the i th prime We will show that for n A k such that n and n have property, n/µ k n nµ k Let = exp k/5 Recall that a positive integer s is squarefull if and only if for every prime p dividing s, p 2 also divides s Also recall that a positive integer is squarefree if and only if it is not divisible by any perfect square Let P denote the largest-prime-divisor function Property 2: For n A k, the largest squarefull divisor of n is at most /4 Property 3: For n A k, the largest squarefree divisor d of n with P d satisfies d e k/3 Property 4: For n A k, P gcdn, sn /2 Let n = pm, n = p m where p and p are the largest prime factors of n and n Note that properties 2, 3, and 4 imply that p p, p m, and p m Property 5: For n A k, mm ek Property 6: For n A k, p e 3k/4 Write m = m 0 m where m 0 is the largest divisor of m such that m 0 is either a squarefull number or twice one, then m is odd, squarefree, and coprime to m 0 Let q = P m, and write q + = q 0 q where q 0 is the largest divisor of q + such that q 0 is either a squarefull number or twice one 4

9 Property 7: For n A k, q 0 is at most 6 Property 8: For n A k, P σm For each property i, let P i = n A min{n,n }>expk n,n pass i- n or n fails i n Then, P i k=k+ n A min{n,n }>expk n,n pass 8 n,n pass i- n fails i k=k+ n + n = P n>p n n,n passes 8 k=k+ s i k, say Finally, we are left with n A k where both n and n have properties -8 We proceed to estimate P = n = n + n = s k, say, so that n A min{n,n }>expk n P + 8 P i i= k=k+ To optimize the sum of the expression above and P K, we choose K = 5935 In this paper, we will repeatedly use ɛ, ι, and c to denote different constants In all cases, the explicit formula for these constants are clear from the context and are used in computation The naming of constants follows the convention that ɛ 0 and ι as k The letter p, q, and r are reserved for prime variables We use ζ and ϕ to denote the Riemann zeta function and the Euler phi function, respectively We also use πx; d, a to denote the number of primes p congruent to a modulo d with p x 5

10 3 Lemmas Lemma 3 For all k, d, v Z +, z > 0, { v min + z, 3 } 2 zvez Proof If z, then If < z 2, then If z > 2, then zvez z<v<ez zvez v v <v2e ve v = 3 2 v = 28 3 < 3 2 v ez z + z v dv = + z Lemma 32 For all x > 286, we have Hx := px p log log x + B + 2 log 2 x 3 and H x := px p log log x + B + 2 log 2 + D, 32 x where B = and D = p /pp = Proof The inequality 3 is from [], and 32 follows immediately Lemma 33 : For all z > 286, z<pez p log z + log 2 z Proof From equations 37 and 38 of [], we have z<pez p = p p log + log z + B log z 2 log log z B + 2 log 2 z pez pz = log + + log z 2 + log z log 2 z log z + log 2 z 6

11 Lemma 34 : For all w Z +, x 0 such that Hx < w +, we have ωd w P dx d squarefree d Hxw w! Proof By the multinomial theorem, we have ωd w P dx d squarefree d j=w j! = Hxw w! Hxw w! = Hxw w! /p px j + Hx w + w + Hx w + + Hx 2 + Hx w + + w + w + Hx w + w Hx2 w Lemma 35 For all d Z +, d, x 3, p mod d d<px p 2log log x log log2 /d + / logx/d ϕd Proof From the Brun-Titchmarsh theorem [5], we have Using partial summation, we have p mod d d<px πx; d, p 2 ϕd logx/d + 2 ϕd 2 ϕd = 2 ϕd 2x ϕd logx/d x 2d + log logt/d logx/d dt t logt/d x 2d + log logx/d log log2 /d logx/d 7

12 Lemma 36 For x, S 0 x := S x := ζ3/2 x, sx s squarefull s x s squarefull s 3 x For x 2, S x := s x s squarefull log log s s 3 log log x + 3 log 2 x log log x + log 2 + 2ζ3/2 + x 2x log x Proof Since every squarefull number can be written as a product of a square and a cube, S 0 x a 2 b 3 x = bx /3 ax/b 3 /2 bx /3 x b 3 /2 = x bx /3 b 3/2 ζ3/2 x We can verify S x 3/ x for x 4000 using S = ζ2ζ3/ζ6 For x > 4000, we have S x a 2 b 3 x bx /3 bx /3 a 2 b 3 b 3 b 3 x 2/3 + x /2 ζ3/2 x /2 3 x a x/b 3 /2 a 2 + b x /3 b 3 a 2 b 3 b 3 /2 x + + ζ2 x bx /3 + + ζ2/2 x 2/3 a b x /3 b 3 + ζ2 b3/2 x + 2x 2/3 + ζ2 x By partial summation and our inequality for S 0, we also have S x = lim z = lim = z x xsz s squarefull log log s s S0 z S 0 x log log z z log log t S 0 t t 2 t 2 dt log t 8 z x log log t S 0 t S 0 xd t

13 ζ3/2 = ζ3/2 ζ3/2 = ζ3/2 x x x log log t t 3/2 t 3/2 log t log log t t 3/2 log log t t 3/2 2 log log t t log log x = 2ζ3/2 + x Now, we can improve this estimate: S x = xs<x 2 s squarefull log log x 2 log log s s s x s squarefull 3 log log x + 3 log 2 x dt 2 t 3/2 log t + 2 t 3/2 log t + 2 t log x x x log x + s x 2 s squarefull s + 2ζ3/2 t 3/2 dt log t dt t 3/2 log x log log s s log log x + log 2 + x 2x log x log log x + log 2 + 2ζ3/2 + x 2x log x Lemma 37 Let f be the multiplicative function defined as in 2 For all j Z +, π j = + f jp + f jp 2 p p 2 + is a finite number p odd Proof First, we have f j p p = hpj p = + p j p = O p 2 Because 0 < f j p a < hp a j, for all a 2, f j p a p a hpa j p a σpa j p a+ja p j p p a Therefore, for each fixed j Z + + f jp p + f jp 2 p 2 + = + O p 2, 9

14 so that log π j = O p 2 p odd = O In order to estimate an upper bound for π j, we choose large integers A, B, and compute pb + f jp + f jp 2 p p pb A a= f j p a p j+ p a + p p A+, and p>b + f jp p j p p p>b exp B p j + p pp + x j x j + x x xx dx Table 3 shows upper bounds of π j for j [, 40], computed with A = 500 and B = 0 6 We pick j = 8 to get the smallest coefficient in section 2 Table 3: Estimates of upper bound of π j j π j π j /2 j j π j π j /2 j

15 4 Main argument 4 For n A k, ωn 4 log k Let u = 4 log k ; as in Lemma 34, we have ωn u n j=u j! qe k q a a= j j! q qe k By equations 34 and 342 in [], we have that for n, n > expk, j=u n n < σn = σn n n n ϕn < eγ + ɛ log log n, j H e k u u + u! u + H e k and similarly, n/n < e γ + ɛ log log n, where e γ = 7807 and ɛ 00 We have which implies n n < e γ + ɛ log log n < e γ + ɛ log logn + log loge γ + ɛ log log n n < eγ + ɛ log log n + eγ + ɛ log loge γ + ɛ log log n 2 log log n < < 2 log k n n n n, for n > expk Therefore, we have where We can compute s k k=k+ P k=k+ + 2 log k 0 6 k=k+ s k, H e k u u + u! u + H e k s k For k > 0 6, H e k log k by Lemma 32, and u + u + H e k = H e k u+ log k log k By Stirling s inequality, u! e u 2πu u

16 Therefore, E := k=0 6 + s k k=0 6 + k= log k elog k πu u + 2 log k elog k π log k 4 log k u 4 log k because u 4 log k and elog k log k For all k > 0 6, elog k log k = e 4 + e log k = h, so we have E π log 0 6 = k=0 6 + k= log kh 4 log k + 2 log kk 4 log h + 2 log t 0 6 t dt Thus, we have P = Amicable pair multiplier For n A k, k > K satisfying property, we will show that where Proof If n, n are even, we have n µ k n nµ k, µ k = 4 log k i=2 p i p i n n = d > 2 d n d> Similarly, n/n > /2 The inequality follows immediately since µ k > 2 If n and n are odd, the inequality on the right is straightforward: n n = sn n = σn n = d n d < p n 4 log k + p p i=2 p i p i = µ k 2

17 If n > n, then clearly n > n/µ k If n n, then n A k where k k, and it follows that n n sn n = σn n = 4 log k p + p p n i= 4 log k + p i p i < i=2 p i p i = µ k 42 For n A k, the largest squarefull divisor of n is at most /4 Write n = sv where s is the largest squarefull divisor of n Assume that s > /4; then by Lemma 3 and Lemma 36, n + n +µ k n = +µ k s v 3 + µ ks /4 2 s> /4 Therefore, s> /4 P 2 By Lemma 36, we can compute k=k+ s> /4 s squarefull 3 + µ k S /4 2 e k <v< ek s s 0 6 k=k+ 3 + µ k S / For k > 0 6, Thus, we have E 2 := 9 k= µ k S / log t t=0 6 e t/0 9 k=0 6 + dt P log k e k/0 43 For n A k, the largest squarefree divisor d of n with P d has d e k/3 For n A k, write n = dv where d is the largest squarefree divisor such that P d Assume that d > e k/3, we have d>e k/3 n d>e k/3 P d< d squarefree d e k <v< ek d d v 3 2 H u u! 3 u + u + H,

18 where u = log e k/3 k log = 3 log Therefore, We can compute 0 6 k=k+ P 3 For k > 0 6, H log k k=k+ 3 + µ k H u u + 2u!u + H µ k H u u + 2u!u + H + B + 5 2k < log k, and u = = k 3 log > k 3 log > k, so u + u + H = H u+ log k < 2 k+ We can estimate E 3 := 3 Thus, we have k= µ k H u u + 2u!u + H log t 3 k= log k e log t t t dt P For n A k, P gcdn, sn /2 e log k k k Let r = P gcdn, sn, and suppose r > /2 Because r σn, there exists q a n, such that r σq a Then /2 < r σq a < 2q a, so q a > /4 It follows from property 2 that a =, implying that q mod r Because r is a prime larger than /2, q > r We can write n = rqv, so that r> /2 n r> /2 r r<qe k q mod r by Lemma 3 From lemma 35, we have r<qe k q mod r q e k qr <v< ek qr v 3 2 r> /2 q 2log k log log2 /r + / logek /r ϕr where c = 2/k log log2 2/ because /2 < r < qr n < e k/2 r r<qe k q mod r q, 2log k + c, r 4

19 We have Therefore, We can compute r> /2 n 0 6 k=k+ 3log k + c P 4 k=k+ r> /2 6 + µ k log k + c 2 For k > 0 6, 2/k log log2 2/ <, so that Thus, we have E 4 6 3log k + c rr /2 6 + µ k log k + c log tlog t + t=0 6 e t/5 2 P = dt For n A k, mm > ek In [9], it is shown that each pair m, m yields at most one amicable pair n, n Assume that mm ek If m, m are both even, then mm ek m,m even m<m ek 2 m ek m even m ek 4 i ek 2 i ek k log + log 2 4 If m, m are both odd, because pm, p m is an amicable pair, exactly one of pm, p m is abundant, call it ˆp ˆm Then, ˆp + σ ˆm > 2ˆp ˆm, so h ˆm > 2ˆp/ˆp + > 2 ɛ, for some small ɛ = 2 ˆp+ because ˆp is big In particular, ˆp > eˆk / 4 log ˆk, by properties, 2, 3 We have mm ek m,m odd Therefore, we have mm ek = ek n e k ˆm ek ˆm odd h ˆm 2 ɛ e k mm e k / ˆm ek πjk log ɛ j 5 mm ek m,m odd + mm ek m,m even

20 Since we are picking up all pairs of m, m such that m, m e k / without any constraint as to in which intervals n and n fall, this part does not need a multiplier In fact, we not only pick up both n and n in the argument, but also may have counted each pair multiple times, thus P 5 e πj /2k log ɛ j + k log L k + log 2 4 k=k+ Choosing j = 8, we can compute 0 6 e k=k+ πj k log ɛ j + k log L k + log 2 = For k > 0 6, we can ignore the parity of m, m and estimate Thus, E 5 e k=0 6 + k log + log 2 < e P t=0 6 + t + e t/5 dt For n A k, p e 3k/4 Suppose that n A k and p > e 3k/4, then m < e k/4 / By property 5, m > e 3k/4 Since n < µ k e k, it follows that p < µ k e k/4 We can write n = p p 2 p j s where s is the largest squarefull divisor and {p i } are in descending order Since n satisfies property 2, we have p p 2 p j = n s > ek µ k 4 > ek µ k > e k/2 Therefore, we can pick i to be the largest such that p p i < ek/2 Let D = p p i Write n = DM, so that gcdd, M = and M < e k µ k /D Furthermore, we have e k/2 < D = p p 2 p i p i < e 3k/4 µ k Recall from [9] that such m satisfy the following congruence σmdm mσm mod σd The number of choices for M < e k µ k /D which satisfy this congruence is at most + e k µ k DσD/ gcdσmd, σd + ek µ k σm gcdd, σd D 2 Every prime dividing D exceeds /2, because p i p j = n p p i s > ek/2 µ k 4 > e k /3, 6

21 and n has property 3, so it follows that p i > > /2 Since gcdd, σd n, n, property 4 implies that gcdd, σd = Also, σm mµ k < e k/4 µ k / Given the choice of m, D, the number of choices for M is at most: Therefore, we have p>e 3k/4 + e5k/4 µ 2 k D 2 e k/2 <De 3k/4 µ k e 3k/4 µ k + e5k/4 µ 2 k e 3k/4 µ k + e5k/4 µ 2 k e 3k/4 µ k + e5k/4 µ 2 k + e5k/4 µ 2 k D 2 D>e k/2 µ k D 2 e k + e k + e k/2 e k/2 t 2 dt p>e 3k/4 e 3k/4 µ k + ek/4 µ 2 k + e3k/4 µ 2 k Thus, the reciprocal sums of such numbers are at most n a6 k e k e µk e k/4 + so that P 6 e k=k+ µk + µ k e k/4 + µ2 k e 3k/4 + µ2 k e k/4, µ2 k e 3k/4 + µ2 k e k/4 Note that the denominator contains e k/4, which is significantly larger than the numerator In particular, we have P log k log t 3 3e e k/4 3e K e t/4 dt k=k+ 47 For n A k, q 0 6 Recall that for n = pm A k, we can write m = m 0 m where m 0 is the largest divisor of m such that m 0 is either a squarefull number or twice one Similarly, we can write q+ = q 0 q, where q 0 is the largest divisor of q + such that q 0 is either a squarefull number or twice one Assume that q 0 > 6 Because q < p, it follows that q < e k/2, so we have n A q 0 >6 n q 0 >6 q mod q 0 q<e k/2 q e k /qve k /q 7 v ι q 0 >6 q mod q 0 q<e k/2 q

22 where ι = + /e k/2 If q 0 > e k/4, then q > e k/4, so that q mod q 0 q<e k/2 q ι q 0 q+ q<e k/2 q + = ι q 0 where ι = + /e k/4 Therefore, we have q 0 >e k/4 q mod q 0 q<e k/2 e k/2 q 0 i= q ι + k/4 i ι + k/4 q 0, q 0 >e k/4 Because q 0 q is even, the definition of q 0 implies that q 0 is either an even squarefull number or twice an odd squarefull number, ie, 2q 0 is squarefull We have q 0 >e k/4 2q 0 squarefull q 0 = 2 i>2e k/4 i squarefull If 6 < q 0 e k/4, then by Lemma 35, we have q mod q 0 q<e k/2 q 2 q 0 + q log k + c, ϕq 0 where c = 8/k 2 log log2 /6 Therefore, 6 <q 0 <e k/4 q mod q 0 q<e k/2 By Lemma 36, we have q q 0 >6 2q 0 squarefull q 0 i = 2S 2e k/4 log loge k/2 log log2 /q 0 + loge k/2 /q 0 ϕq 0 q 0 >6 2q 0 squarefull q 0 ι q log k + c q 0 >6 2q 0 squarefull q 0 2ι S 32, where ι = + /6, and, using equation 34 in [], e γ log log q <q 0 <e k/2 ϕq 0 q 0 >6 2q 0 squarefull q log log q 0 q 0 >6 2q 0 squarefull ϕq 0

23 where e γ = 7807 Thus, we have n + n q 0 >6 e γ 2e γ q 0 >6 2q 0 squarefull i>32 i squarefull Using Lemma 36, we can compute log log q 0 q 0 + log log i i 5 2 log log log log 6 2e γ S S 32 log log 6, q 0 >6 2q 0 squarefull i>32 i squarefull [ + µ k ι 2ι + k/4s 2e k/4 + 2ι S k=k+ +2 log k + c 2e γ S S ] 32 log log 6 s 7 k = For k > 0 6, we have c <, ι, ι, ι < 2, so that 2 + k/4 E log k k= log t e k/8 + 3 e + 2 log k + k/0 2 + t/4 e t/8 + 3 e + 2 log t + t/0 i q 0 4 log k e + 5 k/0 4 log t e t/0 + 5 e t/5 e k/5 dt Thus, we have P For n A k, P σm Assume P σm <, then P q + = P σq P σm < We have ωm 4 log k by property, and m = n 2ek/4, pm 0 L 2 k by property 6 and since m 0 2s /2 by property 2, so that q 2e k/4 L 2 k / 4 log k = Q 0, say 9

24 Let Q i = Q 0 L i k, and consider q [Q i, Q i+ for separate values of i We have q+ = q 0 q as in the last section, and P σm < q [Q i,q i+ n ι i q 0 q 0 6 m 0 m 0 /2 q Q i 6 P q q m 2 2ek/4 Q i+ L 2 k P m 2 <Q i+ where ι i = + /Q i, and z = e k /qm 0 m 2 Since q 0 is always even, = q q = 2ζ2ζ3 pp 3ζ6 0 If n, n are both even, then m 0 is even, and If n, n are odd, then m 0 is odd, and Thus, g i := P σm < q [Q i,q i+ m 0 m 0 = p>2 n + n p>2 = 2ζ2ζ3 m m 0 3ζ6 0 + P σm < q [Q i,q i+ n,n even 4 + µ k ι i 2ζ2ζ3 3ζ6 pp = 2ζ2ζ3 3ζ6 n + n + 2 q Q i 6 P q q m 2 P σm < q [Q i,q i+ n,n odd m 2 2ek/4 Q i+ L 2 k P m 2 <Q i+ since the multiplier for even pair is 3 and for odd pair is + µ k Here, ωq logqi /6 log = u i, say, so as in Lemma 34, we have q > Q i 6 P q q P q < ωq u i H /2 ui q u i! since q runs over odd squarefree numbers Similarly, ωm 2 k/4 +log 2 log Qi+ 2 log log Q i+ = w i, say, so m 2 2ek/4 Q i+ L 2 k P m 2 <Q i+ m 2 P m 2 <Q i+ ωm 2 w i HQ i+ /2 wi m 2 w i! 20 p>q i zpez p, n + n m 2 p>q i zpez p, u i + u i + /2 H, 4 w i + w i + /2 HQ i+ 42

25 We also have p>q i zpez p + log Q i log 2 Q i Note that estimates 4 and 42 are valid only if u i + /2 > H and w i + /2 > HQ i+ For small values of i, u i will be too small to use 4 and for large values of i, w i will be too small to use 42 Let [a, b] be the interval of indices i where the estimates 4 and 42 are valid For q < Q a, we have α = P σm < q<q a n + n + µ k ι 0 where ι 0 = + /Q 0 For q > Q b, we have β = n + n + µ k ι b P σm < q Q b where ι b = + /Q b Thus, we have We can compute s 8 k = m 2 2ek/4 QaL 2 k P m 2 <Q a m 2 zvez z>q 0 + µ k ι 2 0 HQ a /2 w a w a! q > Q b 6 P q q zvez z>q b + µ k ι 2 b H /2 u b u b! P σm < 0 6 k=k+ For k > 0 6, we use a = b = 0, so that n + n s 8 k = v v b α + β + g i i=a w a + w a + /2 HQ a, u b + u b + /2 H, and H /2 < log k 2 logq0 /6 u 0 = log log 5 + B + 5 log k /2 < 84794, 2k 2 > log Q 0 log 6 k/5 k/5 k/4 +log 2 2 k/5 4 log k log 6 k/5 k/5 > 029k /3, 2

26 which implies E 8 k= log k 2π 029 k /3 elog k/ k /3 + 2 log t elog t/ π 029 t /3 029 t / k /3 029 t /3 dt Thus, we have P

27 49 Remaining amicable numbers We are left with amicables n such that both n and n have properties 8 We want to calculate P = n + µ k n n A min{n,n }>K n,n pass -8 k=k+ n,n pass -8 P n>p n By property 8, there exists a prime r σm with r > Thus, there is a prime q m with q mod r But σn = σsn, so there is a prime power q a sn with r σq a Then q a > r/2 > /2 > /4, so by property 2, a = and q mod r Since q > /2, by part 4 we have q n, so q q Also, q sn implies that sn = psm + σm 0mod q Since q n, it implies that q σm, so p is in a residue class a = am, q mod q And since P n > P n, p > q We want to calculate the following: n,n pass -8 P n>p n n r> q mod r q<e k m 0mod q m<e k m q mod r q <e k p amod q q <pe k /m Since q mod r, we have q > r Let ι =, then /ϕr ι/r and /ϕq ι/q If q > ek m, then p {q + a, 2q + a}, and since q + a and 2q + a are of different parity, there is at most one choice of p Writing m = qj, then e k /pq j e k /pq, so /j 3/2 Therefore, j If q ek m n,n pass -8 P n>p n q > ek m, then j ek, so that n,n pass -8 P n>p n q ek m qq r> r> n 3 2 r> q mod r q<e k q mod r q<e k q mod r q<e k q mod r q <e k q mod r q <e k q q mod r q <e k j ek qq j ek qq q p amod q pe k /jq 2ιe k q qj loge k /q qj p 23

28 2ιe k r> q mod r q<e k where z = ek qq Since qq mq e k, z e We have so Thus, jz/e q q mod r q <e k q jz/e j logz/j z/e log z + dt t logz/t = log z log log z t n,n pass -8 P n>p n q ek m n,n pass -8 P n>p n = + log log z < + log k, log z = 2ιe + log k n r> n 2ιe + log k q mod r q<e k r> q q mod r q<e k j logz/j, q mod r q <e k q z/e q q mod r q <e k q By Lemma 35, we have q mod r qe k q 2log k + c ϕr 2ιlog k + c, r where c = k/4 log log log2 / because r < p e 3k/4 Therefore, Thus, r> We can compute q mod r q<e k q q mod r q <e k q 4ι 2 log k + c 2 r s k 4ι3 + µ k 2ιe + log k + 3/2 log k + c k=k+ s k 735 r 2 4ι2 log k + c 2 For k > 0 6, c <, and choose a large factor, say 30, to offset the constants, we have E 30 k= log k + log k 3 24 e k/5

29 Thus, + 2 log t + log t e t/ P 736 Putting everything together, we have the result stated in Theorem : P = P s + P K + P + 8 P i + ε < 4084 This upper bound can potentially be decreased further by extending the parity argument currently used for small amicable numbers to address large amicable numbers, and 2 break down part 48 even further by computing small cases of q 0 and m 0 i= 25

30 References [] J Bayless and D Klyve, On the sum of reciprocals of amicable numbers, Integers A 20, Article 5 [2] P Erdős, On amicable numbers, Publ Math Debrecen 4 955, 08 [3] P Erdős and G J Rieger, Ein Nachtrag über befreundete Zahlen, J reine angew Math , 220 [4] H J Kanold, Über die asymptotische Dicte von gewissen Zahlenmengen, Proc International Congress of Mathematicians, Amsterdam, 954, p 30 [5] HL Montgomery and RC Vaughan, The large sieve, Mathematika , 9 34, [6] Jan Munch Pedersen, Known Amicable Pairs, [7] C Pomerance, On the distribution of amicable numbers, J reine angew Math 293/ , [8] C Pomerance, On the distribution of amicable numbers, II, J reine angew Math , [9] C Pomerance, On amicable numbers Submitted for publication, 204 [0] G J Rieger, Bemerkung zu einem Ergebnis von Erdős über befreundete Zahlen, J reine angew Math , [] J Barkley Rosser and Lowell Schoenfeld Approximate formulas for some functions of prime numbers Illinois J Math, 6: 64 94,