AN ABSTRACT OF THE THESIS OF
PATRICIA CARRIE PIERCE
(Name)
for the
M.S. in
(Degree)
Mathematics
(Major)
August 12, 1966
Date thesis is presented
Title SOME CONDITIONS FOR
A GIVEN POSITIVE INTEGER TO
BE CONTAINED IN A SUM OF SETS OF NONNEGATIVE
INTEGERS
Redacted for Privacy
Abstract approved
In this
(Major professor)
thesis, we study conditions not involving density which
guarantee that a given positive integer is contained in a sum
of nonnegative
integers.
We
of
sets
survey the literature, give more de-
tailed proofs of some known theorems, develop some new theorems,
and make some conjectures.
SOME CONDITIONS FOR A GIVEN POSITIVE INTEGER TO BE
CONTAINED IN A SUM OF SETS OF
NONNEGATIVE INTEGERS
by
PATRICIA CARRIE PIERCE
A THESIS
submitted to
OREGON STATE UNIVERSITY
partial fulfillment of
the requirements for the
in
degree of
MASTER OF SCIENCE
June 1967
APPROVED:
Redacted for Privacy
Associate Professor
of
Mathematics
In Charge of Major
Redacted for Privacy
Chairman of Department
of
Mathematics
Redacted for Privacy
Dean of Graduate School
Date thesis is presented
Typed by Carol Baker
August 12, 1966
ACKNOWLEDGMENT
To Dr. Robert D. Stalley whose patience, understanding and
time were so freely given.
TABLE OF CONTENTS
Page
Chapter
I.
II.
INTRODUCTION
1
EVOLUTION OF A CONJECTURE
8
III.
SYNOPSIS OF MANN'S TRANSFORMATION
30
IV.
TWO THEOREMS
35
V.
RESULTS FOR
BIBLIOGRAPHY
k > 2
SETS
88
101
SOME CONDITIONS FOR A GIVEN POSITIVE INTEGER TO BE
CONTAINED IN A SUM OF SETS OF
NONNEGATIVE INTEGERS
CHAPTER I
INTRODUCT ION
Let
k
be any integer where
and let A. (i =
k > 2,
i
1 ,
,
k)
denote subsets of the set of nonnegative integers.
k
k
Definition
A.
1. 1.
a.Ia.eA
=
1
1
A
,
i=1,,k}.
i=1
i=1
Let
i
1
be any subset of the set of nonnegative
Definition
1. 2.
For
n
a
positive integer,
the number of positive integers not greater than
integers.
A(n)
denotes
n.
k
In this
thesis we study conditions on
A. (n)
not involving
,
1
i =1
density, which guarantee that a given positive integer
n
is in
k
k
Ai,
or equivalently, restrictions on
k
i =1
that
n
is missing from
Definition
gap in
A.
1. 3.
A
1i=1 Ai
Ai( n)
by assuming
i=1
.
positive integer not occurring in
A
is a
2
The complement of
Definition 1.4.
the set of all gaps in
A,
denoted by
A,
is
A.
As a special case of a
result
of Schur (10, p. 275),
Rohrbach
proved, in 1939, the following theorem (9, p. 206).
Theorem
()EA
If
1. 1.
A2, and n is a gap in
0 e
A +A2,
then
Al (n)
+
A2(n)
n-1
<
In 1952, Khinchin gave a proof of
of
Schur's result (6, pp.
.
Theorem 1.1 independent
24 -25).
In 1958, Erdgis and
Scherk stated the following generalization
(4, p. 45).
Theorem
k
A,
,
1. 2.
If
()EA. (i
=
1, 2,
,k), and n is a gap in
then
k
i =1
Ai(n)
<
-(n-1).
i=1
We prove
Theorems 1.1 and 1.2 in Chapter
there that Theorem
1. 2
V.
We
also show
is the "best possible" inequality by using an
example of Erdbs and Scherk (4, p. 45).
If
n
is further
restricted to
k
be the smallest gap in
A.,
i =1
3
then stronger results can be obtained. In 1955, Lin proved the
following theorem (7, pp. 31-40).
Theorem 1.3.
the smallest gap in
A1(n)
If
0
E
A. (i
i
=
1 ,
A2(n)
+
A3(n)
0 <
n < 15,
and
n
is
then
Al +A2 +A3,
+
2, 3),
n-1
<
.
Lin gives a counterexample which shows that the smallest gap
can be made no larger in this theorem (7, p. 41).
In
Chapters II,
III, and IV, we give a proof of Theorem 1.3 that is a more readable
version
ters
of
Lin's proof with more details and some changes.
II and III, we give background
In
Chap-
material to make the proof self
contained. In Chapter III, we describe Mann's set transformation
and establish its properties which are needed in the proof.
ples, to provide insight into the mechanics
of the
proof are supplied
by the author.
k
Definition
1 .
5.
If
n
A.,
is a gap in
i
i=1
fk(n)
=
max
Ai(n)
iLLL=1
i=1
over all collections
{
A.1
1
< i < k 1.
,
Exam-
then
4
k
If
n is a gap in
A.,
/
then
i=1
Ai(n)
<
fk(n),
i =1
for all collections
{
A.
I
k
< i <
1
while
} ,
k
Ai( n)
1
i=1
fk(n),
fk(n)
=
for at least one collection
{
A.
I
1
e
-< i -< k
Thus the central prob-
} .
lem of this thesis is to find information about fk(n), for example,
upper bounds or asymptotic formulas.
In 1958, Erdös and Scherk proved the following
theorem (4, pp.
46 - 55).
Theorem 1.4.
If
OEA. (i
=
1, 2,
,
k),
i
k >- 3,
k
the smallest gap in
) A.,
LL
i
then
i=1
k-1
Z
where
ak
kn-akn k
and
yk
k-1
<
fk(n)
<
2
kn-ykn k
are positive constants.
Erdös and Scherk showed furthermore that
ak
=
(k +1)22k
-3
,
and
1
3k
k+4
22
(k-1):
and n
is
5
They also conjectured the following stronger asymptotic formula
(4, p. 46).
Conjecture
If
1. 1.
0
E
A. (i
r
=
1, 2,
,
k),
and
k > 3,
-
n
is
k
the smallest gap in
,
then
i=1
k-1
fk(n)
holds, where
=
Zkn - (ßk+o(1))n
k
tends to infinity, for some constant
n
Pk
which is
positive. 1/
In 1964,
1.
1
Kemperman proved a theorem which yields Conjecture
He gave the following definition and proved
(5, pp. 376 -385).
the following theorem.
Definition
1.
6.
n
If
is a gap in
,
then
i=1
k
=
'`)k(n)
min
{
Z
(n-
1) -
I
Ai(n)
}
i=1
over all collections
1/
{A,
In the original paper
place of the minus sign,
i
1
1
< i < k 1.
this formula is given with a plus sign in
clearly an error.
6
Theorem
OEA. (i
If
k
1. 5
the smallest gap in
i
1, 2,
=
k), k
,
-> 3,
and n
is
then
) A.,
i
i=1
k-1
lim
(1. 1)
n--"'
We show
1. 5.
k-1
k(n) n k
k2
=
k
co
that Conjecture
1.
follows readily from Theorem
1
Since
k(n)
then equation (1.
1)
=
2 (n-1)-fk(n),
may be written
k-1
lim
-
n
and so
[
00
-fk(n)] n
(n-1)
2
k-1
k
=
lim
-oo
k
n
[ Zn-kn
k
k-1
k-1
n
k2
=
k
k2
Hence, this equation may be written
k -1
k -1
[ -2
where
n
-fk(n)] n
k
=
k
o(1) +k2
tends to infinity. Finally, this equation yields
k -1
k -1
k
2n -fk(n)
=
(o(1) +k2
or equivalently,
2
)n k
k-1
fk(n)
=
2n-(o(1) +k2
2
k-1
)
n
k
7
k -1
This is the equation of Conjecture
1.
1
with
ßk
=
k2
2
.
Since
k- 1
k2
k > 0
and since the hypotheses of Conjecture
as those of Theorem
1.
5, Conjecture 1.
1
1.
1
are the same
is proved.
k
Kemperman also showed that the condition
{
i
I
1
<i <n} C. A.
i=1
of
Theorem
In
1. 5
Chapter
can be generalized (5, pp. 381
V,
we include proofs of
-
387 ).
several further miscellaneous
smaller theorems. Also some conjectures are made.
8
CHAPTER II
EVOLUTION OF A CONJECTURE
Our purpose in this chapter is to investigate in detail the man-
ner in which a particular conjecture, Conjecture
From Conjecture
2.
7
evolves Theorem 4.
1,
formulated.
proved in Chapter IV,
which is a powerful tool in the proof of Theorem
To formulate Conjecture 2.7, we
2. 7, is
1. 3.
start with Conjecture
2. 1.
While Conjecture 2.1 is a very natural extension of Mann's funda-
mental inequality (8, pp. 523 -524), it is seen to be invalid. From
Conjecture
2. 1, we
Conjecture
2.
1
formulate Conjecture
2. 2, a
closely related to Theorem
1.
special case
of
3, which is shown
here
also to be invalid. We then introduce several definitions, and Con-
jectures
2.
3
and 2.4 are reformulations of Conjecture 2.
2
using
these definitions. Following Conjecture 2.4, we define a new set,
the inversion set, and prove several properties of this set. We
introduce the inversion set into our conjecture by making a set substitution when formulating Conjecture 2.
5.
We
thereby not only reduce
our consideration to two, instead of three, sets but we have all the
properties
of the
inversion set available to us. Making use
properties and previous definitions,
and 2.7.
we
of
these
formulate Conjectures 2.6
Our evolution terminates with Conjecture 2.7.
9
The purpose of this sequence of conjectures, Conjecture 2.
through Conjecture 2. 7, is to put Conjecture
easily handled in the proof
of
under a certain restriction.
Theorem
2. 2 in a
4. 1, which is
3
form more
Conjecture 2.7
Each conjecture in our sequence is
implied by the next one and in some cases the conjectures are equivalent. This implication or equivalence is shown in each case. We
include a two page summary of this evolution at the end of this
chapter.
The formulation of Conjecture 2.
3
through Conjecture
2. 6,
as well as the proofs of the equivalences and implications of these
conjectures and the summary page, is the work
of
the author. Also,
for easy referral, we formalize definitions into statements which
are displayed and numbered, which in most instances Lin does not do.
First
gave the
we introduce Conjecture 2.
first proof
(8, pp. 523 -527).
of the following
of the
aß
1
(7, p. 21).
In 1942, Mann
Theorem for Schnirelmann density
The aß Theorem is an immediate consequence
more fundamental theorem which he proves in the
same paper. In this proof, Mann introduces his method
of
set con-
struction which we introduce in Chapter III and use for our purposes
thesis.
in this
Theorem
integer
n,
2. 1.
either
If
0
E
Al
and
0
E
A2,
then for any positive
10
(A 1+A2)(n)
-
n
1
,
or
A 1(m)+A2(m)
(Al +A2)(n)
min
<m <n
->
n
1
m
m A1+A2
Our Conjecture 2.
k
sets where
is the generalization of Theorem 2.
1
to
k > 2.
Conjecture
positive integer
1
2. 1.
If
0
E
Ai (i
i
=
1, 2,
,k),
then for any
either
n,
k
A.)(n)
i
i=1
n
-
1
,
or
k
k
A.(m)
Ai) (n)
i=
1
n
>
1
min
<m <n
i= 1
m
k
mff
Ai
i=1
This conjecture is not true according to an example following
Conjecture
2. 2.
11
Next we formulate Conjecture 2.
2.
1
we set
k
and restrict
= 3
2
(7, p. 21).
In
Conjecture
to be the smallest gap in
n
3
Ai
i
.
Since
i =1
3
A.)(n)
i
(
n-1 <1
i=1
n
,
n
we have
/
3
3
3
A.)(n)
i=1
n
min
1 <m <n
>
Ai(m)
L/ Ai(n)
i=1
i=1
m
n
3
mi
A.
i=1
and so
3
3
A,)(n)
(
i
i=1
A.(n)
>
.
i
i=1
Hence the following conjecture follows from Conjecture 2.
Conjecture
2. 2.
If
OEAi (i
i
3
smallest gap in
/
i=
1
,
then
=
1, 2, 3),
and if
1.
n is the
12
3
3
Ai(n)
<
i=1
and
2
.
i=1
Clearly Conjecture
Conjecture 2.
A.)(n)
(
Conjecture
2. 2 does not imply
is obtained from Conjecture 2.
k.
1
by
2.
1
because
restricting n
3
Since in Conjecture 2.
2
we have
A.)(n)
(
=
n -1, it follows that
i=1
Conjecture 2. 2 differs from Theorem 1.3 only by not having the
hypothesis
0 <
n < 15
Theorem
of
1. 3.
Since Conjecture 2.2 is a special case of Conjecture 2.
following example shows that both are invalid.
Al
=
A2
={ 0,2,8,
A3
=
{0,4,8,9,
A1 +A2 +A3
=
{iI1
{
0, 1, 8, 10, 12, 14
1,
the
Let
} ,
9, 12, 13 },
10, 11
}
.
Then
< i < 14
< i < 38 },
}
3
and so
n
=
15
is the smallest gap in
Ai
i=1
of
Conjecture 2.2 are satisfied, but
.
Thus the hypotheses
13
A1(15)+A2(15)+A3(15)
We cannot
groundwork..
proceed to Conjecture
In the
first place,
2.
3
=
15 >
n-1
.
without some preliminary
we wish to avoid the confusion
which arises in Lin's work with the use of the same set notation for
the possibly infinite sets
A.
i
n
{
0,
2,
1 ,
,
n } (i
=
and the finite sets
Ai
1,2,3).
To this end, we make the following
definition.
Definition 2.1.
In addition
A
=
The set
A1
N,
B
=
is defined by
N
A2
n
N,
C
=
A3
N=
nN
{j 10 < j
< n 1.
and
3
A +B +C
=
(,
n
Ai)
with similar definitions for all possible
N,
i=1
A.'s
i
combinations of the
e. g.
,
A +B
=
(A +A2)
1
We
n
and for the complements of these sets,
and
N
A
=
Al r-, N.
also adopt the convention that any new, possibly infinite,
sets introduced in the discussion to follow will be regarded as their
intersections with the set
the two sets
X
and
integers, and state that
Xm NC
Y
N.
Thus, for example, if we introduce
both subsets of the set of nonnegative
Y,
X
is a subset of
Y,
we mean
N.
In the second
place, to make our notation more concise, we
introduce the following definitions.
14
The set
Definition 2. 4.
I
n
=
is defined by
In
{0,1,2,,n-1}.
The expression
Definition 2. 3.
will be referred to as the excess of
(A +B)(n)
of
E
and will be denoted by
A(n- 1)
A(n -1)
B,C,n) = A(n- 1) +B(n- 1) +C(n -1)
Definition 2. 4.
sentations of n
Let
p
n
=
a
over
g.
,
- (A+B +C)(n).
denote the number of repre-
(n, A, B)
as a sum
(A +B)(n)
This definition can,
course, be extended to three or more sets, e.
(A,
1) -
B(n -1)
+
(A, B, n).
E
B(n-
+
+
where
b,
aEA
and
b E B.
With the necessary groundwork laid, we may now proceed to
a new formulation of our conjecture.
Conjecture
2. 3.
If
A +B +C
A(n)+ B(n) +C(n)
=
In,
then
- (A +B +C)(n) < 0
.
That the hypothesis A +B +C =I is equivalent to the hypotheses of
n
Conjecture
2. 2
as previously stated, namely,
0
E
Ai (i
=
1, 2, 3)
and
,
3
n
is the smallest gap in
i'
is made clear by close examination
i=1
of Definitions 1. 1, 2.
that
n
1
and 2. 2.
is the smallest gap in
(Definition 2. 2).
That is,
A +B +C
But by Definition 2. 1,
A +B +C
=
In
n
implies
by the definition of
In
15
3
A+B+C
=
3
Ai)n
i
(
Hence
N.
I
n
L) A.)i
= (
(Th
and so n is the
N,
i=1
i=1
3
smallest gap in
A.
.
Conversely, if
is the smallest gap
n
i=1
3
3
then
Ai,
in
is the smallest gap in
n
) Ai) r. N.
(
i =1
But
i =1
3
A.),.
N
=
and so
A +B +C,
n
is the smallest gap in
A +B +C.
i=1
Thus equivalence is established for one of our two hypotheses.
Since
0
I
E
and
n
A +B +C
=
it follows by the definition
In,
n
3
of
the set sum (Definition 1.1) that
is in
0
A +B +C
=
)
(
A.)
i r-- N,
i=1
and so
0 EA.
O
EA. ( i
r N.
=
1 ,
Hence
Definition 2.
1.
2, 3)
0
.
Conversely,
if
O
is in each of the sets
Thus, the equivalence
E
A. (i
=
A,
and
B
3
C
by
hypotheses is substantiated.
of
Let us note here that the step from Conjecture 2.
2.
then
3),
1, 2,
2
to Conjecture
leads us from consideration of possibly infinite sets to that of
finite sets.
Conjecture 2.4 is a reformulation
of
hypotheses are the same and imply that
n
is not in A, B,
B
and
For example, suppose
n +0 +0
=
n
is in
A +B,
n E A.
Then
O
E
Conjecture
0
E
contradicting our hypothesis
C
2. 3.
Thus the
or C.
imply that
A +B +C
=
In.
n
16
Hence
A(n)
=
i
s
not in
A(n -1),
B(n)
n
Therefore, we can say that
A, B, or C.
B(n -1),
=
and
C(n)
=
that the conclusions are the same by Definition
Conjecture 2.4.
A +B +C
If
(A,B,C,n)
E
Proceeding to Conjecture
=
< 0
2.
5
In,
This shows
C(n -1).
2. 3.
then
.
involves the set substitution
previously mentioned. Therefore before stating this conjecture,
several properties
we define two new sets and prove
of one of
these
sets needed for future reference. To this end, let
n2 <
0 < n1 <
of our
<
=
{
d. d.
i
i
I
=
the inversion of
i
1,
=
,
D
is defined by
r 1.
Given a set
X,
for any positive integer
is defined to be the set
X
X
X
.
The set
Definition 2. 6.
also denoted
A +B
2. 5.
n-n.i ,
be the gaps in the sum of some fixed two
n
=
three sets, say
Definition
D
nr
ti
.
=
{n-XiXEX },
The set
ti
X
is empty if
X
=
N.
(Recall
our convention regarding the introduction of new sets.)
Two observations should now be made: (1) comparison of
n,
17
Definitions 2.
and 2.6 shows us that
5
D
and (2) corn-
(A +B)`v,
=
parison of Definitions 1.4 and 2.6 shows us that n4
x>
XE
for all
0
Lemma 2.
.
For any set
1.
ti
a)
X
X
=
X
c)
n4 X
d)
OE
e)
ni X+
ti
X.
+
the following properties hold:
X,
.
ti
implies
b)
since
X
O
(If
E
X
.
=
,
ti
then
X+ X
=
<D.)
ti ti
...
f)
X
g)
E
(X+ X)
Proof.
0 < b <
n.
a E X,
then
X
C
ti
ti
X
.
=
X, n)
(X, X,
Thus
=
X.
This proves
b)
X(n-
If
1) + X
(n-1)
then
a
,
n -a,
and so
=
ti
ti
n I X,
(X+
=
X
n -b
a e X.
Consequently,
X
-
ti
)-C-)(n)=
ti
n-1-(X+X)(n) > 0,
In.
n
=
E
b
i
=
a X
If
l
(n-1).
X+ X
a)
n -a
X
ti
equality holding if
YCX.
if, and only if
Y
(n-1)
.
where
bt
X
Hence
X
C
n -(n -a)
=
a
E
X.
and
X.
If
Hence
.
then
nE
X.
Therefore
n -n
= 0
E
ti
X.
18
and
nE X+ X
Hence
xE X.
x
d)
By (c),
e)
If
nIX+X
YC
n$ X +X
if
or
then
and
X +Y
Y
C
ti
n
X.
ti
ni X+X
By (c),
n4 X+ Y.
nl X+ Y.
ti
yX
and assume
yE Y
OE(X+X)~.
Then
contrary to our hypothesis,
Thus,
Y
Y =
n-x,
ni X +Y.
X.
This
(e).
f)
1
of
xE X
implies
C X+ X implies that
then
X,
x +yE X +Y,
=
ti
X+YC X+X.
Therefore, our assumption yiX is false.
proves
where
x +(n -x)
X+X
By (b),
X,
Conversely, let
X,
=
n X +X.
Therefore,
XE
n
contrary to the definition
x
=
Then
.
ti
Therefore,
Thus,
ti
Suppose
c)
< n -x <
For all
n-
Therefore,
1.
such that
1
X(n-1)
(n-1).
> X
g)
<
n -x <
By (f)
n-1,
X(n- 1)
ti
X(n-
x < n -1,
<
1
1
we have
For all n -xE
X(n- 1).
>
we have
Hence
< x < n-
1
.
-
1) =
X(n-1).
,
ti
E
such that
xE X
(X, X, n)
ti
=
X(n- 1)
=
X(n-
=
ti
n- 1-(X+X)(n),
(X+56
ti
+
X(n- 1)
-
(X+X)(n)
1) +
X(n- 1)
-
(X+X)(n)
Therefore,
X
19
since
X(n -1)+ X(n -1)
=
n -1.
ti
ndX +X.
By (c),
Hence
ti
ti
(X+ X )(n)
(X+ b(n-1) < n-1.
=
Therefore,
ti
E
n-1-(n-1)
(X, X,n) >
= O.
ti
However, we have also to show that if
ti
E
(X, X, n)
= 0
ti
and so
(X, X, n)
E
ti
But if
.
X+ X
=
X+ X
then (X+
In,
=
ti
X )(n)
then
In,
=
n-1,
= O.
This completes the proof of Lemma 2. 1. Observe here that
ti
The proof of Lemma 2. 1, and also
(a) implies that D = A +B.
that of Corollary 2.
1
which follows, is not done by Lin (7, pp. 22 -23).
The next result is an important one in that it justifies the set substitu-
tion that we wish to make in Conjecture
Corollary
Proof.
if
n4X +Y,
Lemma
2.
1
2. 1.
X+ Y
If
then
(c),
X+ Y
If
Y
=
In,
C X'.
ti
=
I
then
n
,
2. 5.
then
X +Y
X
This completes the proof of Corollary 2.
material
-
X +Y
=
ti
CX
and
ti
X+ X
By Lemma 2.
n4X +Y.
Therefore,
n1X +X. Hence,
Y
ti
C. X+ X.
ti
+X
1
=
1
(e),
By
In.
n
and gives us the
we need to state Conjecture 2. 5, but before we
=
terminate
our discussion of the inversion set, let us prove another property
In.
n
20
which is needed in Chapter IV (7, p. 23).
Theorem
to satisfy
A
2. 2.
ti
X+ X
=
I
necessary and sufficient condition for
is that, for every
n
where
h
0 < h <
X
n,
we have
X+ {O,h}
Proof.
If
X
ti
+X
X
for any h.
By
then
In,
ti
+X+ {O,h}= In+ {O,h },
hypothesis,
which implies that
0 <
so that
h < n,
Consequently,
n -h E In.
n
In+ {0,h }4 In.
and
=
X.
Now suppose
X +
{
n
0, h
ti
ti
X+X+ {0,h}= In+ {0,h}= X+X
contradicting our previous result
X+ {O,h}
h < n.
+
{
0, h
=
}
n -h < n,
(n -h) + h E In +
n
=
=
Then
X.
In,
Hence
In.
X+
Then, for all
{
0, h } 4 X
0 <
i < n,
for all
we have
h
where
0 <
n -i < n.
Consequently, our hypothesis tells us there exists an
that
Hence
{0,h}
X.
Conversely, let
0 <
In
}
0 <
x+ (n -i)
< n
and
x+ (n -i) 4 X.
That is
x+ (n -i)
X
E
=
X
X
such
X.
21
i
Thus
ti
X+ X
also satisfies
x +n -x
=
for by Lemma
In,
We now
2.
2. 5.
(2. 1)
E
A +B +C
=
(A, B, (A +B)
C
Conjectures
E
ti
X+ X.
ti
,
then
In,
=
n) < 0
.
that Conjecture
X
(A +B)
=
and
A +B
and
2. 4 is
2. 4
(A, B, C, n)
=
and 2.
5.
5.
Conjecture 2.4,
of
Corollary
Y
=
C.
=
2. 1,
X +Y
=
In,
Consequently, we
In.
n
Hence, we have
Consider now the conclusions
We have in
A(n- 1) +B(n-
implied by Con-
of
A +B +(A +B)
the hypothesis of Conjecture 2.
of
not
satisfies the hypothesis
In,
C
.
x +n -x<n. Consequently,
(d),
1
X
First observe that the hypothesis
where we are letting
have that
0 <
A +B +(A +B)"'
If
We wish to show now
2. 5.
E
formulate the next conjecture in our evolution.
Conjecture
jecture
x+n-x X+
=
1) +C(n-
Conjecture 2.4
1)- (A +B +C)(n) <
0
,
and in Conjecture 2. 5, we have
ti
E
ti
ti
(A, B, (A +B) ,n) =A(n- 1)+B(n-1)+(A+B) (n-1)-(A+B+(A+B) )(n)<
However, since
A +B +C
=
A +B +(A +B)
ti
=
In,
we have
O.
22
(A+B+C)(n)
Also, since
C(n -1)
C (A +B)ti
C
< (A +B)
E
ti
(A+B+(A+B)
=
)(n)
nlC
and
,
ti
=
and
n-
1
.
n (A +B)
ti
,
we have
Hence,
(n -1).
(A, B, C, n) <
E
(A, B, (A +B)
ti
,
n)
and the implication is established.
Let us progress another step in our evolution by exploring
(2.
D
more thoroughly. Recalling our previous observations,
ti
ti
(A +B)
and D = A +B, we may write
1)
=
(2. 2)
E
(A, B, (A+B)
ti
ti
ti
n)=A(n-1)+B(n- 1)+(A+B) (n- 1)-(A+B+(A+B) )(n)
,
ti
ti
=A(n- 1)+B(n- 1)-(A+B)(n)+D(n-1)+75(n-1)-(D+D)(n)
=
Since
(A +B)(n)
and subtracted
ti
=
2.
1
(g),
(2. 3)
E
(D, D, n)
ti
+
E
D(n)
=
ti
(D, D, n).
ti
D(n -1),
we have
on the right side of (2. 2) in
D(n)
the desired result.
ti
(A, B, n)
and
D(n)
ti
E
Observe here that
Thus (2.
= O.
E
(A, B, (A +B)
2)
ti
,
D
ti
+D
=
merely added
order to obtain
In
and so, by Lemma
becomes
n)
=
E
(A, B, n).
To formulate Conjecture 2. 6, we substitute (2. 3) in Conjecture 2.
Conjecture 2.6.
If
A +B +(A+B)
ti
=
In,
then
5.
23
(A, B, n) < 0
E
Clearly Conjectures
2.5 and 2.6
.
are equivalent.
Let us concentrate now on the last step of our evolution
which involves the use of Definition 2. 4. How Definition 2.
4
applicable here is made clear by considering three subsets
of the
set
{
1, 2,
To be more
n- 1 }.
,
precise,
is
we introduce the follow-
ing definition.
Definition 2.7. The sets
A' , B'
1)
A'
_
{n -all<
a < n -1,
aEA },
2)
B'
=
{b1l <b <n -1,
bEB },
3)
L
=
{17:1.173-=
n-a,
aEA,
and
,
beB
L
are defined by:
}.
The proofs of the following two lemmas concerning these sets
are not given by Lin.
Lemma 2.2.
The sets
A' ,B'
,
and L
are mutually dis-
joint.
Proof.
b)
A'(
a)
bEB'
L
=
We
1),
have three statements to prove:
and
Assume A'
and an element
c)
r
B'r
B' 4
n -a E A'
L
4).
a)
A'
r.
B'
= (1),
= r1).
Then there exists an element
such that
n -a= b,
or
n
=
a +b.
This contradicts
Assume
b)
,
A'
n -a EA'
and an element
implies
a
Assume
c)
B'
definition
Lemma 2. 3.
1, 2,
We show
Suppose
and
1
-1
,
1
< b <
Either
XE
1, 2,
{
n-1.
=
A'
{
v
v
2. 2, shows
,
L
that
=
and so
XE
n -a
But this
.
A'
n
L
=
bEL,
.
then
n -1.
=
v
B'
L
=
are all subsets
A'v
or
,
,
n-
or
,n -1
v B' v
1 } .
.
contradicting the
,
x
x
=
1, 2,
{
of the
b
b
n-11.
set
L C {1, 2,
B'
=
,
where
where
.,n -1 }.
bE B
bEB and
latter case, there are only two possibilities.
xE {1, 2,
11, 2,
B'
b
A +B,
Then either
n-11.
n- 1 }C A'
=
rL
(n, A, B)
and L
,
and so
n -a
1, 2,
B'
,
In the
Consequently, if
so
B'
=
first that A' v
B'
,
+ p
=
Hence
A.
of
b
and so it follows that
} ,
< b < n -1,
b
B(n- 1)
+
By Definition 2. 7, A'
{
such that
is a gap in
n
n -a
Then there exists an element
.1).
bEB
If
A(n- 1)
Proof.
L #
.
_
such that
Consequently,
B.
of
B'
Then there exists an element
13.
n -a E L
rm
and an element
bEB'
L
A'
contrary to the definition
a
=
Hence
n4 A +B.
24
L.
},
b
=
n -a,
then
xe
and so
A'v
B'
bEA'
v
L,
.
and
This proves
This result, in addition to Lemma
25
A' (n- 1)+B' (n-
(2. 4)
Definitions 2.4 and 2.
1) +L(n- 1) =
n-1.
yield
7
A' (n -1)
=
A(n-
1)
,
B' (n -1)
=
B(n -1)
,
L(n -
=
p(n,A,B).
1)
Making these substitutions in (2. 4), we obtain
A(n- 1) +B(n- 1) +p (n, A, B)
=
n -1,
and the proof is complete.
Using Lemma 2. 3, we have
E
(A, B, n)
(2. 5)
However, since
=
A(n-
=
n- 1-p (n, A, B)-(A+B)(n).
n 4 A +B,
1) +
B(n- 1)- (A +B)(n)
we have
n-1,
(A+B)(n)
+
(A+B)(n-1)
(A +B)(n)
=
n-- 1- (A +B)(n -1).
=
or
(2.6)
Substituting (2.
6) in (2. 5)
gives
26
E
(A,B,n)
(2. 7)
Now
recall that
A +B,
and so
since
nEA +B,
we let
=
n-1-p(n,A,B)-[n-1-(A+B)(n-1)]
=
(A+B)(n-1)-p (n,A, B).
0 <
(A +B)(n -1)
(2. 8)
Substituting (2.
8)
nr
<
=
gaps. Thus,
r
has
A +B
n1 < n2 <
=
n
be the gaps in
(A +B)(n)
r -1
r,
=
or,
.
in (2.7) gives
(2. 9)
E
(A, B, n)
Hence, by substituting (2.
=
r - 1 -p (n, A, B).
in Conjecture 2. 6, we complete our
9)
evolution (7, p. 26).
Conjecture 2.7.
gaps,
r
If
A +B +(A +B)
ti
=
In,
and if
A +B
has
then
p(n,A,B)
>
r-1
.
Since substitution of equivalent quantities was our only modifi-
cation, clearly Conjecture 2.6 is equivalent to Conjecture 2.7.
The theorem proved in Chapter IV, Theorem 4.
2.
7
under the restriction r<
Conjecture
2.
2
5.
1,
is Conjecture
Hence, since we have shown that
is implied by Conjecture 2. 7, Theorem 4.
1
shows
27
that Conjecture
2. 2 is
valid under the added restriction that the sum
of some fixed two of our
three sets,
A, B,
and C,
has less than
six gaps.
In the following
transformation.
of
We
chapter is a brief discussion of Mann's set
bring into this discussion only those properties
this transformation which are needed in the proof of Theorem 4.
so that this work is self contained.
1,
28
SUMMARY OF THE EVOLUTION
Conjecture
2.
1
Arguments
Statement
If
0
E
A. (i
1,
=
-
k), then for
,
Generalization of
Theorem 2. to k
sets where k > 2.
1
any positive integer
either
n,
k
A.)(n)
i
(
i=1
or
- 1,
n
k
Ai) (n)
i=1
n
min
1 < m < n
>
k
A.
mt
i=1
2. 2
OEA.(i= 1,2,3), and if
If
Conjecture 2. 1
for k = 3 and n
the smallest gap in
n
3
is the smallest gap in
1A.(n)
Ai
.
i=1
Ai)(n)
(
i=1
If
3
3
3
3
Ai'
i=1
then
2.
/
i=1
A +B +C
=
In,
Definitions 2.
then
A(n)+B(n)+C(n)-(A+B+C)(n)
<
0.
1, 2. 2
29
2. 4
If
E
2.
5
If
E
2.
6
If
E
2.
7
If
A+B+C
(A,B,C,n)
(A, B, (A+B)
(A, B, n) <
0
In,
n) < 0
ti
then
Definition
Corollary
2. 6
2. 1
.
=
In,
then
Lemma 2. 1 (g)
Definition 2. 5
=
In,
and if
Definition 2. 4
Lemma 2. 3
Equation (2. 9)
.
A+B+(A+B)
has
=
,
3
.
ti
ti
A +B +(A +B)
A +B
p
< 0
A+B+(A + B)
Definition 2.
then
In,
=
ti
r
(n, A, B) > r -1
gaps, then
.
30
CHAPTER III
SYNOPSIS OF MANN'S TRANSFORMATION
It is not our intent
here to scrutinize minutely Mann's trans-
formation, but only to dwell on it briefly so that it may be employed
in Chapter IV without loss of continuity in this work.
We
introduce
only those properties which are needed for our purposes and of these
properties, only those whose proofs are not immediately available
in the
literature
will
(2, pp. 10 -18; 5, pp. 4 -6; 6, pp. 524 -526)
be discussed in detail.
The sets
and
A, B,
were defined in
D
Chapter II. We emphasize here that in the following definition we
are assuming that at least one step in the transformation is possible.
Definition
a)
Let
B
be any integer in
b)
d. D
(Mann's Transformation)
3. 1.
=BO
Bj
=
=B,
BO v
The subscript
(1 < i <
r -1),
T=
TO
J+
if
1
and a subscript
J
there exist
aEA,
T*
vT v
k4 T.
=
J
such that
nk-a
=
e.
j+1
+
e. +1(J >0)
v B.
B*
Bi v
iET +1
=1) and
=
di,
i#T
J
0
v
T
31
Let
c)
{e.
Bi*+1 =
+
diIiET3+1 }.
There-
This defines Mann's transformation up to uniqueness.
fore, to arrive at a uniquely determined construction, choose
e.
as the least element
This
such that
eE B.
process terminates at BJ,
j =
least element
we choose the
0,
when
el+dt
(3. 1)
is solvable for some
equation (3.
1)
dt E D,
n
E
=
el
such that the equation
E
e2EB1
B
nv-a
is in
t
T
=
and
Notice that
a E A.
1
nt-a,
.
struction, if possible, with the sets A,
where
Thus for
is empty.
may be rewritten
as well as
v
.
TJ +1
A +B,
el+dv
so that
is not empty
T +1
+1
We then continue the conD,
and B1
=
B
Bi,
.
The following properties are readily verified.
Lemma
T*
j+1
3. 1.
a)
'
'
b)
B. r` B3+1
c)
Tj
-
T+1 -
0'
,T;
B*
+1
has the same number of elements as
32
element in
d)
No
e)
n4A+B+1
f)
nteA +Bj+1
is of the form
B +1
n -a,
where
aEA;
'
tETj+1
and only if,
if,
]
Lemma
T
If
.
Let
3. 2.
exists and if
s
Proof.
n > d
If
t
be the least subscript, if any, not in
s
s
nt
then
,
0 <
A +B,
ns -dt
say
script not in TJ,
3.
1
be
teTJ,
Corollary
i
where
i <
Proof.
=
3
nss -dtt
.
s
Since
uETJ
.
.
aEA,
-dt - a +b*,
Thus
or
s
ns-dt
is a gap
is the least sub-
Hence, by Lemma
If
0
further.
E
A
and
b *e BJ
TJ
were not in
one step
1
.
t
n.
<
u < s.
nu,
for if t
carried at least
> d
s
ntEA +BJ (7, p. 11).
Therefore, in either case,
}
and so
n
we must have
ns- dtEA +BJ.
(f),
then
ns- dteA +B CA +BJ,
Consequently, either
in
ds,
>
Thus
0
E
B,
our construction could
nt eA +BJ
then
n.
1
.
d
s
for all
r.
Suppose
n.
=
d
s
for some
i <
r.
Then, as the
33
first step in Mann's construction, choose el
el+ds
We
ni
n1-a,
=
a
= OE
= O.
are immediately led to the false conclusion that
for all
ds
where
i
with
B
c,
e
T
Hence
.
1
r.
i <
The proof of the following lemma is not given by Lin.
Lemma 3.3.
all elements in
equation n
=
Proof.
B
r
BJr,
bE
b
for some
and so
j > 0
,
BJn BÇ
B1
v
v
where
b
j >
E
B.
0,
of
b
BJ
and
satisfying the
then
B,
e+di,
v
BJ
,
By
iET+1
BJ
.
Hence
bEB1
v
.
then
b*=e+d,
for some
v
Biv Be-)
=
a E A.
by the definition of
b*EB1v
If
=
B
are distinct values
B
where
a +b,
If
BJr,
We have
iET +1
by the definition of
B +1.
Then
b-ej+l+di=nk-a,
Suppose
b*
=
b
,vBJ
34
by the definition of
j+ 1
and so
'
a +b
This contradicts
where
..
v
being a gap in
Consequently,
b E B.
B1
nk
BJ C
BJ
nk
=
b *E BJ
B.
n
b
=
n -a.
n -a
Hence, all elements in
where
a EA.
B
b*
b
=
and so
B
This proves
By Lemma 3. 1(d), no element in
form
Hence
A +B.
B3(- B=
v
1
BJ
are
B
BJ
131'1._.)
BJ
of the
is of the
form
That these elements are distinct is immedi-
ate, for in the first place the
B'4' s
are disjoint by Lemma
3. 1(b),
and in the second place, if
ej+l+di
where
in
iETj
BJ m B
where
+1
e
k E T +1,
are distinct values
then
of
b
The phrase
"
i
=
k.
Hence all elements
satisfying
n
=
a
+
b,
a A.
E
Definition
"
and
ej+1+dk'
=
3. 2.
is the least element in
B.
e
corresponds to k"
such that
kE T
+1.
This concludes our discussion of Mann's transformation.
means
.
35
CHAPTER IV
TWO THEOREMS
In this
chapter, we prove two theorems, Theorem 4. 1 which
is Conjecture 2.
7
under the restriction r
< 5,
and Theorem 1.3
which we restate as Theorem 4. 2. The proofs of Theorems 4.
and 4.
2
are reorganized and clarified, and in the case
4. 1, examples
of
1
Theorem
are introduced. However, the underlying theory in
most instances is still that
We begin by
of Lin (7, pp. 27 -40).
turning our attention back to Conjecture 2.7. Our
purpose here is to lay the background for the proof of Theorem 4.
by proving three lemmas pertaining to Conjecture 2.
7
now
1
that
Mann's set transformation is available to us. We again state
Conjecture 2.7.
If
ti
A +B +(A +B)
=
In,
and if
A +B
has
gaps, then
r
p
(n, A., B)
>
r-1
.
Notice that this conjecture is trivially true for
p
(n, A, B) >
r
>
1
0
is always satisfied.
r
=
1,
since
Therefore, we assume that
in the argument to follow.
The necessity of Mann's transformation lies in Lemma 3. 3,
for we construct
r -1
elements
bB
that are distinct solutions
36
of the equation
n
=
where
a +b
some of which are in
a EA,
BJnB.
Lemma 4.
The set
1.
is not empty; in
T
particular,
r -1ETJ.
Note that the hypothesis of Conjecture 2.7 is assumed in
Lemmas 4.
1,
Proof.
(A +B)
ti
=
D,
4.2 and
By
4. 3.
hypothesis,
ti
A +B +(A +B)
- - r -1)
every gap n.
in
(1 < i <
1
=
In.
A +B
Thus, since
can be expressed
asasum
(4.
a+b+dt
1)
where
a E A,
bEB,
dt D,
E
dt
ni - a
=
ni
0,
=
b+dt
,
or
.
Therefore, at least one step in Mann's construction is possible.
Let us emphasize here that we can say no more than this. That is,
since we choose
bEB
to be the smallest
ni-a
and since the
T
s
=
b
such that
b+dt,
are disjoint by Lemma
3. 1(c), it is
entirely
J
possible that the construction can be carried no further. However,
37
we have shown that at
not empty.
If no
least
T
subscript
where
i
r- l TJ
E
r -1
< i <
1
is missing from
If, on the other hand,
.
scripts missing from T3,
1) ,
TJ
is
This concludes the first part of the proof.
then, of course,
By ( 4 .
is not empty, and hence,
1
let
v (l < v
a+b+dv
there exist sub-
be the least such subscript.
s
there exists a subscript
TJ,
n1
=
<r -1)
such that
,
and so
d
By the definition of the set
dr-1
<
v-< n
1
-<
n
.
s
we have
D,
di (1
-< i <- r-1)
.
Hence,
dr-1
-<
dv
-< n1 Ç ns
n
> d
and so
r-1
s
since equality cannot hold by Corollary
Lemma
3.
1
3. 2,
nr
EA +BJ,
,
3. 1.
and therefore,
Consequently, by
r - 1 T3
E
by Lemma
(f).
Lemma 4.
2.
Conjecture 2.7 is true if at most one subscript
38
i
(1<i<r-1)
Proof.
BJn
.
subscript
If no
then the equation
bE
TJ
is not in
n
=
Assume there exists one subscript
3. 3, we
bE
BJ
n B.
solution
We want to show
but
bE B,
BJn
b4
have
a +b +ds
=
a
n
of
subscript t
for suppose
=
ns
s
Again by Lemma
where
a +b
=
such that
t < r -1)
(1 <
.
Then, since we also
b +dtE BJ.
b +dsEBJ,
we would have
nt,
T.
By the hypothesis of Conjecture
B.
a+b+d t
b +dt4 BJ,
4
that we also have at least one more
2.7, we know that there exists
Then
s
distinct solutions
r -2
then have
sETJ,
and so
contrary to our assumption. Consequently, we assert that
is another distinct solution of
n
this, suppose that b'
where
and so
a +b'
=ns,
s
pose further that
b +dt
=
=
contradicting
b +dt
=
n -a'
a +b
b' B.
E
ns
s
where
,
where
fact that
n
=
b +dt
p (n,761-.,53) >
with
a +b
=
a +d
s
BJ, and hence,
4
r -1
a
.
a
=
a
+
a'
Then
EA.
b
=
b
+dt
b =b +dt
To show
ds.
Then also
being a gap in
and we reach the same contradiction. Hence
tion of
J,
and Conjecture 2.7 follows.
by Lemma 3. 3,
B
T
distinct solutions where
r-1
has
a +b
is missing from
r -1)
(1 < i <
i
b' =ns -a,
Sup-
A +B.
a' +b
=
nt
is a solu-
Distinctness follows from the
b +dt4 BJmB,
and so
39
Lemma 4.
s
Suppose there are subscripts not in
3.
be the least subscript,
and
where
TJ
aEA, bEB, dkED, dk
a)
kE
TJ
b)
if
e
c)
if
t
and
=
nt
.
Let
,
k > s;
k (Definition 3. 2), then
then all subscripts
s,
Let
Then
O.
corresponds to
=
TJ
any subscript, not in
t
+b+dk
a
T.
k, k +1,
b > e;
,
r -1
are in
ti
TJ
and
.
Proof,
a)
Suppose
k
a
ft
Ti..
+b +dk
Then, since
=
nt
,
Hence
This contradicts the definition of TJ
TJ +1 is not empty.
kETJ. Since t{TJ, then nt 4 A +B by Lemma 3. 1(f). Since
.
nt4A +BJ,
then nt
<
ds
by Lemma 3.2
and Corollary 3.
so
dk<nt<ds
Hence
k > s
.
by the definition of the set D.
1,
and
40
Since
b)
Then
Suppose
e +dk E B
b
t
Suppose
E
b < e.
k
nt
t TJ.
contradicting our hypothesis
TJ,
sponds to
=
correspond to k.
e
Then
e.
=
a +e +dk
and so
we let
by (a),
kE T
Hence
This contradicts our hypothesis that
by Definition 3. 2.
Hence
b
J
e
e
and so
,
b
/
e.
correb > e
follows.
c) Let the subscript
definition of the set
D,
satisfy
i
k < i < r -1.
Suppose
d. < dk.
t
=
Then, by the
Then, by
s.
hypothesis,
a+b+dk
and so
dk
<
ns,
=
n
,
strict inequality holding by Corollary
di
<
dk
<
3. 1.
Hence,
ns
and so
n.
>
i
d
s
.
and therefore,
Thus,
nEA +BJ by Lemma
3. 1(f).
This proves (c), thus completing the proof of Lemma 4. 3.
3. 2,
iE
TJ
by Lemma
41
We now
state Conjecture 2.7 under the restriction
r
< 5,
and proceed with its proof.
Theorem
r
4. 1.
ti
A +B +(A +B)
If
Proof.
=
In, and if
n
has
A +B
gaps, then
< 5
p(n,A,
r
=
2, 3, 4,
r-1.
proof into four parts corresponding to
We divide the
respectively.
5
Then
r =2.
Suppose
i)
B) >
=
1
by Lemma 4.
r -1 T
E
Hence no subscripts are missing from
and Theorem 4.
T
1.
follows
1
by Lemma 4. 2.
ii) Suppose
r
=
Then
3.
2 =
r -1 TJ
E
by Lemma 4.
Hence there is at most one subscript missing from
rem 4. again is valid by Lemma
1
iii) Suppose
If
r
=
4. 2.
at most one subscript is missing from
again is valid by Lemma 4.
subscripts
1
subscript not in
and
T
2
1,
3
=r-1
E
T.
then Theorem 4.
TJ,
1
Hence, assume the remaining two
2.
are not in
then
and Theo-
T
Then, again by Lemma 4.
4.
1.
s
==
1.
T.
Thus, if
Since
r
and so by hypothesis, there exists a subscript
=
k
4,
s
is the least
then
nl In,
such that
E
42
a+b+dk
4.3(a), we have
By Lemma
n1
=
.
and
kE T
k > s.
all the subscripts greater than or equal to
since
2f
and
T
k
>
we have
s,
k
a+b+d3
=
n1,
a+b+d
=
n3
are in TJ.
k
4.3(c),
Hence,
Then
3.
=
By Lemma
or equivalently,
where
b +d3
and
are values
b +d1
,
satisfying
bEB
of
as shown in the proof of Lemma 4. 2. However, b +d3
are not in BJ n
correspond to
of
b
b > e
since the subscript
B
Then
3.
satisfying
n
=
e +d3
E
BJ
(Th
s
B,
=
1
and
b+d3
<
a +b,
namely
our theorem for
iv)
Let
r
e +d3,
r
=
=
5.
e +d3
e
is a value
b+d
and so we have at least three distinct values of
=
Let
b +d1
Hence
by Lemma 4. 3(b).
e + d3 <
n
TJ.
and
Also, we have that
by Lemma 3. 3.
a +b
4
n =a +b,
and
b +d3,
b +d1.
bEB
satisfying
This establishes
4.
Then
4
=
r-1 TJ
E
again by Lemma 4.
Hence there are at most three subscripts missing from
TJ.
1.
If
43
not more than one of these three subscripts is missing from
Theorem 4.
follows from Lemma 4.
1
again. Hence we assume
2
T.
there are at least two subscripts missing from
the least, and
to consider,
s
Suppose
subscript not in
be any
t
=
and
1
s = 2.
Furthermore, since
Let
and
n2 E In
i
Then
s.
t
=
k
=
We have two
=
= 3
cases
14ETJ.
and so
we have by hypothesis
n2,
.
iETJ, kETJ, and
the only subscript larger than
4,
be
s
such that
k
a' +b' +dk = n3
Consequently, by Lemma 4.3(a),
t
n3 E In,
and
a +b +di
i
T.
Let
s = 2.
that there exist subscripts
Hence
TJ,
i >
in
s
s, k
>
TJ, and
so
a+b+d4
=
n
a' +b' +d4
=
n3
2'
.
Again we have that
b+d4, b+d2, b'+d4, b' +d
4'
3
are not in
TJ
.
BJ
B
since the subscripts
2
and
3
are not in
However, as before these four elements are values
s.
of
be B
44
satisfying
and
n
=
but are not necessarily distinct.
a +b,
correspond to the subscripts
e'
and
1
4
Let
e
respectively.
Then
e' +d4
e+d
B
n
satisfying
n
are in
B,
=
and therefore, are two distinct values of
a +b
4. 3(b), w e have
b'
Furthermore, by Lemma
by Lemma 3. 3.
>
e'
and
e'
b >
and so
,
b+d4 > e' +d4
b' +d4
Consequently, if
b'
b
that
e
1
e+d
1=
L b +d4
b+d4
.
e' +d4, b +d4, b' +d4
and
or
e' +d4
>
,
then our four distinct solutions are
,
e +d
since
b
e +d
e+d
# b' +d4.
To show
this, we note
implies that
+d4
1= b'
a+e+d
,
=
n2
or
a'+e+d1
and so either the subscript
2E
TJ
=
n3,
or the subscript
3
E
T
,
45
or both. Now suppose
b
e' +d4
and so our four values of
b'
=
Then, we have
.
< b +d4 <
b' +d3
< b +d2,
satisfying n
bEB
=
e' +d4, b+d4, b' +d3, b+d2
Observe here that the solution
b
since we do not know that
# b +d2
e +d
This completes the proof for
For
and
i
=
5,
a' +b' +dk
than
2
=
=
are in TJ,
3
and
or k
4
By
e
1
# b' +d3.
2.
hypothesis, there exist
E
,
{
2,
3
} ,
i >
s, k
>
s.
Sup-
Then, by Lemma 4. 3(c), all the subscripts greater
2.
Suppose
= 3
=
n1
=
nt, t
contradicting our assumption that at least
two subscripts are missing from
k
s
iETJ, kETJ and
and so, by Lemma 4. 3(a),
i
or that
such that
k
a+b+di
pose
.
is not available to us,
e +d1
we argue as follows.
s = 1,
subscripts
r
=
are
a +b
=
i
4.
=
3.
Let
Then
e
and
respectively. Then
3,4
e'
T.
E
T
Hence, either i
and so
t
=
2.
= 3
or i =4.
Hence,
correspond to the subscripts
46
e+d3, e'+d4
are in BJ
n
=
and therefore, we have two distinct solutions of
B,
by Lemma 3. 3.
a +b
Let
k
=
Then we have
3.
a +b +d3
=
n1
a' +b' +d3
=
n2
,
,
where, as before,
b'+d3, b'+d2
b+d3, b+dl,
are values
of
bE B,
but not in
B
satisfying
B,
Also, by Lemma 4. 3(b), we have b'> e,
b' +d3
> e +d3
b+d3 > e + d3
Hence, if b
n=
a +b
b'
,
=
a +b.
and so
b > e,
,
.
our four distinct values of
bE B
satisfying
are
e+d3,
for again
n
e' +d4
b +d3
e' +d4, b+d3, b' +d3,
and
e' +d4
4
b' +d
,
3
On the other hand,
47
if
b
b'
=
then
,
e+d3< b+d3
and these four values of
that
e' +d4
b
=
1
Now let
<
b+d1,
are our solutions. Again we observe
b
is not usable since we do not know that
e' +d4
b+d
b'+d2
<
or that
k
=
e' +d4
b' +d2.
Then
4.
a+b+d3
=
n1,
a' +b' +d4
=
n2,
with
b+d3
b' +d4
>
>
e+d3,
e' +d4.
Here, we have that
b+d3, b+d1, b' +d4,
are not in BJ
e +d3
e' +d4),
r B.
Hence, if
e' +d4
e' +d4, (recall that
then
e +d3 <
or if
e +d3 <
b' +d2
< e +d3,
then
e' +d4
< b
+d4< b' +d2,
48
e' +d4
< e+d3 < b+d3 <
b+d1,
and so, in either case, we have at least four distinct solutions of
n
=
This completes the proof for
a +b.
Suppose i
4.
=
a ' +b' +dk
e
=
e +
are in BJn
B
and satisfy
again by Lemma 4. 3(b),
to consider,
For
k
k
=
=
n
b > e
=
nt, t
correspond to
e'
and
=
3.
Then we have
a +b +d4
Let
i
s = 1,
{
E
2,
and
4
3
} .
respectively. Then
k
e' +dk
d4,
a +b
=
n1
again by Lemma
b'> e'
and
3. 3.
We have
.
Also,
three cases
2,3,4.
2, we have 2, 4E TJ
and so
a+b+d4
=
n1
,
a'+b' +d2
=
n3
,
t
=
3.
Thus, we have
and as before,
b+d4,
are in
B,
b+dl, b'+d2, b'+d3
and of the form n -a,
but are not in
BJ
n
B.
Hence,
49
we have
a)
e' +d2
b)
e+d4 , e' +d2,
b+d4 # b' +d2
c)
b +d4
or
=
=
For
k
b+d4 , b' +d2, if
e' +d2
<
k
=
b > e
<
e+d4
;
e' +d2
and
and
< b +d
if e +d4 < e' +d2
b+d4
b' +d2,
then
=
b -b'
d2 -d4 > 0,
=
a +b.
we have
3,
=
3,4ET
and
2
and so
t
=
2.
We need only
in the preceding argument
3
and the result follows.
2,
For
e+d4
<
b' +d2
interchange the subscripts
for
e' +d2
if
b+d4 < b+d
Hence, in all three cases, we have four distinct solu-
.
of n
<
since if
b' +d2,
b'
e+d4
;
e +d4 <
b >
tions
<
k
=
and b'
we have
4,
>
e.
t
2
=
or t
=
3,
e
=
e'
and both
Thus, we have
a+b+d4
=
n
a' +b' +d4
=
nt
,
and
b+d4, b+d1, b' +d4,
are not in
Br-
B.
b' +dt
In addition, we have that
From these five values
distinct ones as follows:
of
b
satisfying
n
=
e +d4
a +b,
E
B
r'
B.
we find four
50
a)
e+d4 < b+d4 < b' +dt
<
b+d
b)
e+d4
<
b+d4
b' +d4
<
b'+dt,
c)
e+d4
<
b' +d4
<
<
r
if b< b'
;
=
b> b'
.
and therefore, Theorem 4.
5,
=
;
b
if
b+d4 < b+d
This concludes the proof for
b'
if
1,
1
is established.
The example following Conjecture 2.
serves to show that Theorem 4.
example, n
A +B
=
=
15, A +B +(A +B)
(A +A2)
nN
ti
=
{
A(n) +B(n) +(A +B) (n)
>
ti
=
2
in Chapter II, also
cannot be improved. In this
1
and r
I
n
=
since
6,
0, 1, 2, 3, 8, 9, 10, 12, 13, 14
}
.
However,
n-1.
At this point, let us emphasize the significance of Theorem
4.
1
relative to the proof
of
Theorem
Theorem 4.
1. 3.
1
implies the
following statement.
If
A +B +C
=
and if the sum of some fixed two of these
I
n
three sets has less than six gaps, then
A(n)+B(n)+C(n)
<
n-1
.
Therefore, we may make the assumption in the proof
Theorem
1. 3
that none
of the
sets
A +B, A +C, and B +C
of
have less
than six gaps. We now restate Theorem 1.3 as Theorem 4.
proceed with its proof.
2, and
51
Theorem 4.
2.
A +B +C
If
=
A(n)+B(n)+C(n)
Proof.
By
A
=
Corollary
(B+C)
thereby maximizing
A +B,
> 5,
The proof of Theorem 4.
0 <
> 5,
n < 15,
n
<
then
15,
.
may assume that
(A+C)
A +C,
0 <
C
,
=
(A+B)
.
,
Theorem 4.
By
and B +C
1, we
may
all have more than
These two assumptions imply that
A(n)
B(n)
< n -1
A(n) +B(n) +C(n).
assume that the sets
five gaps.
=
and
n
2. 1, we
B
,
I
C(n)
> 5
B(n)
2
> 5,
C(n)
> 5.
consists in showing that A(n)
and our hypotheses,
A +B +C
=
I
> 5,
and
n
are contradictory.
By Lemma 2. 1(g), we have
ti
(A+B)(n) +(A+B) (n)
=
n-1.
Combining this result with our assumptions, A(n)
and
C(n)
> 5,
B(n) > 5,
we conclude that
(4.2)
From
> 5,
5 <
(4. 2) and the
B(n)
< (A +B)(n) =
hypothesis
0 <
n- 1 -C(n)
n < 15,
< n -6.
we have two
results:
52
(A +B)(n) < 8,
(4. 3)
with similar inequalities for
and
A +C
and
B +C,
n > 11.
This last result establishes Theorem 4. 2 for
By hypothesis,
our three sets.
1
n < 11
.
and must, therefore, be in one of
EA +B +C
Without loss of generality, we assume that
lEA.
Let
Then
A +B
(4. 4)
=
{O,l,a2,, ah}
B
=
{0,b1,
-
,
,
bk },
h>
k
5,
>5.
contains at least the following numbers:
1
where
A
l +bk <
can sharpen (4.
<
bl
n -1,
2)
<
1
+b1 < b2 <
or
bk
1
+b2 <
< n -2.
<
bk
< l +bk,
Hence, since
l +bk4 B,
we
to
6 <
B(n)+
1
<(A+B)(n)
<
n-6,
and so
n
This completes the proof for
>
12.
n < 12.
Suppose that equality holds for all the signs "<"
of (4. 4).
53
Then
b =m (1 < m < k),
and so
- -
m
B
A + B
{0,1,'',k
=
0,1,,k+l}.
Q
Furthermore, we assert that a5 +k
elements
a5 +m
suppose that
Thus, n
=
m
(1 <
a5 +k
<
Then k
and
a5+k < n
But
not only the elements
a5 +k -3 (a5 +k -3
>
0, 1,
k +2),
n,
and therefore, that the
A +B.
> n -a5 > 0
To show
and so
this,
n -a5 E B.
contrary to our hypothesis. Hence
a5 +(n- a5)EA +B
a5+k < n.
<
are also in
k)
n.
>
},
a5
,
-> 5
k+1,
imply that
contains
A +B
but also the elements
a5 +k -2, a5 +k -1,
and a5 +k (a5 +k
< n).
Hence,
(A+B)(n)
contradicting
all the signs
inequalities
k+5 > 10,
(4. 3), (A +B)(n) < 8,
"
<
"
in (4. 4).
"<" in (4.
can again sharpen (4.
(4. 5)
>
7 <
B(n)
2)
4)
and so equality cannot hold for
We conclude
that at least one
must be strict inequality. Hence, we
to
+ 2 <
(A+B)(n)
_=
n- 1-C (n)
<
n-6
and so
n > 13.
Consequently, Theorem 4.
of the
2
is valid for
0 <
n < 13.
,
54
Furthermore from
C(n)
>
(4. 5) and our
assumptions B(n)
and
> 5
5, we establish the following inequalities:
7 <
(A+B)(n)
< 8
,
(4. 6)
10 <
cases to consider,
We have two
These two cases incorporate
here that
B(n)+C(n)
(4. 6) and our
give that either
B(n)
(A +B)(n)
n
13
=
assumptions,
or
= 5
generality, we assume that
< 11
C(n)
B(n)
=
=
.
and (A +B)(n)
= 7
and
n
B(n)
> 5
=
14.
and
We
8.
observe
C(n)
> 5,
Hence, without loss of
5.
or equivalently
5,
=
k
=
5,
in the argument to follow. However, before we proceed to this
argument, let us display here all the assumptions upon which this
argument is based. They are:
1)
A+B+C
2)
12
3)
A
4)
A(n)
5)
1 E
6)
at least one of the inequalities
=
I
;
n
<n <15;
=
(B +C)
.,
,,..
,
> 5,
A;
strict inequality.
B
B(n)
(A +C)
=
=
5,
,
5 <
C
=
C(n) <
(A +B)
6
;
It
<
,N,
;
and
II
in (4. 4) must be
55
Case I.
(A +B)(n)
signs
=
"<
Let
(A +B)(n)
we have that
7,
(4.7)
where
0 <
and
0 < s < 5
strict inequality holds for
bs
<
b0
=
l+bs
+bs
and
1
+b5.
b
m
We
bs+1
-<
bk
=
b5
such that
s
'
(A +B)(n)
which are not in
A +B
one of the
=
B,
7,
bs+1
1,',$),
=
+
j-1
(j
=
1,
5-s),
,
and so
since
B
=
{
0, 1,
,
s,
A+B
=
{
0
, 1 ,
,
s+ l ,
1
+b5
bs +1
-s
=
is the
t
Then, since
bs +1
>
bs+
bs +1,
=
.
l' bs+ 1+ 1,
bs+1'
'
,
+1
bs+ 1+4 -s 1,
+5
-s
largest element in
bs+ 1
we have
bs
} ,
A +B.
Let
.
t > 1.
We
there
namely,
also have
=m (m
bs+ l -
<
Hence, since
0.
are exactly two elements in
1
Since
13 < n < 14.
Thus, there exists a subscript
in (4. 4).
"
and
= 7
also have
56
=
bs+ 1
bs+t
bs+2 - bs+ 1+1
=
bs+j-5
Consequently,
B
A +B=
{
s+t
=
s+t+ 1,
> s+ 1,
s+j j-1= s+t+(5-s)-
and
B
=
b
=
A +B
0, 1,
,
{0,1,
1
=
4+t
can be written in the form
s, s +t,
,
,s +1,s +t,
4 +t
,5
} ,
+t }.
Let us now examine the elements in
closely. For instance, we know that for
m=
A +B
i
=
2,
little more
a
,
5
and
1,,5,
ai+bm
< 1+b5
if
ai+bmEA+B;
a.+b
1
m
>
n
if
a.+b m 4A+B;
ai+b0
=
ai+b5
for
.
ai +b5 ->2+b 5
where
0 < v < 5,
i
ai A+B;
>
+B.
E
n,
Therefore, there exists
such that
a
subscript y,
57
ai+bv+l
ai+bv
< 1+b5 <
a.+b
-< 1+b5 < n-1
Thus,
(4. 8)
{ ai+bv+
n,
>
1
and so
ai +bv+ 1<
n< ai+bv+
1
or
bv+ 1 < bv+
1
Hence v
=
s
.
by (4. 7), and so by (4. 8), we have
ai +s
==
ai+bs
<
1+b5
5+t,
=
(4. 9)
ai+s+t
=
ai+bs+
1
n
>
'
and so
n-(s+t)
(4. 10)
We note
here that
<
a.
- 5+t-s
<
(i
=
2, 3, 4, 5).
(4. 9) gives
a2+s+3
<
a5+s
a2+s+t
>
n,
< 5+t <
or
n-1,
a2+s+3
>
n-t+3,
58
and so
n-t+3
(4. 11)
Hence, since
<
a2+s+3
< 5+t <
n-1
.
we have
13 < n < 14,
16-t
< 5 +t <
13,
and so
(4. 12)
6 <
t < 8.
Furthermore, we assert that a.1->
suppose
a.
1
Then, since
< s +t.
s +t (i
=
2,3,4,
5),
for
we must have
a. EA +B,
i
- a.l-< s +1,
2 <
and so
Then it follows that
s > 1.
< s +2 -a. <
1
-
ï-
and so, by the construction of the set
s,
B,
we have
s +2 -a.
E
B.
Hence
s+2
=
a.
+
(s+2-a.) EA+B,
i
i
and, by (4. 12),
number in
-
a.
> s +t
(i
s +2< s +t,
A +B
=
contradicting our assumption that no
lies between
s +1
and
2, 3, 4, 5),
Hence, this result and (4.
10) give
s
+t.
Consequently,
59
s+t
(4. 13)
-< a. -< 5+t-s
(i
2, 3, 4, 5).
=
1
desired contradiction for the case
We now have the
for suppose
and
bE B
therefore,
a2 +b
< 5 +t
by (4. 9),
= 7,
and
Consequently,
a2 +b A +B.
E
s
Then
s.
b <
(A +B)(n)
+t < a2+b < a5+s < 5+t,
and so
s+t
<
a2+b
However, by assumption,
s +t < y < 4 +t.
Hence,
ti
a2 +bE B,
I
B+B
n
sis, since by assumption
proof
of
Theorem
4.
2
for
4+t.
contains all numbers
B
1
Therefore,
<
0 , a.
y
such that
and so
}
+ B =
B
by Theorem 2. 2, contradicting our hypotheB +B
n
=
_=
13
B +A +C
or n
Before we proceed to the case
=
=
In.
n
This completes the
14, and (A +B)(n)
(A +B)(n)
=
8,
=
7.
let us make
some further observations about the preceding proof, and look at
some examples. First, we observe that (4.
13)
gives
s+t<a2<2+t-s,
since
a2
<
a5 -3,
and so we have the added restriction that
0 < s <
1.
60
We could,
from this point, proceed with a proof by considering all
the possible cases reamining under the restrictions
6 <
t <
8
0 < s <
and
1
However, this is not our purpose. We wish
(4. 12).
of the
merely to get some insight into the mechanics
proof. Therefore, suppose
s =
Then, from
t - 6.
and
1
preceding
n
=
13,
we have
and so
In.
B +B
0, 1, 7, 8, 9, 10
B
{
B +
{0,8}
For n
B
=
B +
B,
=
let
14,
=
s = 0,
{O,8,9,
10, 11, 121-,
{0,8}
=
B,
B +B
I
and again we conclude that
t
=
Then
8.
by Theorem 2. 2.
We
observe
Yl
here that we could compute the set
B
} ,
directly to show that
B
+B # In, and thereby bypass Theorem 2.2.
Let us now consider the last case in the proof of Theorem 4. 2.
Case
Let
IL
5 <
and so
n
-=
14.
(A +B)(n)
C(n)
_
With
arising from (4. 4).
=
(n- 1)(A +B)(n)
Then we have
8.
+B)(n)
=
8,
_
(n -1) -8
=
n -9
we have two
If only one of the signs
n< !i
,
possibilities
in (4. 4) is
strict
61
inequality, then, as in the case
(A +B)(n)
7, we have
=
,4+t},
B
=
{0,1,
A+B
=
{0,1,,s+1,x,s+t,,5+t},
A+B
=
{0,1,,s+l,s+t,°°,5+t,x},
,
s, s+t,
and so either
or
where the extra element
(A +B)(n)
arises from our assumption
x
8.
=
On the
other hand, if two of the
"<
signs of (4.4)
"
strict inequality, then we have that there exists
where
0 <
u <
5
and
0 <
Hence, instead
not in
B,
(A +B)(n)
_=
u
u
of two, we
namely,
B(n)+
3 =
l +bG,
8,
l+bu
<
u
<
bu+ 1
-<
1
u,
bk
have three elements in A +B
and
+bu,
1
+b5,
which are
and so
satisfying our assumption. Clearly we can-
not have more than two of the signs
inequality, for then
subscript
such that
{ s,
bu
a
are
"<"
in (4.4) being
strict
(A +B)(n) > 8.
Therefore, we have three cases to consider:
one of the signs
i)
l+bs
<
x
<
bs+1
" <P?
in (4. 4) is
strict inequality and
62
one of the signs
ii)
1
+b5
<x<n;
iii)
x
=
l +b
<II
in (4. 4) is
two of the signs
-
< 5,
IT
s).
u
are strict inequalities and so
<n
treat these three cases separately.
We
Suppose one of the signs
i)
strict inequality and
and
(0< u
u
II
n
strict inequality,
in (4.4) is
<n
and that the extra element arising from our assumption
satisfies
Let
x
=
we have
1
+bs < x
=
{
0, 1,
A +B
=
{
0, 1,
where
a +b
a > a2
1
+b5
8
s, s +t,
,
s +l,
b
E
' ,
°
x, s +t,
4 +t
} ,
,
5 +t
Then since
B.
.
s +l
Consequently, a>
b < s.
}
<x< s +t,
s +1 -b> s +l
-s>
1,
.
Since the set
since
'
,
and
a EA
and so
=
Then we have
bs +1
B
b < s +t
and so
<
(A+B)(n)
B
is still the
is the same as in the case
(A +B)(n)
largest element in
we have as before
that there exists a subscript
ai+bv
We conclude in the same
where
v,
< 1+b
<
0 < v < 5
ai+bv+1
manner that
v
A +B,
=
,
(i > 2).
s, and so
= 7
such that
and
63
a.+
s
a.+b
=
1
1
ai+s+t
(4. 14)
-
< 1+b
s
ai+bs+1
=
>
=
5
5+t
,
n,
n-(s+t) <a.<
1- 5+t-s (i ->2).
Letting
i
first, and
in the
= 5
and noting that
a5
n-t+3
in the second, inequality (4.14),
= 2
we have
a2 +3,
>
i
a2+s+3
<
<
a5+s
< 5+t <
n-1,
and so
(4.15)
<t<8.
7
Then, using (4.
and (4. 15), we have
14)
a.>3(i>2).
-
(4. 16)
1
Since
a2
<
a
<
lbs+ 1
and so
the case
bs
s +l
t
=
>
>
a +bs+
1
>
n,
This result, together with (4. 15), eliminates
8.
7,
it follows from (4. 14) that
x < bs
+1'
b
s
=
0.
B
=
Consequently, if
{0,8,9,10,11,12
b
s
=
0,
then t
=
8,
1,
and so
B +
Hence,
B +A +C
=
B+ B
{0,8 }=
In
B.
by Theorem 2. 2, contradicting our
64
hypothesis. Therefore,
Furthermore,
b
1
and
assert that
we
for suppose first that
l +bs
would lie between
a.+1
>
s
<
ai
<
1 E
a.
i
bs +1
and
l +bs
B.
where
-> b s +l - 1,
-
i > 2,
Then not only
-2.
ai
but
and both be in A +B,
bs+
1
contrary to our assumption, for since
l +bs <
a,
ai
> 3
<
ai +1
<
bs +1
1 E
Suppose now that
.
we have
B,
ai
< l +bs.
Then, since
by (4.16), we have
-(l+b s )
and so adding
-< -a.i-<
-3,
we obtain
b +2,
s
-<b s +2-a.i-<b s -1
1
2<b+3-a.<b
- s i- s
Hence, both b +2 -a.
s
i
and
.
are in
b +3 -a.
i
s
,
b +2
=
a.+(b +2-a.)
b +3
=
a.+(b
+3-a.)
i s
i
s
i
s
and so both
B,
i
and
s
are in
A +B,
member
ai
>
bs+
of
1
- 1.
a
A +B
contradiction since by assumption
between
bs +1
and
bs
+1'
x
is the only
Consequently,
65
Hence we have the following results.
a
ai
a2,
>
-> b s+1 -1
-
(i > 2),
x==a+b<bs+1'
s+1
x
> b
s
+1,
and so we conclude that
x
a2
=
bs+1 -1
=
'
(4. 17)
a.>
b
i- s+1
Combining (4.
t
=
7,
a5
8
>
a3 +2,
-
with (4. 17) we obtain
14)
s +t +2 <
since
(i > 3).
a5
and so
< 5 +t -s,
s < 1.
Consequently,
s
=
1
and
respectively are the only remaining cases to consider.
Therefore let
s
=
Then
1.
B
:
-:
{
0, 1, 8, 9, 10, 11
and so
B
{0,8}= B,
}
(t
=
7),
66
and
B
=
0, 1, 9, 10, 11, 12
{
(t
}
8),
=
and so
{0,91= B.
B +
In
either case, we reach the same contradiction
B +A +C
=
B +B
# In,
n
which concludes the proof for (i).
and
1
n<n
Suppose of the signs
ii)
+b5
<
in (4. 4) one is
strict inequality
Then again we have
x < n,
B
=
{
0, 1,
,
s, s+t,
A+B
==
{
0,
,
s+
,
4+t
} ,
and so
Furthermore,
for suppose
x
if
1
where
a +b
Then
+be B.
largest number in
and so
=
1 ,
A +B.
1 ,
aE
a +(1 +b)
=
s+t,
A,
5+t, x
,
b e B,
l +x >
n,
} .
then
l +b
since
x
But since xEA +B, we have
4
B,
is the
x < n -1,
Hence, since both inequalities cannot be satisfied
x +1 < n.
at the same time, we conclude that
the construction of
B,
that
b
=
1
bs
Thus it follows, by
+b B.
or
b
=
b5
.
We
treat these
two cases separately.
Suppose
a+bs
=
2+b5,
b
=
bs.
and so
Suppose also that
a +bs
+1 > n.
x
= 2
+b5.
Then
Consequently, it follows that
67
a+s
=
6+t,
a+s+t
>
n,
and so
6 +2t >
or
4 +t
t
=
Furthermore, since b5 +2
> 5.
b5 < n -3,
and so
t
5 <
We know
Then
that
a <
- a,
a.
<
al=
< 7.
where
and
1
x
where
Next suppose that
a.1
tion
it follows that
a +s
However
= 2 +b5,
s
> a
satisfies
we have
< n -1,
Hence,
t > 7.
(4. 18)
a22 > a.
14,
0 < s <
14 <
a+bE
=
{
b, 1 +b
a contradiction.
} ,
Then, since by assump-
i > 3.
ai +s
4.
for suppose first that
i > 2,
>
where
n,
ai
<
n-1.
Hence,
a.+s
1
-< a.+4
,
1
and so
- a.1-<
11 <
But these two
,
A +B,
a
-
Hence
a. +s > 14
-
(i > 3).
and
- a. -<
1
contradiction. For example, suppose
Then, since
11 +3
=
a. +s
nEA +B.
>
n,
we have
imply that
13,
11 <
since all the numbers less than or equal to
nEA +B
i > 3.
results,
13
s
=
a.
4,
=
s
11
and so
are in
for some
3
E
B.
1
We conclude
that
a.1-< a
where
i > 2.
-
68
In addition, we
assert that
where
a. > s +t
1-
i > 2.
-
We
proved this before in a different situation so let us be brief. Suppose
Then
a. < s +t.
2
and so
s +2
,
It follows that
s > 1.
1
and so
<a.
- l-<s +1
-a.1 B.
E
< s +2 -a. <
- s,
Thus
s+2
a.
=
+
(a+2-a. ) A+B
E
,
1
1
but
s+2
since
Hence
t > 5.
Therefore
a.
1->
s +t
<
and we have a contradiction.
s +2EA +B,
(i > 2)
-
s+t,
.
Hence, from these two results,
a.<
1- a
and
-> s +t,
a.1
obtain
< a (i
s+t < a.1-
(4.19)
-
-> 2).
Consequently,
2s +t
since by assumption a+s
a2 +s
<
=
<
2+b5,
a5+s -3
< a +s
and so
-3
<
b5
=
4 +t,
we
69
0<s
(4.20)
Consequently, we have
We
s
0,
=
or
assert further that
Then since
< 1.
s = 1.
a2 +s +t
for suppose
n,
>
a2 +s +t < n.
we have
a2 +s +t A +B,
E
a2+s+t
<
x
=
a+s
=
6+t.
Lin claims strict inequality here, but we are unable to show that
this follows from his method. Hence, we use an alternate method
for obtaining a contradiction.
s+t
<
a2
By (4. 19), we have
6+t-(s+t)
<
=
6-s
,
and so
6-2s
Therefore, s= 0
5
<t <6.
>
t
> 5.
and so using this result and (4. 18), we have
Hence, we have
5<s +t <a2 <6 -s =6.
Our problem is reduced to considering the cases,
5 <
a2
< 6
and
s
=
0,
t= 6,
a2
_
=
6.
0,5,6,7,
s
=
0,
t
Therefore, we have
8, 91 (t
=
5)
,
=
5,
70
and
B
Thus, in either case,
a2
=
{0,6,7,8,9,10}(t=6).
or
= 5
a2
=
6,
we have
nEA +B,
a
contradiction. Consequently
a2 +s +t
(4. 21)
We now use (4. 21) to
sharpen
>
n
.
(4. 18).
n < a2 +s +t < a5 -3 +s +t < a -3 +s +t
We have
= 3
+2t,
and so
6
<t<7.
This result, together with (4. 20), shows that we need only consider
the four cases
contradiction.
s =
0,1 and
t
6,7
=
respectively to get our final
For
Let us then look at these four cases.
we have
B
=
{0,6,7,8,9,
10} (s
B = {0, 1, 7, 8, 9, 10
(s
}
=
0),
=
1),
and so
B +
For
t
=
7,
{0,9}=
B.
we have
B = {0, 7, 8, 9, 10, 111 (s
=
0),
{0, 1, 8, 9, 10, 11
=
1),
B
=
}
(s
t
=
6,
71
and so
B+ {0,8}= B.
Hence, in all cases, we have
the assumption
b
=
Now suppose
a +bs > 3 +b5.
0
<v <5,
and
bs
b
=
x
= 2. +b5
x > 1+b5.
Let us again determine the subscript
such that
1+b5 < ai+bv+1 (i > 2).
<
situation, we have
ai+bv+1
x
'
and so
ai
x-bv+1
<
1+b5-bv,
or
by +1 -bv
By our
Thus we must reject
.
and also suppose
bs
ai+bv
In this
B +B V In.
assumption
x >
1
+b5,
>
x -(1 +b 5).
we have
x-(1+b5)
> 1,
and so
bv+l-bv
>
1.
v,
Then
where
72
Hence
v
s,
=
and so
ai +b
contradicting
a +b
-< 1+b 5'
Therefore,
-> 3+b5.
s
s
b =b
and
s
x
> 2 +b
5
are
impossible.
Let
s
b
b5
and
b5
x =-2+b
we conclude that
4 +t
=
and
= 6
x
have that
a3
s +2 = s
+t,
a
A +B
x
where
- a. -< 8,
3 <
1
=
a2
=
Consequently,
2.
lie between
and so t
Since
2 +b5 = 8.
=
it follows that
A +B,
Then
5.
and since no numbers in
+2EA +B,
s +t,
=
=
i > 3.
and
Hence, we have
2.
is the
s +l
largest number in
particular, we
In
satisfies
3<a3 <6,
and so
12.
9
Hence,
x.
and
a3 +b5EA +B
>
x,
This shows that we cannot have
b
Suppose
case
b
=
b
=
and
bs
a3 +b5
b5
and
x
> 2 +b5,
bv+l-bv
and so
v
=
s.
Hence,
x >
>
2 +b5.
contradicting our choice
=
b5
Then
we have
x-(1+b5)
5)
and
> 1,
x
= 2
a > 3.
+b5.
As in the
of
73
ai +b
< 1+b5
ai+bs+1
,
x (i > 2).
s+1
Consequently,
x
since
and
a > a2
a+b5
=
a2+bs+1
>
ai
ai +b5
and
> a
a2 +bs
> l +bs
bs +1
and
b5_
and
a2
=
bs
>
a2+bs+1
=
and so
bs +1,
where
n,
x'
and so
b5 > bs
+1'
a+b5
Hence, a
->
.
s = 4.
-
i > 3.
It
follows that
Therefore, since
and since there are no numbers in
+l'
we conclude that
a2 +bs
A +B
> bs +1
=
between
b5
But
this is a contradiction for then
b5
<
a2 +bs
This completes the proof for
iii)
and so
Then
x
<
a5 -3 +bs
b - b5 and
Suppose two of the signs
1+b
u,
where
u
<
r
s,
1
+b5 -3
x >
1
=
b5 -2.
+b5.
"<" of (4.4) are strict inequalities
u L 5
and
0 < s < u <
4.
74
B
=
{0, 1,
A+B
=
{
0,
,b s' bs+ 1'
.
1 ,
,
°
bu,
.
bs+1, bs+1'
'
bu+ 1'
.
.
,b51,
bu' bu+l, bu+1,
,
1+b5
1.
}.
Let
t
If we
=
bs+ 1 bs
=
bu+ 1 -bu
> 1.
consider the numbers
a. +b0,
where
0
w
1,
>
i > 2,
<v<
ai +bl,
then as earlier we can find a subscript
a.i +b
i
<
v+l
1+b5
>
and conclude in the same manner that
v
If
v,
where
such that
5,
ai+bv
either
ai +b5,
,
-s
=
v
=s,
or
v
n,
by
=u.
then
ai+bs
< 1+b5
(4. 22)
ai+bs+1
>
,
n,
+1
-b
-bv
> 1.
Consequently,
75
and if
v
=u,
then
a +b
i
u
-< 1+b5'
(4. 23)
ai+bu+1 > n (i
>
2).
Therefore, either
(4. 24)
1+b5 -bs,
n-bs+1
<
ai
-<
n-bu+1
<
ai
< 1+b5 -bu
or
(4. 25)
Observe here that in either case,
v
=
ai+bs
s
<
(i > 2).
or
v
=
we have
u,
1+b5,
(4. 26)
ai+bu+1
due to our assumption
0 < s <
u < 4
>
n (i > 2),
.
Due to the construction of the set
bs
=
s,bs+1
=
s+t, bu
B,
=
we may write
t+u-1.
Then
bu+l
=
t+u-l+w,
b5
=
5+(t-1)+(w-1) =t+w+3
<
n-2
=
12,
76
and so
t+w < 9.
Using these identities, (4. 24) and (4.25) become
14-(s+t)
<
i- t+w+4-s,
a.
<
14 - (t+ u +w - 1) < a < t +w +4 - (t +u - 1) ,
1-
and so either
2t +w
>
11,
t+2w
>
11.
or
By the definition of
Now suppose
and so
we have
Hence
and
t
w > 5,
a
t
< 6
7
w > 2.
we have
t > 5,
t +2w > 11,
again a contradiction.
-w < 1,
<t +w
< 9.
that
a.
i
(4. 28)
i > 2.
2t +w > 11,
and
and so
(4.27)
where
t > 2
contradiction. Similarly, since
and so
t +w > 7,
We have
we have
Then since
t +w < 6.
w < 6 -t < 1,
w,
We
-> max (t, w, b s +1 ),
may suppose that
i
==
2.
Further suppose that
77
a2
<
Then
t.
2+bs
<
a2
+bs
-< bs +1
'
contradicting our assumption that no number in
and
l +bs
for
a2
<
bs
Hence
+1.
a2
for then
w,
2+bu
<
a2+bu
and again there is no number in
Hence
a2
<
=
s+l.
Thus, adding
bu+1
<
Hence,
2 <
a2 <
'
between
A +B
The last possibility,
a2 > w.
bs+l
similar contradiction
We have a
t.
>
lies between
A +B
a2
s+l,
<
l +bu
bu
+1'
implies that
bs +1
-(s+l)
or
and
<
a2
<
-2.
we obtain
s +2,
1
< s+
2-a2
s,
<
and so
a2+(s +2 -a2)
is in
A+B.
have
w > 5,
Hence,
s+2
=
s+t,
0 < s <
s +2
and so
t
=
3,
the proof of (4. 28).
<
s+l
<
By (4. 27), we
2.
and by what we have just shown
5<w<a
since
=
a2
>
w.
Hence,
4,
and we have a contradiction.
This completes
78
From
that
(4. 28), it follows
i-> 4,
a.
(4. 29)
where
i > 2,
If we
since by (4. 27) either t
> 4
w > 4.
can show that
(4. 30)
0, ai }+ B
{
for some fixed
=
B,
then, by Theorem 2. 2,
i > 2,
is the contradiction that we wish to obtain.
to show that for some fixed
exist an
or
e .E B
i
where
i,
B+33'
In.
n
This
Consequently, we wish
i > 2,
there does not
such that
a.+e.
i
< n
i
and
a.+e. B,
i
i
thereby showing that
{0,ai +B =B
}
for that fixed value of
we assume that for each
such that
Hence, we assume the contrary. That is,
i.
i
where
-
i > 2,
there exists an e.
1
E
B
79
a.+e.
i i
<
n,
(4. 31)
$ B
a.+e.
i i
Then, since
a.1 +e.1
of the following
a.+e.
a)
1
=
l+b
at least one
;
s
ai+ei
=
1+b5
;
c)
a.+e.
=
l+b
.
1
u
contradiction to our assumption by producing a value
of
for which these three equations all fail to hold.
This equation fails to hold for each
a)
ai+ei
ai +ei
Suppose
b)
ei
Hence
+e.
E
e.
1
B
for by (4. 28),
i > 2,
Hence,
i- b s +1
a1. >
1
i
three equations holds:
b)
We obtain a
for each
a.1 +e. EA+B,
1
1
1
i
and
B
.
=
b
s
and so
or
=
1
2 +b5 =
1
=
b
u
>
,
l+bs
.
Then
+b5.
1+b5-ai
=
e.
bs+l
>
<
b5
.
for if this were not the case, then
ai +(1 +ei) EA +B
since
2
+b5
<
n.
We
80
consider these two subcases separately.
ai +ei
Suppose
and
= 1 +b5
e.
bs.
=
Then since by (4. 26),
a +bs < 1+b5,
.
J
where
J
and by assumption
> 2,
ai +bs
a.
since
at least for
A(n)
> 5,
j =
aj+bss
(4. 32)
1
+b5,
largest element in
we conclude that
is the
=
2,3,4,
Hence,
A +B.
we have
< 1+b5,
J
a
contradiction to the supposition a.+e.
Therefore, we have shown that if
e.
=
b
i
ai +ei
= 1
+b5
fails to hold for
Next suppose that
and
2 < i < 3.
ai +ei
i
=
s
1
,
if i
+b5
=
2, 3, 4.
then the equation
2,3,4.
=
1
=
and
+b5
ei
=
bu,
for
i
fixed
Then
ai+ei
=
ai+bu
=
a4+bu
>
1+b5
,
and so
Hence, by (4.23),
v # u,
and so
1+b5
v
=
.
s.
Thus, by (4.22), we have
81
a4+bs+ 1
n,
>
and
a4+bs
strict inequality holding
< 1+b5,
Using this result and (4. 28), we
by (4. 32).
have
bs+1
where
0 < j <
e4 <bs,
a4 +e4
>
a4+b.
<
J
By (4.31),
s.
- b5,
<
<
n
Then
contradicting
n,
>
a4 +e4
e4 > bs.
for suppose
a4 +bs+ 1
a4
<
a4 +e4t B.
and
e4 > bs+
a4 +e4
<
Hence
and so
l'
Thus, it follows
n.
that
bs+
a4 +e4
and so
=
1
a4 +e4
since
+bu,
< b
1
+bu
1
is the only member of A +B
satisfying the above inequalities and which is not in
a4 +e4
=
then for
l +bu,
n > 2+b
a
=
contradiction. Hence,
fails to hold for
i
=
c)
Suppose
=
i
i
ai+a4+e4 > bs+ 1+a4+e4
=
b
u
,
>
n,
then the equation ai +ei = 1 +b5
Thus for both subcases of case (b) we see
= 1
a. +e.
i
bu)
e.
if
2, 3.
that the equation ai +ei
1+
But if
we have by (4. 28) that
2 < i < 3
ai+(
B.
=
fails to hold for
+b5
1
+b
u
,
where
i
=
i
2, 3.
=
2, 3.
Then
82
a2+e2
and so since
we have
a3 > a2,
a3+e3,
=
By (4. 29), a. > 4,
e2 > e3.
so we also have
e3
We wish to show
that
e2 > bs
+1'
that
<e2<b u.
Then, since
a2-1
But, by the definitions of
=
Therefore, suppose first
e2 < bs
+1'
a2 +e2
bu-e2
bu-bs+1
=
we have
bu-bs+1
<
and
bu
l +bu,
=
bs
we have
+1'
t+u-1-(s+t)
=
u-1-s,
and so
a2-1
<
u-1-s.
Thus
4<a2<u-s<4,
which is a contradiction. Hence,
Next suppose
e2
=
bs +l'
e2 < bs
+1'
Then repeating the same argument
with equality, we conclude that
4 < a2
=
u-s
<
4,
83
and so
a2=4,
Consequently, since
and so
1
<
t
e2
since
< 4
are between
A +B
between bs +1
2
Hence, since
6 <
a2 +b
a2
1
=
=
e2 =b1 =t,
and again no numbers in
a2EA+B
Also since all the numbers
s +t.
bl +l
we have
0,
=
< b3
we have
=
b1 +2 < b4
=
bu =b1 +3 < 7.
we have
4,
<
s
are in B,
bu
+b4
=
and
b1 < b2
-<
and
bs +1
4
s +l
and
=
=
u=4.
=0,
s
a +b2
=
2 +b4 <
a2 +b3
= 3
+b4 < a2 +b4
=
4 +b4 < 11 < n,
and therefore,
l+b
u
,
2+b
u
,
are all greater than or equal to
dicting
u
=
4.
We conclude that
3+b
l +b
u
u
e2
4+b
,
u
and are in A +B,
bs +1.
We now have
O<e3 <e2<s,
_
and so
s > 1,
u>
2,
l<1+e3<1+e2<s+1.
Hence
contra e2 < bs+1
84
Consequently,
(4. 26)
1
Hence, since
+e3 E B.
1+
e3 <
s,
we have by
that
2+bu
Thus
2 +b
bu+1
and
u
E
A +B.
bu
+1'
=
a
3+
3)-< a3+bs <
(1+e
Since there exist in
A +B
1+b5.
no numbers between
we must have
2+bu
_
bu+1'
and so
=
w
bu+ 1-bu
=
2.
Thus, by (4. 27) and (4. 28), we have
5
<t <7,
a2 > s +t
since
s > 1.
> 5,
Recalling that
b5
=
t+w+3,
bu
=
t+u-1,
we have
b5-bu - t+w+3-(t+u-1)=w-u+4
since
u > 2.
Thus,
< w+2
=
4.
85
bu
>
b5 -4,
and so
5 +bu >
Hence, since
a2
> 5,
we have
a2+bu
and so, by (4.23), v
=
1+b5.
s.
>
5+bu > 1+b5,
Therefore, by (4. 22), we have
a2 +bs +1
a2 +s
<
a5 +s
n,
>
<
1
+b5.
The proof can now be completed with the help of the following
three inequalities, validity
already been established
of which has
(see 4.22 and 4.28).
a2
-> bs+l
a2+(l+bs)
(4. 33)
a2+bs+1
We show
>
'
<
b5,
n.
that
{O,a2 } +A +B =A +B
.
86
Recall that
2 +b
u
=
b
Therefore, the set
u +1
A +B
Consequently, if
=
{
0, 1,
satisfies
z
,
bs +1, bs
,
+1,
then
- - b s +1,
0 < z <
l +b5
} .
z eA +B.
Thus,
by (4. 33),
- a2+z < a2+( l+bs
<
bs+ 1
Hence not only
but
z
satisfies bs +1 <
z <
1
is in
a2 +z
<
A +B.
Then again
+b5.
)
1
+
b5
.
Now suppose
z eA +B
z
and, by (4. 33),
we have
a2+z
Hence
{
a.+e.
=
O,
a2 }+ A +B
# I
A+B+(A+B)
We
a2+bs+ 1 > n.
Thus, we have shown that
a2 +z 4A +B.
and so
>
A +B,
contradiction.
a
,
n
=
conclude that we must reject the assumption that
l +bu,
where
i
=
2, 3.
This means that this equation
x
fails to hold for
i
=
2,
or
i
=
3,
or both of these values for
i.
Hence in all three cases (a), (b), and (c) our equation fails to hold
for
i
(4. 30).
=
2,
or
i
=
3,
and we have obtained our contradiction to
This completes the proof of Theorem 4. 2.
87
The example to show that Theorem 4.
2
given in Chapter II following Conjecture 2. 2.
say,
of
cannot be improved was
Theorem 4.
2
course, that it is not possible to find three sets, for
such that
A +B +C
=
In
and
A(n) +B(n) +C(n) < n -1,
as we will see in the next chapter.
does not
n > 15,
88
CHAPTER
RESULTS FOR
In this final
k
V
SETS
> 2
chapter, we turn our attention to the proofs
several miscellaneous theorems, some
of which
of
were mentioned
in Chapter I. Also included here is a discussion of the possibility
generalizing Theorems 4.
of
We
1
and 4.2 to
sets where
k
k > 3.
close the chapter with a few unanswered questions.
Theorem
1.
1
is a necessary tool in the proof of Erdös and
Scherk's generalization, Theorem 1.2 (4, p. 45). To keep this
work self contained, we restate Theorem
1.
1
as Lemma
5.
1
and
proceed with its proof.
Lemma
5. 1.
If
OE
A1,
OEA2 and n is a gap in
Al +A2,
then
A1(n) +A2(n)
Proof.
1
n
<
=
n -a <
n-1.
Suppose
<
n -1.
aEA1
and
Furthermore,
n -a
a +(n -a) EA1 +A2,
a
<
1
A2,
n-
1 .
for if
Then
n -a E A2,
contrary to the hypothesis. Hence
A1(n-1)
< A
2(n-
1)
=
n-1-A2(n-1)
and so
A1(n-1)+A2(n-1)
<
n-1.
then
89
However,
n #A2.
if
OEA1,
Thus,
A1(n)
=
niA1 +A2,
and
OEA2
and
A1(n -1)
A2(n)
OA,
then
A2(n -1),
=
and
and
therefore,
A1(n)+A2(n)
<
n-1,
which concludes the proof.
Now we
restate Theorem 1.2 as Theorem
with its proof.
a
5.
1
and proceed
Erdos and Scherk indicated the lines along which
proof of this theorem would proceed, but the details were supplied
by the author.
Theorem
If
5. 1.
OEA. (i
=
,k),
1,
and n is a gap in
k
A.,
then
i
k
i=1
)
Ai(n)
<
i=1
Proof.
where
1
By Lemma 5. 1, since
< j <
m
<
k,
OEA.,
OEA
m
then
A.(n)+A m (n)
- n-1.
<
Thus,
k-1 k
)
[
A.(n)+A
j=1 m=j+1
m
(n)] < (n-1)[ 1+2+
+
(k-1)]
,
,
and
n A. +A
m
,
90
and so
k
Ai(n)
(k -1
<
Z
(k-1)(n-1)
.
i=1
Finally,
k
Ai (n) < 2
(n-1),
1
i=1
and the proof is complete.
That this is the "best possible" inequality is illustrated by the
following example of Erdös and Scherk (4, p. 45).
and let A. (i
i
=
1,
contain
,k)
such that
n21 < <
x n- 1.
hypotheses
of
Theorem
0
The sets
n be odd
and the set of integers
Ai
i
x
thus defined satisfy the
for each contains
5. 1,
Let
0
and
n is a gap
k
in
y A..i
Furthermore,
i=1
A (n)
i
=
n21
n-1- n21
2
2
'
and so
k
Ai(n)
=
2 (n-1).
i=1
The above example shows that Theorem
5.
1
cannot be improved
91
without restricting
more severely than to be odd. Let us then
n
discuss briefly some other possible restrictions which could be made
on
n.
We
question, for instance, if Theorem
proved if we restrict
n
to be even.
5.
could be im-
1
Let us look at the example
corresponding to the one above and let n be even. Then we let
A. (i
=
I,
,
k)
contain
and the set of integers
0
2 < x < n-1. Again the hypotheses
Furthermore,
Ai(n)
=
of
n-1-2 = 2
Theorem
-1
5.
x
such that
are satisfied.
1
,
and so
k
Ai(n)
=
-(n-2).
i=1
k
Although we don't know that
A.(n)
is maximized, for
n
even,
i=1
with this example, it does point out that Theorem
greatly improved under the added restriction that
We have
already seen some
of the
obtained under the restriction that
n
5.
1
n
cannot be
is even.
results which have been
is the smallest gap in
k
EA.,
for instance, the results
of
Erdös and Scherk (4, p. 46),
i =1
and Kemperman (5, p. 376)
of
discussed in Chapter
I,
and the results
Lin (6, pp. 27, 31) proved previously in this work. However,
92
there are still many questions unanswered relative to this restriction,
and some of these are posed later in this chapter.
Another possibility for improving Theorem
restrict
n
so that
n
is the
sth
5.
1
Z A.,
gap in
would be to
where
s > 1.
i =1
What could be done, for instance, if
were the second gap, or
n
the third? We see that there are many questions yet to be answered.
The material in the remainder of this chapter is original with
the author.
Before we terminate the discussion of Theorem
serve here that with Theorem
Theorem
4. 2 is
Corollary
5.
1
we ob-
5. 1,
available, a weaker version
of
immediate, namely, the following corollary.
If
5. 1.
A +B +C
=
I
n
and
0 < n < 11,
then
A(n) +B(n) +C (n) < n -1.
Proof.
In
order to maximize A(n) +B(n) +C (n), we assume that
A
By
theorem
4. 1, we
=
(B +C)
,
B
=
(A +C)
,
C= (A +B)
.
may also assume that
A(n)
> 5,
B(n)
> 5,
C(n)
> 5,
and so
(5. 1)
We note
A(n) +B(n) +C(n)
> 15.
here that these are the same assumptions made in the proof
93
of
Theorem 4.
2.
Since the hypotheses of Theorem
A(n)+B(n)+C(n)
Combining this result with (5.
1)
<
5.
3
1
are satisfied, we have
(n- 1).
yields
15 < A(n)+B(n)+C(n) <
2 (n-1),
and so
11 <
This contradicts
n < 11
n.
and the corollary is proved.
It seems that extending Corollary 5.
of
Theorem
4. 2 is a
1
to the stronger result
very difficult process. In searching for some
way to improve on the proof of Theorem 4. 2, and in particular,
searching for some way to bypass Theorem
4. 1, and
therefore
Mann's transformation, the author noticed that the inequality
A(n) +B(n)> (A +B)(n)
is implicitly contained in Theorem 4. 2.
For instance, using Theorem 4.
1, we
assumed that A(n) +B(n) >10,
but using other results also came to the conclusion that
and eventually showed that this was impossible.
In
these observations, we prove the following result.
Theorem
5. 2.
If
A +B +(A +B)
=
In,
then
(A +B)(n)< 8,
connection with
94
if, and only if,
A(n) +B(n) < (A +B)(n)
ti
A(n)+B(n) + (A+B) (n)
If A +B +(A +B)
ti
=
1
(g),
which states:
then
In,
(A +B)(n) +(A +B)
Therefore, if A(n) +B(n)
=
n-1.
The proof is based on Lemma 2.
Proof.
n -1
<
ti
(n)
ti
n -1
.
then
< (A +B)(n),
(A +B) (n) +(A +B)
=
(n) > A(n) +B(n) +(A +B)
Conversely, if A(n) +B(n) +(A +B)
ti
<
ti
(n).
then
n -1,
ti
A(n)+B(n) <n-1-(A+B) (n),
and so
A(n) +B(n) < (A +B)(n)
.
This completes the proof.
Let us illustrate this theorem by altering slightly the sets
Al
and
A2,
without changing the set sum
example used following Conjecture
2. 2 in
and
A
=
B
=
{
0, 1, 8, 10, 14
{0,2,9,13}.
} ,
Al +A2,
Chapter II. Let
in the
n
=
15
95
Then
A+B
(A+B)
=
{0, 1, 2, 3, 8, 9, 10, 12, 13, 14 },
=
{
ti
0,4,
8, 9, 10, 11 },
and so
A+B+(A+B)
ti
=
In,
n
and
A (n)+B(n) < (A+B)(n)
7 =
=
9.
Hence,
ti
A(n)+B(n)+(A+B) (n)
by Theorem
5. 2,
A1(n) +A2(n)
>
r
(Al +A2)(n),
in the sense that
imply that
>
and consequently,
Also observe in both examples
n -1.
Hence, Theorem 4.
6.
=
r
n-1,
whereas, in the original example,
A1(n) +A2(n) +(A1 +A2) (n)
that
<
A +B +(A +B)
=
and Theorem
1
I
n
5. 2
and A(n) +B(n)
are not related
< (A +B)(n)
for this is clearly not the case. However, the
< 5,
converse is true, which we state in the following corollary.
Corollary
gaps,
5. 2.
If
A +B +(A +B)
ti
=
I
n
and
A +B
then
Proof.
By
A(n)+B(n)
<
(A+B)(n)
Theorem 4.
1,
under our hypotheses,
.
has
r<-
5
96
ti
A(n)+B(n)+(A+B)
Consequently, by Theorem
<
n-1
.
5. 3,
A(n) +B(n) < (A +B)(n),
and the proof is complete.
here that Theorem
We note
sets where
case
k
=
k
For instance,
ti
A+B+C+(A+B+C)
If
can be generalized to
k
The proof is essentially the same as for the
> 3.
3.
5. 2
for k
=
=
we have the statement:
4,
then A(n)+B(n)+C(n)
In,
-< (A+B+C
)
(n)
if, and only if,
ti
A(n)+B(n)+C (n)+(A+B+C ) (n)
suppose
k
=
r
and 4.
4,
2
A +B +C +E
sets where
k
< 5,
k > 3.
ti
=
In
and if
A +B +C
ti
0?
What is the largest positive integer
p
In
has
is
n) <
=
Therefore,
then the following questions naturally arise:
E
2.
to
A +B +C +(A +B +C)
If
1.
where
1
.
investigate the possibility of generalizing
We wish now to
Theorems 4.
n-1
<
and
(A, B, C, (A+B+C)
0< n<
p
,
imply that
such that
r
gaps,
97
A(n)+B(n)+C(n)+E(n)
We
answer
"
no" to the
<
n-1
?
first question by exhibiting a counter
example and can only make a conjecture concerning the second
question. Observe here that the first question is not a generalization
of
Theorem 4.
Conjecture
2.
1
5
to
sets where
k
k > 3, but a
under the restriction
is implied by Theorem 4.
We
1.
r
generalization
which, for
< 5
k
=
of
3,
were unable to produce a counter
example for the generalization of Theorem 4.
1
to
k
=
4
sets.
Consider the following sets as a counter example for the first
where
question,
n
Let
16.
=
A
=
{0,1,9,11,13,15
B
=
{
C
=
{0,4,9,
=
{
,
0, 2, 9, 10, 13, 14 },
10, 11, 12
}.
Then
A+B+C
(A+B+C)
M
=
0, 1, 2, 3, 4,
E
=
=
In
{
and
A+B+C+E
Hence
r
=
2,
but
n
.
0, 8
5,6,7,9,
},
10, 11, 12, 13, 14, 15
} ,
98
e(A, B, C, E, n)
=
A(n- 1)+B(n- 1)+C (n- 1)+E(n- 1)- (A+B+C+E)(n)
=
5+5+5+1-15
contradicting the conclusion
of
> 0,
the first question. Also observe
here that
15
=
A(n)+B(n)+C(n)
(A+B+C)(n)
>
14,
=
in keeping with the generalization of Theorem
We make the following
5. 2
to k
=
4
sets.
conjecture concerning the second
question and feel quite strongly that it is valid.
Conjecture
If
5. 1.
A +B +C +E
I
=
A(n) +B(n) +C(n) +E(n)
where
A(n) > 1,
B(n) > 1,
C(n)
> 1,
and
n
and
<
0 <
attempts on the part
5.
1
of the
then
n -1,
E(n)
> 1.
The preceding counter example used for the
shows that Conjecture
n < 16,
cannot be improved.
first question also
Unfortunately, all
author to prove this conjecture have met
with failure, but on the other hand, all attempts to find a counter
example for
n < 16
have also met without success.
Another question which deserves consideration concerns the
following generalization of Theorem 4.
3.
2
and Conjecture
What is the largest positive integer
which perhaps depends upon whether
k
g(k),
5. 1,
where
k
>
3,
is even or odd, such that:
99
k
If
Ai)
(
r\
N
=
and
In
n
0 <
then
n < g(k),
i=1
Ai(n)
n-1,
<
illL=1
i=1
where
A.(n) >
-
(i
1
=
,k)
1,
?
The best the author can do with this question is state that it
appears that it is always possible to find a counter example for
n
=
2k,
where
by forming the sets in a manner which
k > 4,
k
appears to maximize
/
This method of set construction
A.(n).
i
i=1
is as follows
We
form the first
k -1
sets by selecting as few
numbers as possible in each set in the closed interval [ 0, 2k -1 -1]
such that the set sum includes all the numbers in the interval, and
select as many numbers as possible in the interval
[
2k- 1 +1,n -1]
k
so that the
(I Ai)
,ThN
=
0I0<j<2k-1-1} ..,{j12k-1+1<j<n-1}.
k
i =1
Then the
kth
set is
{
0, 2k
-1
},
and so
A.)N
(
=
In.
This
i
i=1
procedure was used in constructing the counter example for the first
question in this chapter. For instance, suppose
this construction, with
n
=
2k,
we have
k
= 5.
Then using
100
i
A
=
{0,
B
=
{
1, 17,
0, 2, 17, 18, 21, 22, 25, 26, 29, 30 },
{O,4, 17, 18, 19, 20, 25, 26, 27, 28 },
C
0, 8,17,18,19,20,21, 22, 23, 24
E
=
{
F
=
{0,16}.
1
19,21,23,25,27,29,31},
} ,
A
We
also have
A+B+C+E+F= In,
n
but
A(n)+B(n)+C(n)+E(n)+F(n)
In our concluding
example for
g(k)
we
k
-=
5,
n
.
are seeking is not
at least not for odd
k > 4,
n-1
remark, we wish to emphasize that the
largest positive integer
where
>
=
30.
k,
g(k)
=
2k,
for we have a counter
However, again we have not been
able to construct a counter example for
n < 30
and
k
=
5.
101
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3.
Dyson, T. J. A theorem on the densities of sets of integers.
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Erdös, P. and P. Scherk. On a question of additive number
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