# On the General Erdős-Turán Conjecture

Authors: Georges Grekos; Labib Haddad; Charles Helou; Jukka Pihko
Journal: International Journal of Combinatorics
 (2014)

---

## Abstract

The general Erdős-Turán conjecture states that ifA is an infinite, strictly increasing sequence of natural numbers whose general term satisfies a
n
≤
c
n
2, for some constant c
>
0 and for all n, then the number of representations functions of A is unbounded. Here, we introduce the function ψ
(
n
), giving the minimum of the maximal number of representations of a finite sequence A
=
{
a
k
:
1
≤
k
≤
n
} of n natural numbers satisfying a
k
≤
k
2 for all k. We show that ψ
(
n
) is an increasing function of n and that the general Erdős-Turán conjecture is equivalent to l
i
m
n
→
∞
ψ
(
n
)
=
∞. We also compute some values of ψ
(
n
). We further introduce and study the notion of capacity, which is related to the ψ function by the fact that l
i
m
n
→
∞
ψ
(
n
) is the capacity of the set of squares of positive integers, but which is also of intrinsic interest.


---

## Body

## 1. Introduction

In 1941, Erdős and Turán conjectured [1] that if A is an additive basis of the set N
=
{
0,1
,
2
,
…
} of natural numbers, that is, if every n
∈
N is the sum of two elements of A, then the number of representations functions of A, defined by
(1)
r
A
,
n
=
a
,
b
∈
A
×
A
:
a
+
b
=
n
,
n
∈
N
,
is unbounded. As is well known [2], if A
=
{
a
1
<
a
2
<
⋯
<
a
n
<
⋯
} is a basis of N, then its general term satisfies a
n
≤
c
n
2, for a constant c
>
0 and for all positive integers n. So, a few years later, Erdős stated a more general conjecture, known as the general Erdős-Turán conjecture, to the effect that if A is an infinite (strictly increasing) sequence in N whose general term satisfies a
n
≤
c
n
2, then the number of representations functions of A is unbounded (e.g., see [3]).In a previous paper [4], we defined an order relation between infinite (strictly increasing) sequences A
=
{
a
n
:
n
≥
1
} and B
=
{
b
n
:
n
≥
1
} of natural numbers, by setting A
≪
B if a
n
≤
b
n for all n. We called B anErdős-Turán set if the relation A
≪
B implies that the number of representations functions of A is unbounded. Then, denoting by S the set of squares of positive integers, the general Erdős-Turán conjecture amounts to the following.Conjecture 1 (GET).
For any positive integerc, the set c
·
S
=
{
c
s
:
s
∈
S
} is an Erdős-Turán set.Moreover, we showed that (GET) is in fact equivalent to the statement.Conjecture 2 (GET).
S is an Erdős-Turán set.The present paper is a sequel to [4], to which we refer for broader background and further motivation, but the necessary prerequisite material is here recalled as needed, to make the present text self-contained and clear.In Section 3, we define the capacity of an infinite set (identified to a sequence)A of natural numbers as the infimum of the supremum of the number of representations functions of P, for all infinite sequences P such that P
≪
A; that is,
(2)
cap
A
=
inf
⁡
P
≪
A
s
P
,
where
s
(
P
)
=
sup
⁡
n
∈
N
r
(
P
,
n
)
.
It provides a measure of the deviation of A from being an Erdős-Turán set, and it allows to characterize the Erdős-Turán sets as those whose capacity is infinite. The notion of capacity is also of intrinsic interest, and we give several of its properties. We thus prove that if the capacity of A is finite, then there exists a positive integer n such that, for any infinite sequence P
=
{
p
n
:
n
≥
1
} satisfying P
≪
A, we have s
(
{
p
1
,
…
,
p
n
}
)
≥
cap
(
A
). And if the capacity of A is infinite, then, for any natural number e, there exists a positive integer n such that, for any infinite sequence P
=
{
p
n
:
n
≥
1
} satisfying P
≪
A, we have s
(
{
p
1
,
…
,
p
n
}
)
>
e. Moreover, if two infinite sequences A and B are “close”, in the sense that there exists a positive integer d such that |
a
n
-
b
n
|
≤
d for all n, then we have
(3)
cap
(
A
)
4
d
+
1
≤
cap
(
B
)
≤
(
4
d
+
1
)
cap
(
A
)
.
We also establish that the range of the capacity function is the interval [
2
,
∞
] of N
¯
=
N
∪
{
∞
}. There are therefore sets of any finite capacity ≥2, and those are the ones for which the established properties are most significant.In another paper [5], we introduced a function σ
(
n
), giving the minimum of the maximal number of representations of a set A
=
{
a
1
<
a
2
<
⋯
<
a
n
} of n natural numbers forming a basis of the interval [
0
,
a
n
] of N. We showed that σ
(
n
) is an increasing function of n and that the original Erdős-Turán conjecture is equivalent to li
m
n
→
∞
σ
(
n
)
=
∞. Here, in Section 4, we introduce the analogue for the general Erdős-Turán conjecture of σ
(
n
), namely, the function ψ
(
n
), giving the minimum of the maximal number of representations of a set A
=
{
a
1
<
a
2
<
⋯
<
a
n
} of n natural numbers satisfying a
k
≤
k
2 for all k. We show that ψ
(
n
) is an increasing function of n and that
(4)
lim
⁡
n
→
∞
⁡
ψ
(
n
)
=
cap
(
S
)
,
so that the conjecture (GET) is equivalent to the statement.Conjecture 3 (GET).
Considerlim
⁡
n
→
∞
ψ
(
n
)
=
∞.We also give the results of some computations of values ofψ
(
n
).We note that there is an extensive literature related to the Erdős-Turán conjectures, for example, [6–16].


## 2. Notation and Preliminaries

Notation 1.
(1) The set of natural numbers isN
=
{
0,1
,
2
,
…
}, while the set of positive integers is N
*
=
{
1,2
,
…
}.
(2) A subsetA
=
{
a
1
<
a
2
<
⋯
<
a
n
<
⋯
} of N, finite or infinite, is identified with the strictly increasing sequence of its elements (denoted by the corresponding lowercase letter).
(3) For anyn
∈
N, let r
(
A
,
n
)
=
|
{
(
a
,
b
)
∈
A
×
A
:
a
+
b
=
n
}
|, where |
E
| denotes the cardinality of a set E. Further, let s
(
A
)
=
sup
⁡
{
r
(
A
,
n
)
:
n
∈
N
}, considered as an element of N
¯
=
N
∪
{
∞
}.
(4) For anyt
∈
N, let A
[
t
]
=
A
∩
[
0
,
t
] denote an initial segment of A.
(5) For anyt
∈
N, let A
[
t
[
=
A
∩
[
t
,
∞
[ denote a terminal segment of A.
(6) For anyt
∈
N, let t
+
A
=
{
t
+
a
:
a
∈
A
} denote a translate of A.
(7) For anyt
∈
N
*, let t
·
A
=
{
t
a
:
a
∈
A
} denote a homothetic of A.
(8) The set of all infinite subsets ofN is denoted by P.
(9) A partial order relation≪ is defined in P by A
≪
B if a
n
≤
b
n for all n.
(10) An elementC of P is called an Erdős-Turán set if, for any A
∈
P, the relation A
≪
C implies that s
(
A
)
=
∞.
(11) The class of all Erdős-Turán setsC is denoted by C
(
ET
).
(12) The set of squares of positive integers is denoted byS; that is, S
=
{
n
2
:
n
∈
N
*
}.The following results, some of which are in [4], are used several times in the sequel.Lemma 4.
LetA
,
B
∈
P and t
∈
N. (1)
IfB
⊂
A, then A
≪
B.
(2)
In particular,N
≪
A, and A
≪
A
[
t
[.
(3)
Also,A
≪
t
+
A, and A
≪
t
·
A when t
>
0.ForB
⊂
A, b
n
=
a
k
n with 1
≤
k
1
<
k
2
<
⋯, so n
≤
k
n and a
n
≤
b
n for all n.Lemma 5.
For any subsetA of  N and any t
∈
N, one has s
(
t
+
A
)
=
s
(
A
).We haver
(
t
+
A
,
n
)
=
r
(
A
,
n
-
2
t
), for all n, and the result follows.Lemma 6.
Assume thatA has at least two elements. (1)
Ifa
n
+
1
>
2
a
n  for all n, then s
(
A
)
=
2.
(2)
Ifa
1
>
0 and a
n
+
1
≥
2
a
n  for all n, then s
(
A
)
=
2.Proof.
Ifr
(
A
,
x
)
>
2 for some integer x, there would exist indices i
<
m
≤
n
<
j such that a
i
+
a
j
=
a
m
+
a
n
=
x. Therefore
(5)
a
n
+
1
≤
a
j
≤
a
m
+
a
n
≤
2
a
n
.
Moreover, if a
i
>
0, then the inequality becomes
(6)
a
n
+
1
≤
a
j
<
a
m
+
a
n
≤
2
a
n
.
In either case, this yields a contradiction.
Thuss
(
A
)
≤
2, and since |
A
|
≥
2, s
(
A
)
=
2.Example 7.
LetA
=
{
2
n
:
n
∈
N
}. Then, by Lemma 6, we have s
(
A
)
=
2.Notation 2.
For any subsetsA and B of N, and any n
∈
N, let (1)
R
(
A
,
B
;
n
)
=
{
(
a
,
b
)
∈
A
×
B
:
a
+
b
=
n
}, r
(
A
,
B
;
n
)
=
|
R
(
A
,
B
;
n
)
|,
(2)
s
(
A
,
B
)
=
sup
⁡
{
r
(
A
,
B
;
n
)
:
n
∈
N
}, in N
¯,
(3)
R
(
A
,
n
)
=
R
(
A
,
A
;
n
); so r
(
A
,
n
)
=
r
(
A
,
A
;
n
)
=
|
R
(
A
,
n
)
|, s
(
A
)
=
s
(
A
,
A
),
(4)
An element(
a
,
b
) of R
(
A
,
B
;
n
) will be called an (
A
,
B
)-representation of n.The following properties can be directly verified.Lemma 8.
For all subsetsA and B of  N, and all n
∈
N, one has(1)
r
(
B
,
A
;
n
)
=
r
(
A
,
B
;
n
),
(2)
s
(
B
,
A
)
=
s
(
A
,
B
),
(3)
R
(
A
∪
B
,
n
)
=
R
(
A
,
n
)
∪
R
(
B
,
n
)
∪
R
(
A
,
B
;
n
)
∪
R
B
,
A
;
n,
(4)
max
⁡
(
r
(
A
,
n
)
,
r
(
B
,
n
)
,
r
(
A
,
B
;
n
)
)
≤
r
(
A
∪
B
,
n
)
≤
r
(
A
,
n
)
+
r
(
B
,
n
)
+
2
r
(
A
,
B
;
n
),
(5)
max
⁡
(
s
(
A
)
,
s
(
B
)
,
s
(
A
,
B
)
)
≤
s
(
A
∪
B
)
≤
s
(
A
)
+
s
(
B
)
+
2
s
(
A
,
B
),
(6)
∑
k
=
0
n
‍
r
(
A
,
B
;
k
)
·
r
(
A
,
B
;
n
-
k
)
=
∑
k
=
0
n
‍
r
(
A
,
k
)
·
r
(
B
,
n
-
k
)
≤
(
n
+
1
)
s
(
A
)
s
(
B
)   (both sums being equal to |
{
(
a
,
b
,
a
′
,
b
′
)
∈
A
×
B
×
A
×
B
:
a
+
b
+
a
′
+
b
′
=
n
}
|
).Lemma 9.
LetF and G be nonempty subsets of N such that F is finite with largest element m and G
=
{
g
1
<
g
2
<
⋯
} is either finite or infinite, satisfying (1)
g
1
>
2
m,
(2)
g
n
+
1
>
2
g
n, for all n.Then s
(
F
∪
G
)
=
max
⁡
(
s
(
F
)
,
2
).Proof.
SetH
=
F
∪
G. Since m is the largest element in F, we have
(7)
F
+
F
⊂
0,2
m
.
And, since g
1
>
2
m, we also have
(8)
H
+
G
⊂
2
m
,
∞
.
For n
∈
[
0,2
m
], we therefore have
(9)
r
(
H
,
n
)
=
r
(
F
,
n
)
≤
s
(
F
)
.
Givenf
∈
F, set g
0
=
f. We have g
1
>
2
g
0. Thus s
{
f
}
∪
G
=
2, due to assumption (
2
) and Lemma 6.
Iff
+
g
=
f
′
+
g
′, with f
,
f
′ in F, and g
≤
g
′ in G, then f
=
f
′ and g
=
g
′; otherwise, we would have g
<
g
′ and
(10)
f
-
f
′
=
g
′
-
g
>
g
>
2
m
,
which is impossible. This means that any integer n has at most one (
F
,
G
)-representation. Moreover, if it has an (
F
,
G
)-representation (
f
,
g
), then it cannot have a (
G
,
G
)-representation; otherwise we would have s
{
f
}
∪
G
>
2.
Hence, in view of Lemma8, and since H
×
H is the union of four pairwise disjoint product sets, with H replaced by F or G, we have
(11)
r
H
,
n
=
r
F
,
n
+
r
G
,
n
+
2
r
F
,
G
;
n
=
max
⁡
(
r
(
F
,
n
)
,
r
(
G
,
n
)
,
2
r
(
F
,
G
;
n
)
)
≤
max
⁡
(
s
(
F
)
,
2
)
.
The result then follows.


## 3. The Capacity Function

Definition 10.
For anyA
∈
P, the capacity of A is the element of N
¯ defined by
(12)
cap
(
A
)
=
inf
⁡
{
s
(
P
)
:
P
∈
P
,
P
≪
A
}
.Remark 11.
For anyA
∈
P, the capacity of A is an attained minimum. More precisely, there exists P
∈
P such that
(13)
P
≪
A
,
cap
(
A
)
=
s
(
P
)
=
cap
(
P
)
.Indeed, ifcap
(
A
)
=
∞, then, for every P
≪
A in P, we have
(14)
cap
(
A
)
=
s
(
P
)
=
cap
(
P
)
=
∞
.Ifcap
(
A
) is finite, then at least one of the s
(
P
)’s, for P
≪
A in P, is finite, and the set
(15)
{
s
(
P
)
:
P
∈
P
,
P
≪
A
,
s
(
P
)
<
∞
}
⊂
N
has smallest element m. Taking one of those P
≪
A such that s
(
P
)
=
m, we have
(16)
s
(
P
)
≥
cap
(
P
)
≥
cap
(
A
)
=
s
(
P
)
,
in view of Lemma 12, (
3
).Lemma 12.
For anyA
,
B
∈
P and any t
∈
N, one has(1)
2
≤
cap
(
A
)
≤
s
(
A
),
(2)
cap
(
A
)
=
∞ if and only if A
∈
C
(
ET
),
(3)
ifA
≪
B then cap
(
B
)
≤
cap
(
A
),
(4)
ifB
⊂
A then cap
(
B
)
≤
cap
(
A
),
(5)
cap
(
A
[
t
[
)
≤
cap
(
A
),
(6)
cap
(
t
+
A
)
≤
cap
(
A
),
(7)
ift
>
0, then cap
(
t
·
A
)
≤
cap
(
A
).Proof.
These properties follow from the definitions and Lemma4.The following notion ofcaliber was introduced and studied in [4, item 5.5].Definition 13.
For anyA
∈
P, the caliber of A is the element of
(17)
R
¯
+
=
{
x
∈
R
:
x
≥
0
}
∪
{
∞
}
defined by
(18)
cal
(
A
)
=
lim
⁡
inf
⁡
n
→
∞
a
n
n
2
.We recall the following result from [4, item 5.9] that will be needed in the proof of the lemma following it.Lemma 14.
For anyA
∈
P, one has
(19)
s
(
A
)
≥
1
2
·
cal
(
A
)
.Lemma 15.
For anyA
∈
P, one has
(20)
cap
(
A
)
≥
1
2
·
cal
(
A
)
.Proof.
For anyP
∈
P such that P
≪
A, we have, from the definition,
(21)
0
≤
cal
(
P
)
≤
cal
(
A
)
.
Moreover, by the previous lemma, we have
(22)
s
(
P
)
≥
1
2
·
cal
(
P
)
≥
1
2
·
cal
(
A
)
.
Therefore cap
(
A
), which is the infimum of all such s
(
P
), satisfies
(23)
cap
(
A
)
≥
1
2
·
cal
(
A
)
.Remark 16.
It follows from Lemma15 that if A
∈
P has zero caliber, that is, if cal
(
A
)
=
0, then cap
(
A
)
=
∞; that is, A is an Erdős-Turán set. Thus, for example, if
(24)
A
=
n
2
-
ϵ
:
n
∈
N
*
,
with
0
<
ϵ
<
2
,
where [
x
] is the integer part of a real number x, then cal
(
A
)
=
0 and therefore A is an Erdős-Turán set. In particular, any arithmetic progression is an Erdős-Turán set.
The infinite setsA of natural numbers for which cal
(
A
)
=
0 are the only known Erdős-Turán sets, because, in [4], we proved that the general Erdős-Turán conjecture (GET) is equivalent to the statement that there exists a subset A of N such that A is an Erdős-Turán set and cal
(
A
)
≠
0 [4, item 5.16].Remark 17.
For anye
,
s
∈
N
¯ such that 2
≤
e
≤
s, there exists A
∈
P such that cap
(
A
)
≤
e and s
(
A
)
=
s.
In particular, there existsA
∈
P such that cap
(
A
)
=
2 and s
(
A
)
=
∞, which is a nontrivially finite capacity set. We provide, below, an explicit example of such a set.Indeed, sincee
≥
2, there exists P
∈
P such that s
(
P
)
≤
e (e.g., P
=
{
2
n
:
n
∈
N
} for which s
(
P
)
=
2, as in Example 7). Then, by [4, item 2.3], there exists A
∈
P such that P
≪
A and s
(
A
)
=
s. Hence cap
(
A
)
≤
s
(
P
)
≤
e, while s
(
A
)
=
s.In particular, takinge
=
2 and s
=
∞ yields the latter special case. We next proceed to explicitly construct such a set A
∈
P having s
(
A
)
=
∞ and cap
(
A
)
=
2, following the procedure described in the proof of [4, item 2.3].We start with the set(25)
P
=
2
n
:
n
∈
N
=
p
n
:
n
∈
N
*
,
where
p
n
=
2
n
-
1
for
n
∈
N
*
.
For any n
∈
N
*, let m
,
r be the unique natural numbers such that
(26)
n
=
2
m
+
r
,
0
≤
r
<
2
m
,
i
.
e
.
,
2
m
≤
n
<
2
m
+
1
,
r
=
n
-
2
m
,
and let
(27)
a
n
=
p
2
m
+
1
+
r
=
2
2
m
+
1
-
1
+
r
.
Then
(28)
a
n
+
1
=
a
n
+
1
>
a
n
,
if
2
m
≤
n
<
2
m
+
1
-
1
,
while if n
=
2
m
+
1
-
1
=
2
m
+
(
2
m
-
1
), then
(29)
a
n
+
1
=
a
2
m
+
1
=
2
2
m
+
2
-
1
>
a
n
=
2
2
m
+
1
-
1
+
2
m
-
1
.
Thus
(30)
A
=
{
a
1
<
a
2
<
⋯
<
a
n
<
⋯
}
∈
P
.
Moreover, for n
=
2
m
+
r, with 0
≤
r
<
2
m,
(31)
p
n
=
2
2
m
+
r
-
1
≤
2
2
m
+
1
-
2
<
a
n
=
2
2
m
+
1
-
1
+
r
.
So P
≪
A, and therefore
(32)
cap
(
A
)
≤
s
(
P
)
=
2
,
i
.
e
.
,
cap
(
A
)
=
2
.
Finally, if
(33)
k
=
2
p
2
m
+
1
+
r
=
2
2
m
+
1
+
r
,
with
0
≤
r
<
2
m
,
then
(34)
k
=
p
2
m
+
1
+
q
+
p
2
m
+
1
+
r
-
q
=
a
2
m
+
q
+
a
2
m
+
r
-
q
,
mmmmmmmmmmmmmmmmm
for
0
≤
q
≤
r
,
so that r
(
A
,
k
)
≥
r
+
1. In particular, taking r
=
2
m
-
1, we get
(35)
r
A
,
2
2
m
+
1
+
2
m
-
1
≥
2
m
,
for
any
m
∈
N
.
Hence s
(
A
)
=
∞.Proposition 18.
IfC
∈
P and e
∈
N are such that cap
(
C
)
>
e, then there exists an integer m
∈
N
* such that, for any A
∈
P satisfying A
≪
C, one has s
(
{
a
1
,
…
,
a
n
}
)
>
e for all n
≥
m.Proof.
We argue by contradiction. Assume that, for everyn
∈
N
*, the set
(36)
F
n
=
{
A
∈
P
:
A
≪
C
,
s
(
{
a
1
,
…
,
a
n
}
)
≤
e
}
is not empty. Clearly, F
n
+
1
⊂
F
n. For m
,
n
∈
N
*, let
(37)
F
n
,
m
=
{
(
a
1
,
…
,
a
m
)
:
A
∈
F
n
}
.
We have F
n
,
m
≠
∅ and F
n
+
1
,
m
⊂
F
n
,
m for all m
,
n
∈
N
*, in view of the corresponding properties of the F
n’s. Moreover, each F
n
,
m is finite since it is contained in ∏
k
=
1
m
‍
N
[
c
k
]. We thus have, for every m
∈
N
*, a decreasing sequence of finite nonempty sets (
F
n
,
m
)
n
∈
N
*, which is therefore stationary; that is, there exists some n
m
∈
N
* such that, for all n
≥
n
m, we have F
n
,
m
=
F
n
m
,
m, and therefore the set
(38)
F
m
=
⋂
n
∈
N
*
‍
F
n
,
m
=
F
n
m
,
m
≠
∅
.
Now, consider an m
∈
N
* and fix an n
≥
max
⁡
(
n
m
,
n
m
+
1
). Note that if
(39)
(
a
1
,
…
,
a
m
+
1
)
∈
F
m
+
1
=
F
n
,
m
+
1
,
then
(40)
(
a
1
,
…
,
a
m
)
∈
F
n
,
m
=
F
m
.
Conversely, given any (
a
1
,
…
,
a
m
)
∈
F
m, there is some A
∈
F
n whose first m terms are the given ones in F
m
=
F
n
,
m, and then, since n
≥
n
m
+
1, the first m
+
1 terms of A lie in F
n
,
m
+
1
=
F
m
+
1.
ThereforeF
m consists exactly of the first m terms of the elements of F
m
+
1. Hence the existence, by induction on m, of a set A
∈
P such that A
≪
C and (
a
1
,
…
,
a
m
)
∈
F
m for all m
∈
N
*, so that s
(
{
a
1
,
…
,
a
m
}
)
≤
e for all m
∈
N
*. It follows that
(41)
s
(
A
)
≤
e
<
cap
(
C
)
,
while A
≪
C, contradicting the definition of cap
(
C
).
So the original assumption does not hold; that is, there exists somem
∈
N
* such that F
m is empty, which means that for all A
∈
P satisfying A
≪
C we have s
(
{
a
1
,
…
,
a
m
}
)
>
e, and thus s
(
{
a
1
,
…
,
a
n
}
)
>
e for all n
≥
m.Corollary 19.
LetC
∈
P.(1)
IfC
∈
C
(
ET
), then for every e
∈
N, there exists an integer m
∈
N
* such that, for any A
∈
P satisfying A
≪
C, we have s
(
{
a
1
,
…
,
a
n
}
)
>
e for all n
≥
m.
(2)
Ifcap
(
C
)
<
∞, then there exists an integer m
∈
N
* such that, for any A
∈
P satisfying A
≪
C, we have s
(
{
a
1
,
…
,
a
n
}
)
≥
cap
(
C
) for all n
≥
m.Proof.
IfC
∈
C
(
ET
), then cap
(
C
)
=
∞ and Proposition 18 applies with any e
∈
N to give (
1
).
Ifcap
(
C
)
<
∞, then Proposition 18 applies with e
=
cap
(
C
)
-
1 to give (
2
).Lemma 20.
For every integere
≥
2, as well as for e
=
∞, there exists some set A
∈
P such that cap
(
A
)
=
s
(
A
)
=
e.
Thus the range of the capacity function is the closed interval[
2
,
∞
] of  N
¯.Proof.
For a given integere
≥
2, let
(42)
A
=
{
0,1
,
…
,
e
-
1
}
∪
{
2
2
n
+
e
:
n
∈
N
*
}
.
Thus A is the union of the finite interval
(43)
F
=
N
[
e
-
1
]
=
{
0,1
,
…
,
e
-
1
}
⊂
N
,
for which s
(
F
)
=
e, and of the infinite set G with elements
(44)
g
n
=
2
2
n
+
e
,
n
∈
N
*
.
Since
(45)
g
1
=
2
e
+
2
>
2
(
e
-
1
)
,
where e
-
1 is the largest element of F, and since, for all n
∈
N
*, we have
(46)
g
n
+
1
=
2
2
n
+
e
+
2
>
2
2
n
+
e
+
1
=
2
g
n
,
Lemma 9 applies and gives
(47)
s
(
A
)
=
max
⁡
(
e
,
2
)
=
e
.
Moreover, ifP
∈
P is such that P
≪
A, then, by definition and in view of Lemma 4, we have
(48)
n
-
1
≤
p
n
≤
a
n
,
∀
n
∈
N
*
.
In particular, for 1
≤
n
≤
e, we get
(49)
n
-
1
≤
p
n
≤
a
n
=
n
-
1
,
so that
(50)
F
=
N
[
e
-
1
]
⊂
P
and therefore
(51)
s
(
P
)
≥
s
(
F
)
=
e
=
s
(
A
)
.
It follows that
(52)
cap
(
A
)
=
inf
⁡
{
s
(
P
)
:
P
∈
P
,
P
≪
A
}
=
s
(
A
)
=
e
.
Fore
=
∞, we may take A
=
N. Indeed, N being the smallest element of P for the order ≪, the definition immediately yields
(53)
cap
(
N
)
=
s
(
N
)
=
∞
.Lemma 21.
For anyA
∈
P, one has
(54)
cap
(
A
)
-
2
≤
cap
A
a
2
≤
cap
(
1
+
A
)
≤
cap
(
A
)
.Proof.
As the sequence(
a
n
) is strictly increasing, the inequalities
(55)
a
n
≤
1
+
a
n
≤
a
n
+
1
hold for all n
∈
N
*, and they amount to
(56)
A
≪
1
+
A
≪
A
a
2
.
So, by Lemma 12,
(57)
cap
(
A
a
2
)
≤
cap
(
1
+
A
)
≤
cap
(
A
)
.
Now, letQ
∈
P be such that
(58)
Q
≪
A
a
2
,
i
.
e
.
,
q
n
≤
a
n
+
1
∀
n
∈
N
*
.
There are two possible cases.(i)
Eitherq
n
≤
a
n for all n
∈
N
*, that is, Q
≪
A, and thus s
(
Q
)
≥
cap
(
A
).
(ii)
Or there existsn
∈
N
* such that q
n
>
a
n, and we let m be the smallest such n, so that
(59)
q
n
≤
a
n
for
1
≤
n
≤
m
-
1
,
a
m
<
q
m
≤
a
m
+
1
,
q
n
≤
a
n
+
1
for
n
≥
m
.In this case, we define a sequence (
p
n
)
n
∈
N
* by
(60)
p
n
=
q
n
if
1
≤
n
≤
m
-
1
,
a
m
if
n
=
m
,
q
n
-
1
if
n
≥
m
+
1
.
Since the sequence (
q
n
) is strictly increasing and since
(61)
p
m
-
1
=
q
m
-
1
≤
a
m
-
1
<
a
m
=
p
m
<
q
m
=
p
m
+
1
(assuming m
≥
2 for the first two inequalities, otherwise they are unnecessary), the sequence (
p
n
) is also strictly increasing; that is, P
=
{
p
n
:
n
∈
N
*
} lies in P. Also, from the definition of (
p
n
), we have p
n
≤
a
n for all n
∈
N
*; that is, P
≪
A, so that s
(
P
)
≥
cap
(
A
). Moreover, P
=
Q
∪
{
a
m
} and therefore, by [4, item 2.9], s
(
P
)
≤
s
(
Q
)
+
2. Hence
(62)
s
(
Q
)
≥
s
(
P
)
-
2
≥
cap
(
A
)
-
2
.
Thus, in any case, and for allQ
∈
P such that Q
≪
A
[
a
2
[, we have
(63)
s
(
Q
)
≥
cap
(
A
)
-
2
.
It follows that
(64)
cap
(
A
a
2
)
≥
cap
(
A
)
-
2
.Corollary 22.
For anyA
∈
P and for any n
∈
N
*, one has
(65)
cap
(
A
)
-
2
n
≤
cap
(
A
a
n
+
1
)
≤
cap
(
n
+
A
)
≤
cap
(
A
)
.Proof.
Forn
=
1, the stated inequalities are given by Lemma 21. For general n, the result follows by induction on n.Remark 23.
For anyA
∈
P, there exists m
∈
N
*, such that, for all n
≥
m, we have
(66)
cap
(
n
+
A
)
=
cap
(
m
+
A
)
,
cap
(
A
a
n
)
=
cap
(
A
a
m
)
.Indeed, first, ifcap
(
A
)
=
∞, then, in view of Lemma 12, A
∈
C
(
ET
), and therefore, by [4, item 2.14] and [4, item 2.16], both n
+
A
∈
C
(
ET
) and A
[
a
n
[
∈
C
(
ET
), so that
(67)
cap
(
A
a
n
)
=
cap
(
n
+
A
)
=
cap
(
A
)
=
∞
,
∀
n
∈
N
*
.Second, ifcap
(
A
)
<
∞, then (
cap
(
n
+
A
)
)
n
∈
N
* and (
cap
(
A
[
a
n
[
)
)
n
∈
N
* are decreasing sequences in N
*, since if m
≤
n, then
(68)
A
≪
m
+
A
≪
n
+
A
,
A
≪
A
a
m
≪
A
a
n
,
and therefore, in view of Lemma 12,
(69)
cap
n
+
A
≤
cap
m
+
A
≤
cap
A
,
cap
(
A
a
n
)
≤
cap
(
A
a
m
)
≤
cap
(
A
)
.
Thus, the two sequences are stationary.The following definition is taken from [4, 3.1].Definition 24.
ForA
,
B
∈
P, one sets
(70)
δ
(
A
,
B
)
=
sup
⁡
a
n
-
b
n
:
n
∈
N
*
,
in
N
¯
.
Ifδ
(
A
,
B
)
≤
d for some d
∈
N
*, we say that A and B are d
-close.
More generally, ifδ
(
A
,
B
)
<
∞, we say that A and B areclose.Lemma 25.
LetA
,
B
∈
P and d
∈
N
*. If A and B are d-close, then
(71)
cap
(
A
)
4
d
+
1
≤
cap
(
B
)
≤
(
4
d
+
1
)
cap
(
A
)
.Proof.
LetP
∈
P be such that P
≪
A. Define a sequence (
q
n
)
n
∈
N
* in N by
(72)
q
n
=
max
⁡
n
-
1
,
p
n
-
d
,
∀
n
∈
N
*
.
Since the sequences (
p
n
-
d
)
n
∈
N
* and (
n
-
1
)
n
∈
N
* are strictly increasing, so is (
q
n
)
n
∈
N
*; that is,
(73)
Q
=
{
q
n
:
n
∈
N
*
}
∈
P
.
Since A and B are d-close, and P
≪
A, we have
(74)
p
n
-
d
≤
a
n
-
d
≤
b
n
,
and since n
-
1
≤
b
n (Lemma 4), q
n
≤
b
n, for all n
∈
N
*. So Q
≪
B, and s
(
Q
)
≥
cap
(
B
). Moreover, from the definition of q
n (and using Lemma 4 again),
(75)
p
n
-
d
≤
q
n
≤
p
n
,
∀
n
∈
N
*
,
so that P and Q are d-close. It then follows from [4, item 3.3] that
(76)
s
(
P
)
≥
s
Q
4
d
+
1
≥
cap
B
4
d
+
1
.
Thus(77)
s
P
≥
cap
B
4
d
+
1
,
∀
P
∈
P
such
that
P
≪
A
,
and therefore
(78)
cap
(
A
)
≥
cap
B
4
d
+
1
.
Furthermore, by exchange of A and B, we similarly get
(79)
cap
(
B
)
≥
cap
A
4
d
+
1
.Corollary 26.
For anyA
,
B
∈
P, one has (1)
cap
(
B
)
≤
(
4
δ
(
A
,
B
)
+
1
)
cap
(
A
) and cap
(
A
)
≤
(
4
δ
(
A
,
B
)
+
1
)
cap
(
B
),
(2)
ifcap
(
A
) and cap
(
B
) are finite, then
(80)
|
cap
(
A
)
-
cap
(
B
)
|
≤
4
min
⁡
(
cap
(
A
)
,
cap
(
B
)
)
·
δ
(
A
,
B
)
.Proof.
Ifδ
(
A
,
B
)
=
∞, then all the inequalities hold trivially. We may thus assume that δ
(
A
,
B
)
=
d
<
∞, with d
∈
N
*, so that A and B are d-close.
The inequalities in(
1
) immediately follow from Lemma 25.
Assumingcap
(
A
) and cap
(
B
) are finite, we deduce from (
1
) the two inequalities
(81)
cap
B
-
cap
A
≤
4
d
·
cap
A
,
cap
(
A
)
-
cap
(
B
)
≤
4
d
·
cap
(
B
)
,
of which one is precisely the inequality in (
2
).Lemma 27.
For anyA
∈
P and any x
∈
N, one has
(82)
cap
(
A
)
≤
cap
(
A
∪
{
x
}
)
≤
cap
(
A
)
+
2
.Proof.
LetB
=
A
∪
{
x
}. The case where x
∈
A being trivial, we assume that x
∉
A. Let m
∈
N
* be the smallest positive integer such that a
m
>
x, so that a
m
-
1
<
x
<
a
m, where if m
=
1, we set a
0
=
-
1. Thus
(83)
b
n
=
a
n
if
1
≤
n
≤
m
-
1
,
x
if
n
=
m
,
a
n
-
1
if
n
≥
m
+
1
.
Since A
⊂
B, by Lemma 12, cap
(
A
)
≤
cap
(
B
), which yields the first inequality.
On the other hand, by Remark11, there exists P
∈
P such that P
≪
A and s
(
P
)
=
cap
(
A
). Thus, to prove the second inequality, it is enough to prove the existence of a set Q
∈
P such that Q
≪
B and s
(
Q
)
≤
s
(
P
)
+
2. To this end, we distinguish three cases.
(i) Ifp
m
≤
x and p
n
+
1
≤
a
n for n
≥
m, then P
≪
B, and we may take Q
=
P.
(ii) Ifx
<
p
m, then
(84)
p
m
-
1
≤
a
m
-
1
<
x
<
p
m
≤
a
m
,
and we may take Q
=
P
∪
{
x
}.
(iii) Ifp
m
≤
x and there exists n
≥
m such that a
n
<
p
n
+
1, let k
∈
N
* be the smallest such n, so that
(85)
p
n
+
1
≤
a
n
for
m
≤
n
≤
k
-
1
,
a
k
<
p
k
+
1
.
In particular,
(86)
p
k
≤
a
k
-
1
<
a
k
<
p
k
+
1
,
so that
(87)
p
k
<
p
k
+
1
≤
a
k
<
p
k
+
1
,
and we may take Q
=
P
∪
{
p
k
+
1
}.
In all three cases, we have(88)
P
⊂
Q
⊂
P
∪
y
,
with
y
∈
N
,
and therefore, by [4, item 2.9], s
(
Q
)
≤
s
(
P
)
+
2. Since, in addition, Q
≪
B, we conclude that
(89)
cap
(
B
)
≤
s
(
Q
)
≤
s
(
P
)
+
2
=
cap
(
A
)
+
2
.Corollary 28.
For anyA
∈
P and any finite subset F of N, one has
(90)
cap
(
A
)
≤
cap
(
A
∪
F
)
≤
cap
(
A
)
+
2
F
.Proof.
This follows from Lemma27 by induction on the cardinality of F.


## 4. The Functionψ and an Equivalent Formulation of (GET)

Definition 29.
Forn
∈
N
*, let A
n denote the set of all subsets A of N satisfying
(91)
|
A
|
=
n
,
A
=
{
a
1
<
a
2
<
⋯
<
a
n
}
,
a
k
≤
k
2
for
1
≤
k
≤
n
.
Also, define a functionψ
:
N
*
→
N
* by
(92)
ψ
(
n
)
=
min
⁡
{
s
(
A
)
:
A
∈
A
n
}
,
for
n
∈
N
*
.Lemma 30.
The following properties hold.(1)
The functionψ
:
N
*
→
N
* is increasing
(2)
It has a limitκ in N
¯ satisfying
(93)
κ
=
lim
⁡
n
→
∞
ψ
(
n
)
=
sup
⁡
{
ψ
(
n
)
:
n
∈
N
*
}
≤
cap
(
S
)
.Proof.
For anyn
∈
N
* and any B
∈
A
n
+
1, the set A
=
B
∖
{
max
⁡
B
} is a subset of B lying in A
n, so that s
(
B
)
≥
s
(
A
)
≥
ψ
(
n
). Hence
(94)
ψ
(
n
+
1
)
=
min
⁡
s
B
:
B
∈
A
n
+
1
≥
ψ
(
n
)
.
Thus ψ is an increasing function.
The sequence(
ψ
(
n
)
)
n
∈
N
* is increasing in N and is thus either stationary or unbounded, so that
(95)
κ
=
lim
⁡
n
→
∞
ψ
(
n
)
=
sup
⁡
{
ψ
(
n
)
:
n
∈
N
*
}
exists in N
¯. Moreover, for any A
∈
P such that A
≪
S, the truncated set
(96)
A
∣
n
=
{
a
1
<
a
2
<
⋯
<
a
n
}
,
consisting of the first n elements of A, lies in A
n. Therefore
(97)
s
A
≥
s
A
∣
n
≥
ψ
n
,
∀
n
∈
N
*
,
and thus s
(
A
)
≥
κ. Hence
(98)
cap
(
S
)
=
inf
⁡
{
s
(
A
)
:
A
∈
P
,
A
≪
S
}
≥
κ
.We recall the “diagonal lemma” and part of its corollary from [5, items 2.2 and 2.3] that will be used in the proof of the following proposition.Lemma 31 (diagonal lemma and corollary).
LetP
i
i
∈
I be a family of subsets P
i of  N indexed by an infinite set I. There exists a subset A of  N, called a diagonal of the family P
i
i
∈
I, satisfying the following condition.(DL)
For anyn
∈
N, there are infinitely many indices i
∈
I such that A
[
n
]
=
P
i
[
n
].Moreover,(CDL)
Ifs
(
P
i
)
≤
s, for some s
∈
N
¯ and all i
∈
I, then s
(
A
)
≤
s.Proposition 32.
One hasκ
=
cap
(
S
); that is,
(99)
lim
⁡
n
→
∞
ψ
n
=
sup
⁡
⁡
ψ
n
:
n
∈
N
*
=
inf
⁡
{
s
(
A
)
:
A
∈
P
,
A
≪
S
}
.Proof.
For everyn
∈
N
*, let A
n
∈
A
n be such that s
(
A
n
)
=
ψ
(
n
).
By the diagonal lemma, since(
A
n
)
n
∈
N
* is a family of subsets of N, indexed by the infinite set N
*, there exists a subset A
=
{
a
1
<
a
2
<
⋯
<
a
k
<
⋯
} of N (called a diagonal of this family) satisfying the condition.
(DL) For anyx
∈
N, there exists an infinite subset I
x of N
* such that A
n
[
x
]
=
A
[
x
] for every n
∈
I
x.
Moreover, sinces
(
A
n
)
=
ψ
(
n
)
≤
κ for all n
∈
N
*, we have s
(
A
)
≤
κ, by (CDL) in the previous lemma.
Furthermore, for everyk
∈
N
*, taking x
=
a
k and n
∈
I
x, we get
(100)
A
n
[
a
k
]
=
A
[
a
k
]
=
{
a
1
<
a
2
<
⋯
<
a
k
}
,
and since A
n
∈
A
n, we conclude that a
k
≤
k
2. Therefore A
≪
S.
Thus(101)
cap
(
S
)
≤
s
(
A
)
≤
κ
,
and since, by Lemma 30, κ
≤
cap
(
S
), it follows that cap
(
S
)
=
κ.Corollary 33.
The conjecture (GET) is equivalent to the following condition:
(102)
GET
ψ
:
lim
⁡
n
→
∞
⁡
ψ
(
n
)
=
∞
.Proof.
By [4, item 4.14], (GET) is equivalent to the condition S
∈
C
(
ET
), which, by Lemma 12, is equivalent to cap
(
S
)
=
∞, which, by Proposition 32, amounts to κ
=
∞.Remark 34.
The functionψ is the analogue, for the conjecture (GET), of the function σ, for the conjecture (ET), introduced in [5, item 3.12].
Similarly, the statement (GETψ), equivalent to (GET), is the analogue of the statement (ET σ), equivalent to (ET).Indeed, we just need to recall some notations and results.(i) For anyx
∈
N, B
(
x
) denotes the set of all subsets P of N, called bases of N
[
x
], satisfying P
⊂
N
[
x
]
⊂
P
+
P.(ii) For anyn
∈
N
*, B
(
#
n
) denotes the set of all the subsets P of N, called finite bases of cardinality n, satisfying |
P
|
=
n and P
∈
B
(
max
⁡
P
).(iii) Furthermore,B
(
N
) denotes the set of all subsets P of N, called bases of N, satisfying P
+
P
=
N.(iv) The functionσ
:
N
*
→
N
* is defined, in [5], by
(103)
σ
(
n
)
=
min
⁡
{
s
(
P
)
:
P
∈
B
(
#
n
)
}
,
for
n
∈
N
*
.
It is an increasing function, like ψ.(v) In [5], we proved that
(104)
lim
⁡
n
→
∞
σ
(
n
)
=
Λ
=
inf
⁡
{
s
(
P
)
:
P
∈
B
(
N
)
}
,
in
N
¯
,
and that the conjecture (ET) is equivalent to
(105)
ET
σ
:
lim
⁡
n
→
∞
⁡
σ
(
n
)
=
∞
.(vi) Here, we proved that(106)
lim
⁡
n
→
∞
ψ
(
n
)
=
κ
=
inf
⁡
{
s
(
P
)
:
P
∈
P
,
P
≪
S
}
,
in
N
¯
,
and that the conjecture (GET) is equivalent to
(107)
GET
ψ
:
lim
⁡
n
→
∞
⁡
ψ
(
n
)
=
∞
.Hence the analogy.Lemma 35.
The following properties hold.(1)
IfA
=
{
a
1
<
a
2
<
⋯
<
a
n
} lies in B
(
#
n
), then
(108)
a
k
≤
9
16
k
2
,
f
o
r
1
≤
k
≤
n
.
(2)
For anyn
∈
N
*, we have B
(
#
n
)
⊂
A
n.
(3)
For anyn
∈
N
*, we have ψ
(
n
)
≤
σ
(
n
).Proof.
LetA
∈
B
(
#
n
), so that A
=
{
a
1
<
a
2
<
⋯
<
a
n
} and A
∈
B
(
a
n
). Then, for every 1
≤
k
≤
n, the truncated set
(109)
A
k
=
A
∣
k
=
{
a
1
<
a
2
<
⋯
<
a
k
}
∈
B
(
a
k
)
.
Therefore, by [5, item 3.15], we have
(110)
k
=
A
k
≥
8
a
k
+
9
-
1
2
,
so that
(111)
a
k
≤
k
(
k
+
1
)
2
-
1
≤
9
16
k
2
<
k
2
.
In particular, A
∈
A
n. Hence B
(
#
n
)
⊂
A
n, and therefore
(112)
ψ
n
=
min
⁡
⁡
s
A
:
A
∈
A
n
≤
min
⁡
{
s
(
A
)
:
A
∈
B
(
#
n
)
}
=
σ
(
n
)
.Corollary 36.
One hasκ
≤
Λ; that is,
(113)
lim
⁡
n
→
∞
ψ
(
n
)
≤
lim
⁡
n
→
∞
σ
(
n
)
.Corollary 37.
IfA is an additive basis of N, then
(114)
a
n
≤
9
16
n
2
<
n
2
,
∀
n
∈
N
*
,
so that A
≪
S.Remark 38.
Either Corollary36 or Corollary 37 confirms that (GET) implies (ET), as stated in Section 1.Remark 39.
It would be interesting to find a large lower bound forκ. However, the function ψ seems to increase very slowly. We found by computer calculations that
(115)
ψ
(
n
)
=
2
,
for
2
≤
n
≤
22
,
and that
(116)
2
≤
ψ
(
n
)
≤
4
,
for
23
≤
n
≤
435
.Indeed, we have(117)
ψ
(
1
)
=
1
,
ψ
(
2
)
=
2
,
ψ
(
3
)
=
2
.Moreover, the set(118)
A
=
0,1
,
3,8
,
14,18,40,49,68,91
,
112,124
,
167
,
A
i
=
183,219,253
,
280,300,338,400,425,478
lies in A
22 and satisfies s
(
A
)
=
2, so that
(119)
ψ
(
n
)
=
2
,
for
2
≤
n
≤
22
.Also, the following setB lies in A
435 and satisfies s
(
B
)
=
4, so that
(120)
2
≤
ψ
(
n
)
≤
4
,
for
23
≤
n
≤
435
.
This set B was constructed recursively by setting b
1
=
0 and then systematically taking b
n
+
1 to be the least integer >

b
n satisfying the two conditions
(121)
b
n
+
1
≤
n
+
1
2
,
s
b
1
<
b
2
<
⋯
<
b
n
+
1
≤
4
;
(122)
B
=
0,1
,
2,3
,
5,7
,
11,15,23,31,40,47,48,65,72,98,99
,
B
i
=
114,133,147,156,190,199,222,233,272,304
,
B
i
=
328,356,385
,
411,433,459,506,512,548,565
,
B
i
=
624,635,675,725,753,790
,
869,884,930,956
,
B
i
=
986,1040,1095,1120,1215,1246,1351
,
1469
,
B
i
=
1500,1509,1544,1610,1681,1687,1768,1856
,
B
i
=
1933
,
1983,2118,2180,2260,2323,2403,2461
,
B
i
=
2546,2784,2811
,
2829,2917,2993,3066,3146
,
B
i
=
3267,3307,3402,3683,3708
,
3886,4107,4126
,
B
i
=
4476,4588,4709,4797,4982,5048,5131
,
5274
,
B
i
=
5277,5284,5621,5649,5742,5846,6043,6382
,
B
i
=
6448
,
6512,6656,6794,6909,6975,7140,7289
,
B
i
=
7556,7733,7805
,
7867,8289,8411,8436,8506
,
B
i
=
8963,8996,9123,9271,9519
,
9581,9806,9997
,
B
i
=
10130,10160,10345,10714,10875,10962
,
B
i
=
11359,11771,11862,12286,12456,12705
,
B
i
=
12772,13000
,
13060,13113,13653,13962
,
B
i
=
14101,14261,14430,14434
,
14796,14996
,
B
i
=
15526,15686,15693,15794,16034,16458
,
B
i
=
16702,17076,17188,17645,17946,18074
,
B
i
=
18096,18256
,
18822,19136,19759,19826
,
B
i
=
19883,20325,20536,20592
,
20973,21222
,
B
i
=
21384,21923,22314,22658,22698,23595
,
B
i
=
23821,24157,24176,24387,24696,25329
,
B
i
=
25761,25911
,
26087,26792,26805,26864
,
B
i
=
27698,27954,28365,28927
,
29196,29209
,
B
i
=
29271,30062,30076,30523,30811,31286
,
B
i
=
31331,31566,31626,32570,32669,33026
,
B
i
=
33530,33852
,
34119,34267,35282,36067
,
B
i
=
36292,36551,36698,37252
,
37593,37734
,
B
i
=
38022,38147,38933,39325,39828,40388
,
B
i
=
40597,41219,41446,41897,42215,42445
,
B
i
=
43254,43804
,
44379,44772,45353,45741
,
B
i
=
46107
,
46795,46990,47350
,
47748,48018
,
B
i
=
48649,49221,50286,50445,50745,51407
,
B
i
=
51998,52473,53193,53310,54280,54748
,
B
i
=
54830,55600
,
55801,56742,57178,57777
,
B
i
=
58373,58567,59371,59654
,
60407,60959
,
B
i
=
61315,61742,62228,63216,63689,63914
,
B
i
=
65241,65358,65576,66053,66237,67695
,
B
i
=
67905,68672
,
70034,70476,70928,71882
,
B
i
=
72284,72724,73752,74126
,
74348,74448
,
B
i
=
75116,76825,78307,78802,79048,80039
,
B
i
=
80530,81065,81092,82135,82370,82624
,
B
i
=
84154,84852
,
85583,87485,87668,87681
,
B
i
=
88079,88744,89081,89386
,
89516,90363
,
B
i
=
91090,91158,91744,93948,94054,95516
,
B
i
=
95534,95894,97026,97062,98265,98556
,
B
i
=
99061,100280
,
101882,102085,103455
,
B
i
=
103494,103605,104616,105939
,
106603
,
B
i
=
106658,107830,108787,109544,110211
,
B
i
=
110227
,
110601,112111,113022,113822
,
B
i
=
114018,114724,115905
,
116414,117300
,
B
i
=
117552,117719,118928,119910,122518
,
B
i
=
123182,124383,124536,125662,125931
,
B
i
=
127370,128302
,
128850,130498,130536
,
B
i
=
131365,132111,132196,133477
,
133841
,
B
i
=
134494,134877,135277,135862,136204
,
B
i
=
137804
,
138438,138950,140020,140912
,
B
i
=
142845,143577,144255
,
145242,146902
,
B
i
=
148888,148998,149041,149902,151754
,
B
i
=
152956,153202,154042,155618,157013
,
B
i
=
158103,158332
,
158835,159987,160829
,
B
i
=
161657,162331,162929,163691
,
163874
,
B
i
=
164193,165059,165812,168512,170032
,
B
i
=
170232
,
170771,171764,173767,173838
,
B
i
=
175321,177150,177385
,
177960,178886
,
B
i
=
180725,182329,182774,184137,184302
,
B
i
=
184679,187155,188250,188667
.Remark 40.
In view of the slow increase of the functionψ, it would be interesting to have a probabilistic model of the behavior of ψ, defined on the space A
n. A first step is to determine the cardinality u
(
n
)
=
A
n of A
n, which is, in itself, a nontrivial problem of intrinsic interest.



---

*Source: 826141-2014-11-17.xml*