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
2014
Abstract
Two different positive integers a and b are said to form an amicable pair
if s(a) = b and s(b) = a, where s(n) denotes the sum of proper divisors of
n. In 1981, Pomerance proved that
the count of amicable numbers up to
x is less than x/ exp (log x)1/3 for x large. It follows immediately that
the sum P of the reciprocals of numbers belonging to an amicable pair
is a constant. In 2011, 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
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
Contents
1 Introduction
1
2 Framework Description
2.1 Small amicable numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Large amicable numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
4
3 Lemmas
6
4 Main argument
4.1 For n ∈ Ak , ω(n) ≤ b4 log kc . . . . . . . . . . .
4.2 For n ∈ Ak , the largest squarefull divisor of n is
4.3 For n ∈ Ak , the largest squarefree divisor d of n
4.4 For n ∈ Ak , P (gcd(n, s(n))) ≤ Lk /2 . . . . . .
k
4.5 For n ∈ Ak , mm0 > Le k . . . . . . . . . . . . . .
4.6 For n ∈ Ak , p ≤ e3k/4 Lk . . . . . . . . . . . . .
4.7 For n ∈ Ak , q0 ≤ 16Lk . . . . . . . . . . . . . .
4.8 For n ∈ Ak , P (σ(m1 )) ≥ Lk . . . . . . . . . . .
4.9 Remaining amicable numbers . . . . . . . . . .
. . . . . . . . . . . . . . . .
at most Lk /4 . . . . . . . .
with P (d) ≤ Lk has d ≤ ek/3
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
11
11
13
13
14
15
16
17
19
23
1
Introduction
Let σ denote the sum-of-divisors function, and let s(n) = σ(n)−n denote the sum-of-properdivisors function. Two different positive integers n and n0 are said to form an amicable pair
if and only if s(n) = n0 and s(n0 ) = 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 5th century BC, and was ascribed with many mystical properties. In the 9th century,
Thâbit ibn Qurra found a formula for amicable pairs, which states that if p = 3 · 2n − 1,
q = 3 · 2n−1 − 1, r = 9 · 22n−1 − 1 are primes, then 2n pq, 2n r form an amicable pair. For
example, n = 2 yields the primes p = 11, q = 5, r = 71, which generates the pair (220, 284).
It was not until 1636 that a second pair of amicable numbers (17296, 18416), which
corresponds to n = 4, was found by Fermat. Later, Descartes gave a third pair of amicable
numbers (9363584, 9437056), which corresponds to n = 7.
In the 18th century, Euler generalized Thâbit’s rule: if p = (2n−m + 1)2n − 1, q =
(2n−m + 1) · 2n−1 − 1, r = (2n−m + 1)2 · 22n−1 − 1 are primes, then 2n pq, 2n 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 (1184, 1210), which was discovered by Paganini, a 16-year-old Italian, in 1866.
We now know exhaustively all amicable pairs with the smallest member up to 1014 and
nearly 12 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 A(x) = A ∩ [1, x]. In 1954, Kanold
initiated the quest to explore the asymptotic density of A, showing that #A(x) < 0.204x
for x sufficiently large [4]. In 1955, Erdős was the first to prove that the set of amicable
numbers has zero asymptotic density, showing that #A(x) = o(x) [2]. From 1973 to 1981,
Rieger, Erdős, and Pomerance improved upon this result, provingthat #A(x) is less than
x/(log log log log x)1/2 , O (x/ log log log x), x/ exp (log log log x)1/2 , and x/ exp (log x)1/3
for x sufficiently large [10][3][7][8]. Most recently in 2014, Pomerance showed that #A(x) ≤
x/ exp((log x)1/2 ) as x → ∞ [9].
It follows immediately from the last two results that the reciprocal sum of the amicable
numbers is finite. In 2011, Bayless & Klyve showed a numerical bound for the reciprocal
sum of amicable numbers, which they denote by P :
0.0119841556 ≤ P < 656,000,000.
In this paper, we improve Bayless & Klyve’s result in [1] by showing a smaller upper bound
for the reciprocal sum of amicable numbers.
Theorem 1.1. We have
P =
X 1
< 4084.
n
n∈A
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.
1
2
Framework Description
Let n ∈ A be an amicable number and n0 = s(n) denote the corresponding friend. For some
K > log 1014 , we can write
P =
X 1
=
n
n∈A
X
n∈A
min{n,n0 }≤1014
1
+
n
X
n∈A
1014 <min{n,n0 }≤exp(K)
1
+
n
X
n∈A
min{n,n0 }>exp(K)
1
.
n
From the exhaustive list of amicable pairs with the smaller member less than 1014 , we
can compute
X
1
= 0.011984156739048 . . .
Ps =
n
n∈A
min{n,n0 }≤1014
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 [1] by approximately 10−9 .
An amicable pair (n, n0 ) can fall into one of the following cases: (n, n0 ) are both odd,
(n, n0 ) are both even, or (n, n0 ) are of different parity.
If n and n0 are of different parity, then σ(n) = σ(n0 ) = n + n0 is odd. Since σ(pa ) =
1 + p + p2 + · · · + pa 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 X 1
< 1.5 × 10−14 .
2
2
a
14
a>10
From this point forward, we consider only amicable pairs whose members are either
both odd or both even.
2.1
Small amicable numbers
For amicable pairs (n, n0 ) such that 1014 < min{n, n0 } ≤ exp(K), 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 n0 are both even, then
n0 X 1
1
=
> ,
n
d
2
d|n
d>1
and similarly n/n0 > 1/2, so 1014 < n, n0 ≤ 2 exp(K).
If n and n0 are odd, then min{n, n0 } is an odd abundant number. Recall that an integer
is abundant if the sum of its proper divisors is greater than the number itself.
2
We have
X
P (K) =
n∈A
1014 <min{n,n0 }≤exp(K)

≤
1
2
1
=
n
X
n∈A
n even
1014 <n≤2 exp(K)

1
+ 2

n
1014 /2<n≤exp(K)
where
X
1014 /2<n≤exp(K)
1
<
n
Z
X
n∈A
n odd
1014 <min{n,n0 }≤exp(K)
1
n



X
1
+
n
X
n odd abundant
1014 <n≤exp(K)
exp(K)
1014 /2
1
,
n
dt
= K − log(1014 /2),
t
and
X
n odd abundant
1014 <n≤exp(K)
1
n
is smaller. Let’s find out how small it is.
Let h(n) = σ(n)/n, then h is a multiplicative function. Let fj be the multiplicative
function defined as follow
f (1) = 1
.
fj (pa ) = h(pa )j − h(pa−1 )j
For x ≥ 3, we have
X
h(n)j =
n≤x
n odd
X X
fj (d),
n≤x d|n
n odd
so that
X 1X
X
X h(n)j
=
fj (d) =
fj (d)
n
n
n≤x
n odd
n≤x
n odd
d≤x
d odd
d|n
X
1≤m≤x/d
m odd
X fj (d) X 1
1
≤
.
md
d
m
d≤x
d odd
1≤m≤x
m odd
Here,
X
1≤m≤x
m odd
and
bx/2c
bx/2c
X
1
1
1 X 1
1 + log(x/2)
log x + 2.307
≤1+
<1+
<1+
<
,
m
2k + 1
2
k
2
2
k=1
k=1
X fj (d)
X fj (d)
Y fj (p) fj (p2 )
≤
=
1+
+
+
.
.
.
= πj , say,
d
d
p
p2
d≤x
d odd
d odd
p odd
is finite, as we shall see in Lemma 3.7.
3
Since h(n) > 2 for all n abundant, we have
X
n≤x
n odd abundant
πj (log x + 2.307)
1
1 X h(n)j
.
≤ j
≤
n
2
n
2j+1
n≤x
n odd
After investigating j ∈ [1, 40], we pick j = 18, which yields πj < 6231.87, so that
X
n≤x
n odd abundant
1
≤ 0.01189 log x + 0.02743.
n
Thus, we have
P (K) ≤ 0.52378K − 15.71668.
2.2
Large amicable numbers
For the remaining pairs, i.e., (n, n0 ) such that min{n, n0 } > exp(K), we will consider them
in small intervals and build an ordered list of properties.
Let Ak = {n ∈ A : ek−1 < n < ek }, and let ω(n) denote the number of distinct primes
dividing n.
Property (1): For n ∈ Ak , ω(n) ≤ b4 log kc.
Let µk =
d4 log
Q ke
i=2
pi
pi −1 ,
where pi is the ith prime. We will show that for n ∈ Ak such that n
0
and n0 have property
√ (1), n/µk ≤ n ≤ nµk .
Let Lk = exp( k/5). Recall that a positive integer s is squarefull if and only if for every
prime p dividing s, p2 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 ∈ Ak , the largest squarefull divisor of n is at most Lk /4.
Property (3): For n ∈ Ak , the largest squarefree divisor d of n with P (d) ≤ Lk
satisfies d ≤ ek/3 .
Property (4): For n ∈ Ak , P (gcd(n, s(n))) ≤ Lk /2.
Let n = pm, n0 = p0 m0 where p and p0 are the largest prime factors of n and n0 . Note
that properties (2), (3), and (4) imply that p 6= p0 , p - m, and p0 - m0 .
Property (5): For n ∈ Ak , mm0 ≥
ek
Lk .
Property (6): For n ∈ Ak , p ≤ e3k/4 Lk .
Write m = m0 m1 where m0 is the largest divisor of m such that m0 is either a squarefull
number or twice one, then m1 is odd, squarefree, and coprime to m0 . Let q = P (m1 ), and
write q + 1 = q0 q1 where q0 is the largest divisor of q + 1 such that q0 is either a squarefull
number or twice one.
4
Property (7): For n ∈ Ak , q0 is at most 16Lk .
Property (8): For n ∈ Ak , P (σ(m1 )) ≥ Lk .
For each property (i), let
X
P (i) =
n∈A
min{n,n0 }>exp(K)
0
n,n pass (1)−(i-1)
n or n0 fails (i)
Then,
P
(i)
≤
∞
X
k=K+1
X
n∈Ak
n,n0 pass (1)−(i-1)
n fails (i)
1
1
+ 0
n n
1
.
n
=
∞
X
(i)
sk , say.
k=K+1
Finally, we are left with n ∈ Ak where both n and n0 have properties (1)-(8). We proceed
to estimate
∞
∞
X
X
X
X
1
1
1
(∗)
P (∗) =
=
+ 0 =
sk , say,
n
n n
n∈A
min{n,n0 }>exp(K)
0
n,n pass (1)−(8)
k=K+1
n∈Ak
P (n)>P (n0 )
n,n0 passes (1)−(8)
so that
k=K+1
8
X
n∈A
min{n,n0 }>exp(K)
X
1
P (i) .
≤ P (∗) +
n
i=1
To optimize the sum of the expression above and P (K), we choose K = 5935.
In this paper, we will repeatedly use , 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 → 0 and ι → 1 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 π(x; d, a)
to denote the number of primes p congruent to a modulo d with p ≤ x.
5
3
Lemmas
Lemma 3.1. For all k, d, v ∈ Z+ , z > 0,
X 1
1 3
.
≤ min 1 + ,
v
z 2
z≤v≤ez
Proof. If z ≤ 1, then
X 1
X 1
3
≤
= .
v
v
2
z≤v≤ez
1≤v≤e
If 1 < z ≤ 2, then
X 1
X 1
3
≤
= 1.283̄ < .
v
v
2
z≤v≤ez
If z > 2, then
1<v≤2e
Z ez
X 1
1
1
1
≤ +
dv = 1 + .
v
z
v
z
z
z<v<ez
Lemma 3.2. For all x > 286, we have
H(x) :=
X1
p≤x
p
≤ log log x + B +
1
2 log2 x
(3.1)
and
1
1
+ D,
≤ log log x + B +
2
p−1
2
log
x
p≤x
P
where B = 0.261497 . . . and D = p 1/p(p − 1) = 0.773157 . . . .
H1 (x) :=
X
(3.2)
Proof. The inequality (3.1) is from [11], and (3.2) follows immediately.
Lemma 3.3. : For all z > 286,
X 1
1
1
≤
+
.
p
log z log2 z
z<p≤ez
Proof. From equations (3.17) and (3.18) of [11], we have
X 1
X 1 X1
1
1
=
−
≤ log(1 + log z) + B +
− log log z − B +
2
p
p
p
2(1 + log z)
2 log2 z
z<p≤ez
p≤ez
p≤z
1
1
1
1
1
= log 1 +
+
+
≤
+
.
2
2
log z
2(1 + log z)
log z log2 z
2 log z
6
Lemma 3.4. : For all w ∈ Z+ , x ≥ 0 such that H(x) < w + 1, we have
X
ω(d)≥w
P (d)≤x
d squarefree
1
H(x)w
w+1
≤
·
.
d
w!
w + 1 − H(x)
Proof. By the multinomial theorem, we have
X
ω(d)≥w
P (d)≤x
d squarefree
j

∞
1 X 1 X
1/p
≤
d
j!
j=w
p≤x
H(x)w
H(x)
H(x)2
=
1+
+
+ ...
w!
w + 1 (w + 1)(w + 2)
H(x)
H(x)2
H(x)w
1+
+
+ ...
≤
w!
w + 1 (w + 1)2
w+1
H(x)w
·
.
=
w!
w + 1 − H(x)
Lemma 3.5. For all d ∈ Z+ , d, x ≥ 3,
X
p≡−1(mod d)
d<p≤x
1
2(log log x − log log(2 − 1/d) + 1/ log(x/d))
≤
.
p
ϕ(d)
Proof. From the Brun-Titchmarsh theorem [5], we have
π(x; d, −1) ≤
2x
.
ϕ(d) log(x/d)
Using partial summation, we have
X
p≡−1(mod d)
d<p≤x
1
2
2
≤
+
p
ϕ(d) log(x/d) ϕ(d)
Z
x
2d−1
dt
t log(t/d)
2
≤
ϕ(d)
x 1
+ log log(t/d)
log(x/d)
2d−1
2
=
ϕ(d)
1
+ log log(x/d) − log log(2 − 1/d) .
log(x/d)
7
Lemma 3.6. For x ≥ 1,
√
1 ≤ ζ(3/2) x,
X
S0 (x) :=
s≤x
s squarefull
X
S1 (x) :=
s≥x
s squarefull
3
1
≤√ .
s
x
For x ≥ 2,
X
S∗ (x) :=
s≥x
s squarefull
log log s
3 log log x + 3 log 2
√
≤
+ 2ζ(3/2)
s
x
1
log log x + log 2
+
x
2x log x
.
Proof. Since every squarefull number can be written as a product of a square and a cube,
S0 (x) ≤
X
X
1=
a2 b3 ≤x
X
b≤x1/3 a≤(x/b3 )1/2
1≤
X x 1/2 √ X 1
√
x
x.
≤
ζ(3/2)
=
b3
b3/2
1/3
1/3
b≤x
b≤x
√
We can verify S1 (x) ≤ 3/ x for x ≤ 14000 using S1 (1) = ζ(2)ζ(3)/ζ(6). For x > 14000,
we have
X 1
S1 (x) ≤
a2 b3
2 3
a b ≥x
X 1 X 1
1
+
a2
b3 a a2
b≤x
a≥(x/b3 )1/2
b≥x1/3
!
X 1 b3 b3 1/2
X 1
+
≤
+
ζ(2)
b3 x
x
b3
b≤x1/3
b≥x1/3
1
1
1 X 1
1
≤ 2/3 + 1/2
+
+ ζ(2)
x 2x2/3
x
x
b3/2
1/3
≤
X 1
b3
1/3
X
b≤x
ζ(3/2) 1 + ζ(2)/2 ζ(2)
+
≤ 1/2 +
x
x
x2/3
3
≤√ .
x
By partial summation and our inequality for S0 , we also have
X
S∗ (x) = lim
z→∞
x≤s≤z
s squarefull
log log s
s
(S0 (z) − S0 (x)) log log z
−
= lim
z→∞
z
Z ∞
log log t
1
=
S0 (t)
− 2
dt
t2
t log t
x
8
Z
z
(S0 (t) − S0 (x))d
x
log log t
t
∞
log log t
1
≤ ζ(3/2)
− 3/2
dt
t3/2
t log t
x
Z ∞
2
1
log log t
−
+
dt
= ζ(3/2)
t3/2
t3/2 log t t3/2 log t
x
Z ∞
2
log log t
1
≤ ζ(3/2)
− 3/2
dt
+
t3/2
t log t t3/2 log x
x
∞ 2 log log t
2
= ζ(3/2) − √
−√
t
t log x x
1
log log x
√
.
+√
= 2ζ(3/2)
x
x log x
Z
Now, we can improve this estimate:
X
S∗ (x) =
x≤s<x2
s squarefull
log log s
+
s
X
2
≤ log log x
s≥x
s squarefull
X
s≥x2
s squarefull
log log s
s
1
+ 2ζ(3/2)
s
3 log log x + 3 log 2
√
≤
+ 2ζ(3/2)
x
log log x + log 2
1
+
x
2x log x
log log x + log 2
1
+
x
2x log x
.
Lemma 3.7. Let f be the multiplicative function defined as in 2.1. For all j ∈ Z+ ,
Y fj (p) fj (p2 )
πj =
1+
+
+ ...
p
p2
p odd
is a finite number.
Proof. First, we have
fj (p)
=
p
h(p)j
p
−1
1+
=
1
p
j
−1
=O
p
1
p2
Because 0 < fj (pa ) < h(pa )j , for all a ≥ 2,
fj (pa )
h(pa )j
σ(pa )j
≤
≤
≤
pa
pa
pa+ja
p
p−1
j
1
.
pa
Therefore, for each fixed j ∈ Z+
fj (p) fj (p2 )
1+
+
+ ··· = 1 + O
p
p2
9
1
p2
,
.
so that

X 1
 = O(1).
log πj = O 
p2

p odd
In order to estimate an upper bound for πj , we choose large integers A, B, and compute
Y
Y
fj (p) fj (p2 )
+
.
.
.
≤
1+
+
p
p2
p≤B
p≤B
1+
A
X
fj (pa )
a=1
pa
+
p
p−1
j+1
1
pA+1
!
,
and
Y
p>B
fj (p)
1+
+ ...
p
≤
Y
1+
p>B
Z
∞
≤ exp
B
(1 + p1 )j − 1
!
j
p
1
+
p
p−1
p(p − 1)
! !
j
(1 + x1 )j − 1
x
1
+
dx .
x
x−1
x(x − 1)
Table 3.1 shows upper bounds of πj for j ∈ [1, 40], computed with A = 500 and B = 106 .
We pick j = 18 to get the smallest coefficient in section 2.1.
Table 3.1: Estimates of upper bound of πj
j
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
πj
1.23
1.58
2.10
2.90
4.16
6.21
9.60
15.33
25.24
42.72
74.16
131.79
239.35
443.49
837.30
1608.70
3141.94
6231.87
12541.54
25588.42
πj /2j
0.616851
0.394426
0.262036
0.181057
0.130076
0.097029
0.074986
0.059886
0.049293
0.041714
0.036209
0.032176
0.029217
0.027069
0.025552
0.024547
0.023971
0.023773
0.023921
0.024403
j
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
10
πj
52890.17
110675.79
234319.96
501650.18
1085437.39
2372559.18
5236626.72
11666433.49
26225299.18
59464173.31
135959987.61
313376228.98
727960204.98
1703849881.31
4017349516.24
9539842817.16
22811315060.41
54914363890.79
133067328413.01
324514342521.39
πj /2j
0.025220
0.026387
0.027933
0.029901
0.032349
0.035354
0.039016
0.043461
0.048848
0.055380
0.063311
0.072964
0.084746
0.099177
0.116920
0.138823
0.165974
0.199777
0.242048
0.295144
4
Main argument
For n ∈ Ak , ω(n) ≤ b4 log kc
4.1
Let u = d4 log ke; as in Lemma 3.4, we have

j

j
∞
∞
∞
X 1 X
X
H1 (ek )u (u + 1)
1 X 1 
1 X X 1 
≤
≤
.
≤
n
j!
qa
j!
q−1
u! (u + 1 − H1 (ek ))
k
k
n∈Ak
ω(n)≥u
q≤e a=1
j=u
j=u
q≤e
By equations (3.41) and (3.42) in [11], we have that for n, n0 > exp(K),
σ(n0 )
σ(n)
n
n0
<
=
≤
< (eγ + ) log log n,
n
n
n
ϕ(n)
and similarly, n/n0 < (eγ + ) log log n0 , where eγ = 1.78107 . . . and ≤ 0.01. We have
n
< (eγ + ) log log n0 < (eγ + ) (log log(n) + log log((eγ + ) log log n))
n0
which implies
1
(eγ + ) log log n (eγ + ) log log((eγ + ) log log n))
2 log log n
2 log k
<
+
<
<
,
n0
n
n
n
n
for n > exp(K).
Therefore, we have
∞
X
P (1) ≤
(1)
sk ,
k=K+1
where
(1)
sk
≤
∞
X
(1 + 2 log k) ·
k=K+1
H1 (ek )u (u + 1)
.
u! (u + 1 − H1 (ek ))
We can compute
6
10
X
(1)
sk
≤ 0.788795.
k=K+1
For k > 106 , H1 (ek ) ≤ log k + 1.034656 by Lemma 3.2, and
1
u+1
=
≤
k
k
H
1 (e )
u + 1 − H1 (e )
1−
1 − u+1
1
log k+1.034656
4 log k+1
By Stirling’s inequality,
1 e u
1
≤√
.
u!
2πu u
11
≤ 1.358597.
Therefore,
E
(1)
∞
X
:=
(1)
sk
k=106 +1
∞
X
1 + 2 log k
√
≤ 1.358597
2πu
k=106 +1
e(log k + 1.034656)
u
u
∞
X
e(log k + 1.034656)
4 log k
4 log k
≤ 1.358597
k=106 +1
because u ≥ 4 log k and
For all k > 106 ,
e(log k+1.034656)
4 log k
e(log k+1.034656)
4 log k
E
(1)
=
e
4
1 + 2 log k
√
8π log k
≤ 1.
+
e×1.034656
4 log k
∞
X
1.358597
≤p
8π log 106
≤ 0.730464 = h , so we have
(1 + 2 log k)h4 log k
k=106 +1
∞
X
(1 + 2 log k)k 4 log h
= 0.072910
k=106 +1
Z ∞
≤ 0.072910
106
1 + 2 log t
dt
t1.256301
≤ 0.300429.
Thus, we have
P (1) ≤ 0.788795 + 0.300429 = 1.089224.
Amicable pair multiplier
For n ∈ Ak , k > K satisfying property (1), we will show that
n
≤ n0 ≤ nµk ,
µk
where
d4 log ke
µk =
Y
i=2
Proof. If n, n0 are even, we have
pi
.
pi − 1
n0 X 1
1
=
> .
n
d
2
d|n
d>1
Similarly, n/n0 > 1/2. The inequality follows immediately since µk > 2.
If n and n0 are odd, the inequality on the right is straightforward:
n0
s(n)
σ(n) X 1 Y p
=
=
=
<
≤
n
n
n
d
p−1
d|n
p|n
12
b4 log kc+1
Y
i=2
pi
= µk .
pi − 1
If n0 > n, then clearly n0 > n/µk . If n0 ≤ n, then n0 ∈ Ak0 where k 0 ≤ k, and it follows that
0
b4 log k c+1
b4 log kc+1
Y
Y
s(n0 )
σ(n0 ) Y p
n
pi
pi
≤
=
=
≤
<
= µk .
0
0
0
n
n
n
p−1
pi − 1
pi − 1
0
i=1
p|n
4.2
i=2
For n ∈ Ak , the largest squarefull divisor of n is at most
Lk /4
Write n = sv where s is the largest squarefull divisor of n. Assume that s > Lk /4; then by
Lemma 3.1 and Lemma 3.6,
X 1
X 1
X
X
1
3(1 + µk )S1 (Lk /4)
1
1
+ 0 ≤ (1+µk )
= (1+µk )
≤
.
n n
n
s k−1
v
2
k
n∈Ak
s>Lk /4
n∈Ak
s>Lk /4
s>Lk /4
s squarefull
Therefore,
P (2) ≤
e
s
<v< es
∞
X
3(1 + µk )S1 (Lk /4)
.
2
k=K+1
By Lemma 3.6, we can compute
6
10
X
3(1 + µk )S1 (Lk /4)
≤ 35.876404.
2
k=K+1
For k > 106 ,
∞
X
E (2) :=
3(1 + µk )S1 (Lk /4)
≤9
2
k=106 +1
Z
∞
≤9
1 + 2 log t
√
t=106
e
t/10
∞
X
1 + 2 log k
k=106 +1
√
e
k/10
dt ≤ 1.939049 × 10−37 .
Thus, we have
P (2) ≤ 35.9.
4.3
For n ∈ Ak , the largest squarefree divisor d of n with
P (d) ≤ Lk has d ≤ ek/3
For n ∈ Ak , write n = dv where d is the largest squarefree divisor such that P (d) ≤ Lk .
Assume that d > ek/3 , we have
X 1
≤
n
n∈Ak
d>ek/3
X
d>ek/3
P (d)<Lk
d squarefree
1
d
X
k
ek−1
<v< ed
d
1
3 H(Lk )u
u+1
≤ ·
·
,
v
2
u!
u + 1 − H(Lk )
13
where u =
l
log ek/3
log Lk
m
=
l
k
3 log Lk
m
. Therefore,
∞
X
3(1 + µk )H(Lk )u (u + 1)
.
2u!(u + 1 − H(Lk ))
P (3) ≤
k=K+1
We can compute
6
10
X
3(1 + µk )H(Lk )u (u + 1)
= 3.622428 × 10−154 .
2u!(u + 1 − H(Lk ))
k=K+1
For k > 106 , H(Lk ) ≤ log
√ k
5
+B+
5
2k
l
< log k, and u =
k
3 log Lk
m
>
k
3 log Lk
>
√
k, so
1
1
u+1
=
≤
< 2.
k
H(L
)
u + 1 − H(Lk )
1 − √log
1 − u+1k
k+1
We can estimate
∞
X
E (3) :=
k=106 +1
Z
3(1 + µk )H(Lk )u (u + 1)
≤3
2u!(u + 1 − H(Lk ))
∞
≤3
(1 + 2 log t)
106
e log t
√
t
√t
∞
X
(1 + 2 log k)
k=106 +1
e log k
√
k
√k
dt ≤ 1.904584 × 10−1421 .
Thus, we have
P (3) ≤ 3.7 × 10−154 .
4.4
For n ∈ Ak , P (gcd(n, s(n))) ≤ Lk /2
Let r = P (gcd(n, s(n))), and suppose r > Lk /2. Because r|σ(n), there exists q a ||n, such
that r|σ(q a ). Then Lk /2 < r ≤ σ(q a ) < 2q a , so q a > Lk /4. It follows from property (2)
that a = 1, implying that q ≡ −1(mod r). Because r is a prime larger than Lk /2, q > r.
We can write n = rqv, so that
X 1
X 1
≤
n
r
n∈Ak
r>Lk /2
r>Lk /2
X
r<q≤ek
q≡−1( mod r)
1
q
X
k
ek−1
<v< eqr
qr
1
3 X 1
≤
v
2
r
r>Lk /2
X
r<q≤ek
q≡−1( mod r)
1
,
q
by Lemma 3.1. From lemma 3.5, we have
X
r<q≤ek
q≡−1( mod r)
2(log k − log log(2 − 1/r) + 1/ log(ek /r))
2(log k + c)
1
≤
≤
,
q
ϕ(r)
r−1
where c = 2/k − log log(2 − 2/Lk ) because Lk /2 < r <
14
√
qr ≤
√
n < ek/2 .
We have
X 1
X
1
3(log k + c)
≤ 3(log k + c)
≤
.
n
r(r − 1)
Lk /2 − 1
n∈Ak
r>Lk /2
r>Lk /2
Therefore,
P
(4)
∞
X
6(1 + µk )(log k + c)
≤
.
Lk − 2
k=K+1
We can compute
6
10
X
6(1 + µk )(log k + c)
= 0.050951 . . . .
Lk − 2
k=K+1
For k >
106 ,
2/k − log log(2 − 2/Lk ) < 1, so that
Z ∞
(1 + 2 log t)(log t + 1)
(4)
√
dt ≤ 3.54463 × 10−80 .
E
≤6
t/5
6
−
2
e
t=10
Thus, we have
P (4) ≤ 0.051.
4.5
For n ∈ Ak , mm0 >
ek
Lk
In [9], it is shown that each pair m, m0 yields at most one amicable pair n, n0 . Assume that
k
mm0 ≤ Le k .
If m, m0 are both even, then
X
ek X 1
ek X 1
ek k − log Lk + 1 − log 2
≤
≤
·
.
2Lk
m
4Lk
i
Lk
4
k
k
1≤
k
e
mm0 ≤ L
m,m0
e
m≤ L
k
m even
k
even
m<m0
e
i≤ 2L
k
If m, m0 are both odd, because pm, p0 m0 is an amicable pair, exactly one of pm, p0 m0 is
abundant, call it p̂m̂. Then, (p̂ + 1)σ(m̂) > 2p̂m̂, so h(m̂) > 2p̂/(p̂ + 1) > 2 − , for some
2
small = p̂+1
because p̂ is big. In particular, p̂ > e(k̂−1)/b4 log k̂c , by properties (1), (2), (3).
We have
X
X
ek
1
ek πj (k − log Lk + 2.307)
≤
·
.
1=
Lk
m̂
Lk
2(2 − )j
k
k
e
mm0 ≤ L
k
m,m0 odd
e
m̂≤ L
k
m̂ odd
h(m̂)≥2−
Therefore, we have

1
1
≤ k−1
n
e
X
n∈Ak
k
e
mm0 ≤ L


1 
 X
1 ≤ k−1 
1+

e
0
k
k
mm0 ≤ e
mm ≤e /Lk
X
Lk
m,m0 odd
k
15
X
k
e
mm0 ≤ L
k
m,m0 even



1 .


Since we are picking up all pairs of m, m0 such that m, m0 ≤ ek /Lk without any constraint
as to in which intervals n and n0 fall, this part does not need a multiplier. In fact, we not
only pick up both n and n0 in the argument, but also may have counted each pair multiple
times, thus
∞
X
e πj /2(k − log Lk + 2.307) k − log Lk + 1 − log 2
(5)
+
.
P
≤
Lk
(2 − )j
4
k=K+1
Choosing j = 18, we can compute
6
10
X
k=K+1
e
Lk
πj (k − log Lk + 2.307) k − log Lk + 1 − log 2
+
2(2 − )j
4
= 0.806677 . . . .
For k > 106 , we can ignore the parity of m, m0 and estimate
Z ∞
∞
X
k − log(Lk ) + 1 − log 2
t+1
(5)
√
E
≤e
<e
dt ≤ 3.818819 × 10−77 .
t/5
L
6
k
t=10 +1 e
k=106 +1
Thus,
P (5) ≤ 0.807.
4.6
For n ∈ Ak , p ≤ e3k/4 Lk
Suppose that n ∈ Ak and p > e3k/4 Lk , then m < ek/4 /Lk . By property (5), m0 > e3k/4 .
Since n0 < µk ek , it follows that p0 < µk ek/4 .
We can write n0 = p01 p02 . . . p0j s0 where s0 is the largest squarefull divisor and {p0i } are in
descending order. Since n0 satisfies property (2), we have
p01 p02 . . . p0j =
n0
ek−1 4
ek
>
·
>
> ek/2 .
s0
µk Lk0
µk Lk
Therefore, we can pick i to be the largest such that p01 . . . p0i−1 < ek/2 .
Let D = p01 . . . p0i . Write n0 = DM , so that gcd(D, M ) = 1 and M < ek µk /D. Furthermore, we have
ek/2 < D = (p01 p02 . . . p0i−1 )p0i < e3k/4 µk .
Recall from [9] that such m satisfy the following congruence
σ(m)DM ≡ mσ(m) (mod σ(D)).
The number of choices for M < ek µk /D which satisfy this congruence is at most
1+
ek µk
ek µk σ(m) gcd(D, σ(D))
≤1+
.
Dσ(D)/ gcd(σ(m)D, σ(D))
D2
Every prime dividing D exceeds Lk /2, because
p0i . . . p0j =
n0
p01 . . . p0i−1 s0
>
16
ek/2−1 4
0
·
> ek /3 ,
µk
Lk0
and n0 has property (3), so it follows that p0i > Lk0 > Lk /2.
Since gcd(D, σ(D))|(n, n0 ), property (4) implies that gcd(D, σ(D)) = 1. Also, σ(m) ≤
mµk < ek/4 µk /Lk . Given the choice of m, D, the number of choices for M is at most:
1+
e5k/4 µ2k
.
Lk D2
Therefore, we have
X
X
1≤
n∈Ak
p>e3k/4 Lk
ek/2 <D≤e3k/4 µk
≤ e3k/4 µk +
e5k/4 µ2k
1+
Lk D2
e5k/4 µ2k
Lk
e5k/4 µ2k
Lk
X
D>ek/2 µk
1
D2
∞
1
dt
2
ek/2 t
e5k/4 µ2k 1
1
≤ e3k/4 µk +
+ k/2
Lk
ek
e
e3k/4 µ2k
ek/4 µ2k
+
.
≤ e3k/4 µk +
Lk
Lk
≤ e3k/4 µk +
!
1
+
ek
Z
Thus, the reciprocal sums of such numbers are at most
(6)
X 1
ak
µ2k
µ2k
µk
≤ k−1 ≤ e k/4 +
+
,
n
e
e
Lk e3k/4 Lk ek/4
n∈A
k
p>e3k/4 Lk
so that
P
(6)
≤e
∞
X
(1 + µk )
k=K+1
µ2k
µ2k
µk
+
+
ek/4 Lk e3k/4 Lk ek/4
.
Note that the denominator contains ek/4 , which is significantly larger than the numerator.
In particular, we have
Z ∞
∞
X
(1 + 2 log k)3
(1 + 2 log t)3
(6)
P
≤ 3e
≤
3e
dt ≤ 2.5 × 10−538 .
k/4
t/4
e
e
K
k=K+1
4.7
For n ∈ Ak , q0 ≤ 16Lk
Recall that for n = pm ∈ Ak , we can write m = m0 m1 where m0 is the largest divisor of m
such that m0 is either a squarefull number or twice one. Similarly, we can write q +1 = q0 q1 ,
where q0 is the largest divisor of q + 1 such that q0 is either a squarefull number or twice
one. Assume that q0 > 16Lk . Because q < p, it follows that q < ek/2 , so we have
X 1
X
X
X
X
X
1
1
1
≤
≤ι
n
q k−1
v
q
k
n∈A
q0 >16Lk
q0 >16Lk q≡−1(mod q0 )
q<ek/2
e
/q≤v≤e /q
17
q0 >16Lk q≡−1(mod q0 )
q<ek/2
where ι = 1 + 1/ek/2−1 .
If q0 > ek/4 , then q > ek/4 − 1, so that
X
q≡−1(mod q0 )
q<ek/2
X
1
≤ ι0
q
q0 |(q+1)
q<ek/2
ek/2
q0
1
ι0 X 1
ι0 (1 + k/4)
,
=
≤
q+1
q0
i
q0
i=1
where ι0 = 1 + 1/(ek/4 − 1). Therefore, we have
X
X
q0 >ek/4 q≡−1(mod q0 )
q<ek/2
X 1
1
.
≤ ι0 (1 + k/4)
q
q0
k/4
q0 >e
Because q0 q1 is even, the definition of q0 implies that q0 is either an even squarefull
number or twice an odd squarefull number, i.e., 2q0 is squarefull. We have
X
q0 >ek/4
2q0 squarefull
1
=2
q0
X
i>2ek/4
i squarefull
1
= 2S1 (2ek/4 ).
i
If 16Lk < q0 ≤ ek/4 , then by Lemma 3.5, we have
2
log log(ek/2 ) − log log(2 − 1/q0 ) +
X
1
1
≤
+
q
q0 − 1
ϕ(q0 )
1
log(ek/2 /q0 )
q≡−1(mod q0 )
q<ek/2
≤
1
2 log k + c
+
,
q0 − 1
ϕ(q0 )
where c = 8/k − 2 log log(2 − 1/16Lk ). Therefore,


X
X
16Lk <q0 <ek/4 q≡−1(mod q0 )
q<ek/2
1 
≤
q 
X
q0 >16Lk
2q0 squarefull


1 
 + (2 log k + c) 

q0 − 1 
q0 >16Lk
2q0 squarefull
1
≤ ι00
q0 − 1
X
q0 >16Lk
2q0 squarefull
1
≤ 2ι00 S1 (32Lk ),
q0
where ι00 = 1 + 1/(16Lk − 1), and, using equation (3.41) in [11],
X
X
1
5
1
γ
≤
e log log q0 +
ϕ(q0 )
q0
2 log log q0
k/2
16Lk <q0 <e
X
q0 >16Lk
2q0 squarefull
By Lemma 3.6, we have
X

q0 >16Lk
2q0 squarefull
18
1 
.
ϕ(q0 ) 
X
≤ eγ
q0 >16Lk
2q0 squarefull
≤ 2eγ
X
i>32Lk
i squarefull
log log q0
5
+
q0
2 log log 16Lk
log log i
5
+
i
log log 16Lk
X
q0 >16Lk
2q0 squarefull
X
i>32Lk
i squarefull
1
q0
1
i
5S1 (32Lk )
,
log log 16Lk
≤ 2eγ S∗ (32Lk ) +
where eγ = 1.78107 . . . .
Thus, we have
X 1
1
≤ (1 + µk )ι 2ι0 (1 + k/4)S1 (2ek/4 ) + 2ι00 S1 (32Lk )
+
n n0
n∈Ak
q0 >16Lk
5S1 (32Lk )
+(2 log k + c) 2eγ S∗ (32Lk ) +
.
log log 16Lk
Using Lemma 3.6, we can compute
6
10
X
(7)
sk
= 642.0724 . . . .
k=K+1
For k > 106 , we have c < 1, ι, ι0 , ι00 < 2, so that
∞
X
12(1 + k/4)
3
4 log k
15
(7)
E
≤2
(1 + 2 log k)
+ √
+ (2 log k + 1) √
+ √
k/10
k/10
ek/8
e
e
e k/5
6
k=10 +1
Z ∞
3
4 log t
15
12(1 + t/4)
+ (2 log t + 1) √
+ √
dt
≤2
(1 + 2 log t)
+ √
et/8
e t/10
e t/10 e t/5
106
≤ 6.85011 × 10−35 .
Thus, we have
P (7) ≤ 642.073.
4.8
For n ∈ Ak , P (σ(m1 )) ≥ Lk
Assume P (σ(m1 )) < Lk , then P (q + 1) = P (σ(q)) ≤ P (σ(m1 )) < Lk .
We have ω(m1 ) ≤ b4 log kc − 1 by property (1), and
m1 =
n
2ek/4−1
≥
,
pm0
L2k
by property (6) and since m0 ≤ 2s ≤ Lk /2 by property (2), so that
!1/(b4 log kc−1)
2ek/4−1
= Q0 , say.
q≥
L2k
19
Let Qi = Q0 Lik , and consider q ∈ [Qi , Qi+1 ) for separate values of i. We have q+1 = q0 q1
as in the last section, and
X 1 X
X 1
X
X
1
1
1 X 1
≤ ιi
,
n
q0
m0
q1
m2
p
Q
k/4−1
q0 ≤16Lk
n∈Ak
P (σ(m1 ))<Lk
q∈[Qi ,Qi+1 )
m0 ≤Lk /2
q1 ≥ 16Li
p>Qi
z≤p≤ez
m2 ≥ 2e
Qi+1 L2
k
k
P (q1 )≤Lk
P (m2 )<Qi+1
where ιi = 1 + 1/Qi , and z = ek−1 /qm0 m2 . Since q0 is always even,
Y
X 1
1 1
1
2ζ(2)ζ(3)
=
+ + ...
1+
=
.
q0
2 4
p(p − 1)
3ζ(6)
q
p>2
0
If n, n0 are both even, then m0 is even, and
X 1
2ζ(2)ζ(3)
.
=
m0
3ζ(6)
m
0
If n, n0 are odd, then m0 is odd, and
X 1
Y
2ζ(2)ζ(3)
1
=
.
=
1+
m0
p(p − 1)
3ζ(6)
m
p>2
0
Thus,
X
gi :=
n∈Ak
P (σ(m1 ))<Lk
q∈[Qi ,Qi+1 )
1
1
+
n n0
X
≤
n∈Ak
P (σ(m1 ))<Lk
q∈[Qi ,Qi+1 )
n,n0 even
≤ (4 + µk )ιi
2ζ(2)ζ(3)
3ζ(6)
2
1
1
+
n n0
X
Q
q1 ≥ 16Li
k
P (q1 )≤Lk
X
+
n∈Ak
P (σ(m1 ))<Lk
q∈[Qi ,Qi+1 )
n,n0 odd
1
q1
X
m2 ≥ 2e
k/4−1
Qi+1 L2
k
1
1
+
n n0
1 X 1
,
m2
p
p>Qi
z≤p≤ez
P (m2 )<Qi+1
since the multiplier
l for even pair
m is 3 and for odd pair is 1 + µk .
log(Qi /16Lk )
Here, ω(q1 ) ≥
= ui , say, so as in Lemma 3.4, we have
log Lk
X
Q
q1 > 16Li
k
P (q1 )≤Lk
1
≤
q1
X
P (q1 )<Lk
ω(q1 )≥ui
1
(H(Lk ) − 1/2)ui
ui + 1
≤
·
,
q1
ui !
ui + 1/2 − H(Lk )
(4.1)
since q1 runs over odd squarefree
numbers.
l
m
k/4−1+log 2−log Qi+1 −2 log Lk
Similarly, ω(m2 ) ≥
= wi , say, so
log Qi+1
X
k/4−1
m2 ≥ 2e
Qi+1 L2
k
P (m2 )<Qi+1
1
≤
m2
X
P (m2 )<Qi+1
ω(m2 )≥wi
1
(H(Qi+1 ) − 1/2)wi
wi + 1
≤
·
.
m2
wi !
wi + 1/2 − H(Qi+1 )
20
(4.2)
We also have
X 1
1
1
.
+
≤
p
log Qi log2 Qi
p>Q
i
z≤p≤ez
Note that estimates (4.1) and (4.2) are valid only if ui + 1/2 > H(Lk ) and wi + 1/2 >
H(Qi+1 ). For small values of i, ui will be too small to use (4.1) and for large values of i,
wi will be too small to use (4.2). Let [a, b] be the interval of indices i where the estimates
(4.1) and (4.2) are valid. For q < Qa , we have
X
X
1
1
1 X 1
α=
+ 0 ≤ (1 + µk )ι0
n n
m2
v
k/4−1
n∈Ak
P (σ(m1 ))<Lk
q<Qa
z≤v≤ez
z>Q0
m2 ≥ 2e
Qa L2
k
P (m2 )<Qa
(H(Qa ) − 1/2)wa−1
wa−1 + 1
·
,
wa−1 !
wa−1 + 1/2 − H(Qa )
≤ (1 + µk )ι20 ·
where ι0 = 1 + 1/Q0 . For q > Qb , we have
X
1
1
β=
+
≤ (1 + µk )ιb
n n0
n∈Ak
P (σ(m1 ))<Lk
q≥Qb
X
Q
q1 > 16Lb
1 X 1
q1
v
k
P (q1 )≤Lk
≤ (1 + µk )ι2b ·
z≤v≤ez
z>Qb
(H(Lk ) − 1/2)ub
ub + 1
·
,
ub !
ub + 1/2 − H(Lk )
where ιb = 1 + 1/Qb . Thus, we have
(8)
sk
=
X
n∈Ak
P (σ(m1 ))<Lk
1
1
+ 0
n n
≤α+β+
b−1
X
gi .
i=a
We can compute
6
10
X
(8)
sk
= 306.2117.
k=K+1
For k > 106 , we use a = b = 0, so that
H(Lk ) − 1/2 <
5
log k
log k
− log 5 + B +
− 1/2 <
− 1.847941,
2
2k
2
and
√
log(Q0 /16Lk )
log Q0 − log 16 − k/5
√
u0 =
>
log Lk
k/5
√
√
k/4−1+log 2−2 k/5
− log 16 − k/5
4 log k−1
√
≥
> 0.219k 1/3 ,
k/5
21
which implies
E
(8)
≤
∞
X
√
1 + 2 log k
e(log k/2 − 1.847941)
0.219 · k 1/3
0.219·k1/3
2π · 0.219 · k 1/3
1/3
Z ∞
1 + 2 log t
e(log t/2 − 1.847941) 0.219·t
√
≤
dt
0.219 · t1/3
2π · 0.219 · t1/3
106
≤ 11.2041.
k=106 +1
Thus, we have
P (8) ≤ 317.4158.
22
4.9
Remaining amicable numbers
We are left with amicables n such that both n and n0 have properties (1)–(8). We want to
calculate
∞
X
X
X
1
1
(∗)
≤
.
P =
(1 + µk )
n
n
n∈A
min{n,n0 }>K
n,n0 pass (1)-(8)
n∈Ak
n,n0 pass (1)-(8)
P (n)>P (n0 )
k=K+1
By property 8, there exists a prime r|σ(m1 ) with r > Lk . Thus, there is a prime q|m with
q ≡ −1(mod r). But σ(n) = σ(s(n)), so there is a prime power q 0a ||s(n) with r|σ(q 0 a ). Then
q 0a > r/2 > Lk /2 > Lk0 /4, so by property 2, a = 1 and q 0 ≡ −1(mod r). Since q 0 > Lk /2,
by part 4 we have q 0 - n, so q 0 6= q. Also, q 0 |s(n) implies that
s(n) = ps(m) + σ(m) ≡ 0(mod q 0 ).
Since q 0 - n, it implies that q 0 - σ(m), so p is in a residue class a = a(m, q 0 )(mod q 0 ). And
since P (n) > P (n0 ), p > q 0 .
We want to calculate the following:
X
n∈Ak
n,n0 pass (1)-(8)
P (n)>P (n0 )
X
1
≤
n
X
X
r>Lk q≡−1(mod r) m≡0(mod q)
q<ek
m<ek
1
m
X
X
q 0 ≡−1(mod r) p≡a(mod q 0 )
q 0 <ek
q 0 <p≤ek /m
1
.
p
k
Since q 0 ≡ −1(mod r), we have q 0 > r ≥ Lk . Let ι = LLk −1
, then 1/ϕ(r) ≤ ι/r and
0
0
1/ϕ(q ) ≤ ι/q .
k−1
If q 0 > e m , then p ∈ {q 0 + a, 2q 0 + a}, and since (q 0 + a) and (2q 0 + a) are of different
parity,
there is at most one choice of p. Writing m = qj, then ek−1 /pq ≤ j ≤ ek /pq, so
P
1/j ≤ 3/2. Therefore,
j
X
n∈Ak
n,n0 pass (1)-(8)
P (n)>P (n0 )
1
3 X
≤
n
2
X
r>Lk q≡−1(mod r)
q<ek
1
q
X
q 0 ≡−1(mod r)
q 0 <ek
1
.
q0
k−1
q0 > e m
If q 0 ≤
ek−1
m ,
then j ≤
ek−1
qq 0 ,
X
1≤
n∈Ak
n,n0 pass (1)-(8)
P (n)>P (n0 )
so that
X
X
r>Lk q≡−1(mod r)
q<ek
X
q 0 ≡−1(mod
q 0 <ek
X
r)
k−1
j≤ e qq0
X
1
q0 )
p≡a(mod
p≤ek /jq
k−1
q0 ≤ e m
≤
X
X
X
X
k−1
r>Lk q≡−1(mod r) q 0 ≡−1(mod r)
j≤ e qq0
q<ek
q 0 <ek
23
2ιek
q 0 qj log(ek /q 0 qj)
≤ 2ιek
X
X
r>Lk q≡−1(mod r)
q<ek
where z =
ek
qq 0 .
1
q
X
q 0 ≡−1(mod r)
q 0 <ek
1
1 X
,
q0
j log(z/j)
j≤z/e
Since qq 0 ≤ mq 0 ≤ ek−1 , z ≥ e. We have
X
j≤z/e
1
1
≤
+
j log(z/j)
log z
=
z/e
Z
1
z/e
dt
1
z =
− log log t log(z/t)
log z
t
1
1
+ log log z < 1 + log k,
log z
so
X
n∈Ak
n,n0 pass (1)-(8)
P (n)>P (n0 )
X
1
= 2ιe(1 + log k)
n
X
r>Lk q≡−1(mod r)
q<ek
1
q
X
q 0 ≡−1(mod r)
q 0 <ek
1
.
q0
k−1
q0 ≤ e m
Thus,
X
n∈Ak
n,n0 pass (1)-(8)
P (n)>P (n0 )
1
≤
n
3 X
2ιe(1 + log k) +
2
X
r>Lk q≡−1(mod r)
q<ek
1
q
X
q 0 ≡−1(mod r)
q 0 <ek
1
.
q0
By Lemma 3.5, we have
X
q≡−1(mod r)
q≤ek
where c =
1
k/4−log Lk
X
2(log k + c)
2ι(log k + c)
1
≤
≤
,
q
ϕ(r)
r
− log log(2 − 1/Lk ) because Lk ≤ r < p ≤ e3k/4 Lk . Therefore,
X
r>Lk q≡−1(mod r)
q<ek
1
q
X
q 0 ≡−1(mod r)
q 0 <ek
X 1
1
4ι2 (log k + c)2
2
2
≤
4ι
(log
k
+
c)
≤
.
q0
r2
Lk − 1
r≥Lk
Thus,
(∗)
sk ≤
4ι3 (1 + µk ) (2ιe(1 + log k) + 3/2) (log k + c)2
.
Lk
We can compute
6
10
X
(∗)
sk ≤ 17.1315.
k=K+1
For k >
106 ,
c < 1, and choose a large factor, say 30, to offset the constants, we have
E (∗) ≤ 30
∞
X
(1 + 2 log k)(1 + log k)3
√
e
k=106 +1
24
k/5
Z
∞
≤ 30
(1 + 2 log t)(1 + log t)3
√
e
−77
≤ 3.89548 × 10 .
106
t/5
Thus,
P (∗) ≤ 17.1316.
Putting everything together, we have the result stated in Theorem 1.1:
P = Ps + P (K) + P (∗) +
8
X
P (i) + ε < 4084.
i=1
This upper bound can potentially be decreased further by (1) extending the parity argument
currently used for small amicable numbers to address large amicable numbers, and (2) break
down part 4.8 even further by computing small cases of q0 and m0 .
25
References
[1] J. Bayless and D. Klyve, On the sum of reciprocals of amicable numbers, Integers 11A
(2011), Article 5.
[2] P. Erdős, On amicable numbers, Publ. Math. Debrecen 4 (1955), 108–111.
[3] P. Erdős and G. J. Rieger, Ein Nachtrag über befreundete Zahlen, J. reine angew. Math.
273 (1975), 220.
[4] H. J. Kanold, Über die asymptotische Dicte von gewissen Zahlenmengen, Proc. International Congress of Mathematicians, Amsterdam, 1954, p. 30.
[5] H.L. Montgomery and R.C. Vaughan, The large sieve, Mathematika 20 (1973), 119–134,
[6] Jan Munch Pedersen, Known Amicable Pairs, http://amicable.homepage.dk/knwnc2.htm.
[7] C. Pomerance, On the distribution of amicable numbers, J. reine angew. Math. 293/294
(1977), 217–222.
[8] C. Pomerance, On the distribution of amicable numbers, II, J. reine angew. Math. 325
(1981), 183–188.
[9] C. Pomerance, On amicable numbers. Submitted for publication, 2014.
[10] G. J. Rieger, Bemerkung zu einem Ergebnis von Erdős über befreundete Zahlen, J.
reine angew. Math. 261 (1973), 157–163.
[11] J. Barkley Rosser and Lowell Schoenfeld. Approximate formulas for some functions of
prime numbers. Illinois J. Math., 6(1): 64–94, 1962.
26