# Maximal Midpoint-Free Subsets of Integers

Authors: Roger B. Eggleton
Journal: International Journal of Combinatorics
 (2015)

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## Abstract

A setS⊂Z is midpoint-free if no ordered triple a,b,c∈S3 satisfies a+c=2b and a<b<c. Midpoint-free subsets of Z+ and Z are studied, with emphasis on those sets characterized by restrictions on the base m digits of their elements when 3≤m≤14, and with particular attention to maximal midpoint-free subsets with m∈3,4,7,9,…,13.


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## Body

## 1. Introduction

An ordered triplea,b,c of integers is amidpoint triple of Z if a+c=2b and a<b<c. Themidpoint of this triple is b, itslower endpoint is a and itsupper endpoint is c. For any subset X⊆Z, let Λ(X) denote the set of all midpoint triples a,b,c∈X3. When S⊂X⊆Z, let
(1)AS,X≔a∈X∣∃b,c∈S:a,b,c∈ΛX,BS,X≔b∈X∣∃a,c∈S:a,b,c∈ΛX,CS,X≔c∈X∣∃a,b∈S:a,b,c∈ΛX.
Generically, the members of AS,X∪BS,X∪CS,X are thebalance points for S in X. The balance points comprise thelower endpoint setAS,X, themidpoint setBS,X, and theupper endpoint setCS,X, for S in X. The members of ES,X≔X∖S∪AS,X∪BS,X∪CS,Xare theeccentric points for S in X. Attention to these sets appears to be a new focus, suggested by the underlying geometrical viewpoint. There is an extensive literature associated with treating a+c=2b as specifying an arithmetic progression of length3. A compact discussion and rich bibliography are given in Guy’s survey work [1, Section E10]. For an example of recent work in this field, see Dybizbański [2].IfΛS=∅ it will be convenient to say that S ismidpoint-free; moreover, S is amaximal midpoint-free subset of X if ΛS=∅ and ΛT≠∅, whenever S⊂T⊆X. Hence we have ΛS∪x≠∅ for any x∈X∖S. This implies the following characterization.Theorem 1.
IfS⊂X⊆Z, then S is a maximal midpoint-free subset of X if and only if AS,X∪BS,X∪CS,X=X∖S, or equivalently, if and only if ΛS=∅ and ES,X=∅.Note that ifX is infinite, any maximal midpoint-free subset S⊂X must also be infinite: any pair of elements of S has one midpoint and two endpoints, so precludes at most three elements of X∖S from membership of ES,X; thus ES,X would be infinite if S were finite, but then S∪{x} would be midpoint-free for any x∈ES,X, contradicting maximality of S.In [3] the notions of midpoint triple and maximal midpoint-free subset are studied for several “natural” subsets of the real numbers, but for simplicity in the present note we restrict X to Z and subsets, especially Z+≔{x∈Z∣x≥0} and Z-≔Z∖Z+=-Z+∖{0}. Here, the main focus will be on AS,X, BS,X, and CS,X when S is a maximal midpoint-free subset of X=Z+, Z- or Z.InitiallyS has been defined to be midpoint-free if S3 contains no midpoint triple. There are several semantic equivalents for this condition.Theorem 2.
IfS⊂X⊆Z, then any one of the sets
(2)AS,X∩S,BS,X∩S,CS,X∩S
is empty if and only if all three are empty.Proof.
Consider the contrary. Ifb∈BS,X∩S there is a triple a,b,c∈ΛS; this triple ensures that a∈AS,X∩S and c∈CS,X∩S. The other two cases follow in the same way.TakeX=Z in Theorem 2. Then BS,Z∩S=∅ recovers the “natural” terminology that S is midpoint-free if and only ifit does not contain the midpoint of any two of its members. Equally, S is midpoint-free if and only if it is lower endpoint-free, or alternatively, if and only if it is upper endpoint-free.Corollary 3.
A subsetS⊂X⊆Z is midpoint-free if and only if
(3)AS,X∪BS,X∪CS,X∩S=∅.This yields another semantic equivalent:S is midpoint-free exactly when it is balance point-free.For anyx∈X define themultiplicity of x as a lower endpoint, midpoint, or upper endpoint for S, respectively, as
(4)αx,S≔#b,c∈S2∣x,b,c∈ΛX,βx,S≔#a,c∈S2∣a,x,c∈ΛX,γx,S≔#a,b∈S2∣a,b,x∈ΛX.
The multiplicities for x∈X in particular cases of S⊂X⊆Z will be of interest.


## 2. An Explicit Example, Involving Basem Digit Restrictions

Let us begin with an explicit example to illustrate these notions and sample some of the typical features encountered.For any integerm≥3, let Zm+(0,1) be the set of integers x∈Z+ with regular base m representation in which all digits are restricted to the set {0,1}. When a,b,c are distinct members of Zm+0,1, all digits of the base m representation of 2b lie in the set {0,2}, but base m computation of a+c involves no carry over, so a+c contains the digit1in each place where the base m digits of a and c differ. Hence a+c=2b is impossible. It follows that Zm+(0,1) is midpoint-free.Letxm denote the regular base m representation of x. An easy base m computation shows that
(5)112m=111m+1m=101m+11m.
This corresponds to the identities
(6)m2+m+2=m2+m+1+1=m2+1+m+1.
Note that mm+1+2 is an even positive integer, so z≔(1/2)m2+m+2 is a midpoint for S≔Zm+0,1 when this set is treated as a subset of X≔Z+. Digit considerations make it clear that 2z=112m=am+cm has no other solutions with a,c∈Zm+0,1, so the midpoint multiplicity of z is2. In summary,
(7)z≔12m2+m+2∈BZm+0,1,Z+,βz,Zm+0,1=2.
Again, the general base m computation
(8)112m+110m=222m=2·111m
corresponds to the identity
(9)m2+m+2+m2+m=2m2+m+1,
which implies
(10)2z=m2+m+2∈CZm+0,1,Z+,γ2z,Zm+0,1=1,
when m≥4. Additionally, in the special case m=3 we also have
(11)1123+113=2·1003;1123+1013=2·1103;1123+11113=2·10003.
Hence 1123=14∈AZ3+0,1,Z+∩CZ3+0,1,Z+. There are no other solutions to 1123+c3=2·b3 with b,c∈Z3+0,1, so the multiplicities of 14 as a lower and upper endpoint are α14,Z3+0,1=1, γ14,Z3+0,1=3.Whenm≥5, it can be seen that 24m+cm=2·bm has no solutions with b,c∈Zm+0,1, so 2m+4 is not an endpoint for Zm+0,1, and
(12)α2m+4,Zm+0,1=γ2m+4,Zm+0,1=0.
Now seek a,c∈Zm+0,1={0,1,m,m+1,m2,m2+1,m2+m,m2+m+1,m3,…} such that a+c=2(2m+4) and a<2m+4<c. As m2>2m+4, it follows that a∈0,1,m,m+1. Then it is routine to verify that
(13)c∈3m+7,3m+8,4m+7,4m+8∩Zm+0,1=∅
when m≥5, but note that digit arguments must be sensitive to the magnitude of m; for instance, 4m+7=47m when m≥8, while for smaller m we have
(14)4m+7=m2+m+3=113mwhenm=4,4m+7=m2+2=102mwhenm=5,4m+7=5m+1=51mwhenm=6,4m+7=5m=50mwhenm=7.
It follows that
(15)β2m+4,Zm+0,1=0,
and 2m+4 is not a midpoint for Zm+0,1 when m≥5. This completes the demonstration that 2m+4 is an eccentric point, so the midpoint-free subset Zm+(0,1)⊂Z+ is not maximal if m≥5.We will later return to a more systematic study ofZ3+0,1 and Z4+0,1.


## 3. New Maximal Midpoint-Free Sets from Old

IfS⊂Z and c,d∈Z with c≠0, then cS+d≔{cs+d∣s∈S} is anaffine transform of S. Clearly x,y,z∈ΛZ if and only if cx,y,z+d∈ΛZ when c>0 or cz,y,x+d∈ΛZ when c<0, so any affine transform cS+d is midpoint-free if and only if S is midpoint-free.For example, letZm+0,2 be the set of all integers x∈Z+ with regular base m≥3 representation in which all digits are restricted to the set {0,2}. Then Zm+0,2 is midpoint-free because Zm+0,1 is midpoint-free when m≥3, and Zm+0,2=2Zm+0,1 is an affine transform of Zm+0,1. The last identity shows that Zm+0,2⊂2Z+. Similarly, -Zm+0,2+1=-2Zm+0,1-1 is an affine transform of Zm+0,1 so is a midpoint-free subset of 2Z-+1 when m≥3. Then
(16)Zm+0,2∪-Zm+0,2+1
is a midpoint-free subset of Z, since the positive and negative components of this set are midpoint-free, and a+c=2b has no solution with a∈2Z-+1, c∈2Z+, and b∈Z, because these conditions require a+c to be an odd integer and 2b to be an even integer.SupposeS⊂Z+ is known to be a maximal midpoint-free subset of Z. It turns out that the principle illustrated by the example in the previous paragraph holds strongly for S.Theorem 4.
IfS⊂Z+ is a maximal midpoint-free subset of Z, then the set S(2)≔2S∪-2S+1 is a maximal midpoint-free subset of Z.Proof.
AsS is a maximal midpoint-free subset of Z, Theorem 1 implies
(17)AS,Z∪BS,Z∪CS,Z=Z∖S.
But S⊂Z+, so BS,Z⊂Z+ and CS,Z⊂Z+; hence BS,Z=BS,Z+ and CS,Z=CS,Z+. Therefore, AS,Z=AS,Z+∪Z-.
Givenx∈AS,Z∪CS,Z, there exist b,y∈S such that x+y=2b and b≠y. Let δ∈0,1. Then 2x+δ+2y+δ=22b+δ, so
(18)x<y⟹2x+δ∈A2S+δ,Z;x>y⟹2x+δ∈C2S+δ,Z.
Therefore,
(19)2AS,Z+δ⊆A2S+δ,Z,2CS,Z+δ⊆C2S+δ,Z.
Conversely, suppose u∈A2S+δ,Z∪C2S+δ,Z, so there exist d,v∈2S+δ such that u+v=2d and d≠v. But d=2d′+δ, v=2v′+δ where d′,v′∈S, so u+v≡0mod2 implies u≡v≡δmod2, whence u=2u′+δ with u′∈Z. But S is midpoint-free, so 2S+δ is midpoint-free, and therefore, u∈Z∖2S+δ. Thus u′∈Z∖S. Now u+v=2d and d≠v imply that u′+v′=2d′ and d′≠v′, so u′∈AS,Z∪CS,Z. Therefore, the reverse containments also hold:
(20)u<v⟹u∈2AS,Z+δ;u>v⟹u∈2CS,Z+δ.
Hence
(21)A2S+δ,Z=2AS,Z+δ,C2S+δ,Z=2CS,Z+δ.
Forx∈BS,Z, similar reasoning shows that 2x+δ∈B2S+δ,Z+, so
(22)2BS,Z+δ⊆B2S+δ,Z.
However, this containment can in fact be proper. For instance, if S={2n∣n∈Z+}, then 2BS,Z only contains even integers, whereas {2n+1+1∣n∈Z+}⊂B2S,Z.
SinceS is a maximal midpoint-free subset of Z, no integers are eccentric for S, so no members of 2Z++δ are eccentric for 2S+δ. Then the positive integers eccentric for 2S+δ satisfy
(23)E2S+δ,Z+=2Z++ε∖B2S+δ,Z⊆2Z++ε,
where ε≔1-δ. Since AS,Z=AS,Z+∪Z- it follows that
(24)E2S+δ,Z⊆2Z+ε.
Specifically, all integers eccentric for 2S are odd, and all integers eccentric for 2S+1 are even. Therefore, no integer is eccentric for both 2S and -2S+1, so ES(2),Z=∅. Thus S(2) is a maximal midpoint-free subset of Z.Corollary 5.
IfS⊂Z+ is a maximal midpoint-free subset of Z, the balance point sets for S and 2S+δ with δ∈0,1 satisfy
(25)AS,Z=AS,Z+∪Z-,BS,Z=BS,Z+,CS,Z=CS,Z+,A2S+δ,Z=2AS,Z+δ,C2S+δ,Z=2CS,Z+δ,B2S+δ,Z⊇2BS,Z+δ.It is convenient to refer to the construction in Theorem4 as “doubling” the given set S. Other constructions involving affine transforms of a set are also of interest. For example, since Zm+0,1 is midpoint-free when m≥3, it follows that the disjoint sets mZm+0,1 and mZm+0,1+1 are midpoint-free when m≥3. In fact, their union is midpoint-free. This turns out to be “trivial.” Multiplying a member of Zm+0,1 by m simply shifts its base m digits one place, and a terminal0emerges to occupy the zeroth place; then adding1replaces the terminal0by1. Hence mZm+0,1∪mZm+0,1+1=Zm+0,1.Whenm≥4, the disjoint midpoint-free sets 3Zm+0,1 and 3Zm+0,1+1 are more interesting. Let us verify that
(26)S≔3Zm+0,1∪3Zm+0,1+1
is also midpoint-free. The two component sets are midpoint-free, so any midpoint triple a,b,c∈ΛS must have at least one member in each set. Thus a,b,c∩3Z++δ≠∅ in each case with δ∈0,1. If a,c⊂3Zm+0,1+δ, then b∈3Zm+0,1+ε for ε≔1-δ∈0,1. Then a+c∈3Z++2δ and 2b∈3Z++2ε. But 2δ,2ε=0,2, so a+c≠2b. If a,b⊂3Zm+0,1+δ then c∈3Zm+0,1+ε, while if b,c⊂3Zm+0,1+δ then a∈3Zm+0,1+ε. In each case a+c∈3Z++1 and 2b∈3Z++2δ, so a+c≠2b because 2δ∈0,2. Thus ΛS=∅, as claimed.Generalising the latter example, a “trebling” construction which produces new maximal midpoint-free subsets ofZ will now be studied.Theorem 6.
IfS⊂Z+ is a maximal midpoint-free subset of Z, and all members of Z+∖S are endpoints for S, then S(3)≔3S∪3S+1 is a maximal midpoint-free subset of Z, and all members of Z∖S(3) are endpoints of S(3).Proof.
BecauseS is midpoint-free, each of the affine transforms 3S and 3S+1 is midpoint-free. Assume a,b,c∈ΛS3. Since S(3)⊂3Z+∪3Z++1, there is a δ∈0,1 such that b∈3Z++δ. Then a+c=2b∈3Z++2δ, so a,b,c∈3Z++δ. Thus a,b,c∈Λ3S+δ=∅, so no such triple exists. Hence S(3)≔3S∪3S+1 is midpoint-free.
By hypothesis,AS,Z+∪CS,Z+=Z+∖S. Also AS,Z=AS,Z+∪Z- by Corollary 5, so every x∈Z∖S is an endpoint for S. Suppose x+y=2b and b,y∈S. Two integers from complementary sets cannot be equal, so x,y,b must be different. Also
(27)3x+r+3y+t=23b+s
holds when r,s,t∈{(0,0,0),(1,1,1),(2,1,0),(-1,0,1)}. It follows that
(28)x<y⟹x∈AS,Z,3x+-1,0,1,2⊂AS3,Z,x>y⟹x∈CS,Z,3x+-1,0,1,2⊂CS3,Z.
If s<s′ are consecutive members of S, all members of the interval
(29)s+1,s′-1≔z∈Z∣s+1≤z≤s′-1
are endpoints for S. Then 3s+1<3s′ are consecutive members of S3 and all members of 3s+1-1,3(s′-1+2]=[3s+2,3s′-1] are endpoints for S3. Also each x∈Z- is an endpoint for S, so all members of [3x-1,3x+2] are endpoints for S3. Thus AS3,Z∪CS3,Z=Z∖S3, and ES(3),Z=∅.LetUn≔{x∈Z3+0,1∣0≤x<3n} for each n∈Z+, and let S(3,0)≔S. Iterating the construction in Theorem 6 and combining with Theorem 4 yields the following result.Corollary 7.
IfS⊂Z+ is a maximal midpoint-free subset of Z, and all members of Z+∖S are endpoints for S, then the set
(30)S(3,n)≔⋃x∈Un3nS+x
is a maximal midpoint-free subset of Z, for any integer n∈Z+, and all members of Z∖S(3,n) are endpoints of S(3,n). Moreover, the set
(31)S(2,3,n)≔2S3,n∪-2S(3,n)+1
is a maximal midpoint-free subset of Z.


## 4. Subsets ofZ+ with Base m Digit Restrictions

Fix an integerm≥3. Let 0⊆D⊂{x∈Z+∣0≤x<m}≔[0,m). Then D is adigit subset for base m representations of the integers or, briefly, abasemdigit subset. Let Zm+D be the set of nonnegative integers with base m representation using only digits in D, and let xm,i denote the digit in position i≥0 of the regular base m representation of x∈Z+, so
(32)Zm+D≔x∈Z+∣∀i∈Z+:xm,i∈D.Let us say thatD is midpoint-free as a base m digit subset if 2·max⁡D<m and there is no ordered triple d,e,f∈D3 such that d≠f and d+f=2e.Theorem 8.
IfD is a midpoint-free base m digit subset with g≔max⁡D≥1 and m≥2g+1, then the set Zm+D is midpoint-free.Proof.
Supposea,b,c∈ΛZm+D. Then a+c=2b. There is no carry-over in computing this sum in base m arithmetic since all its digits are less than m/2, so
(33)am,i+cm,i=2bm,i=2bm,i
for every i≥0. Since D is midpoint-free, it follows that
(34)am,i=bm,i=cm,i
for every i≥0, so a=b=c, contradicting the initial choice of a,b,c. Thus ΛZm+D=∅, so Zm+D is midpoint-free.Three early instances of Theorem8, the first of which was independently demonstrated earlier, are of considerable interest.Corollary 9.
EachZm+0,1 is midpoint-free when m≥3.Corollary 10.
EachZm+0,1,3 is midpoint-free when m≥7.Corollary 11.
EachZm+0,1,3,4 is midpoint-free when m≥9.IfD is a midpoint-free base m≥3 digit subset, it is of interest to decide whether Z+ has any members that are eccentric for Zm+D, since this is equivalent to deciding whether Zm+D is a maximal midpoint-free subset of Z+. For x∈Z+, let thesupport for x, as a lower endpoint, midpoint, or upper endpoint for S, be defined by
(35)supp⁡A⁡x,Zm+D≔⋃b,c∈Zm+D∣x,b,c∈ΛZ+,supp⁡B⁡x,Zm+D≔⋃a,c∈Zm+D∣a,x,c∈ΛZ+,supp⁡C⁡x,Zm+D≔⋃a,b∈Zm+D∣a,b,x∈ΛZ+.
The following result is useful for settling whether the eccentric set ES,X is empty in specific cases.Theorem 12.
Supposem≥3 and D is a midpoint-free base m digit subset. If x∈[0,mk) with k≥1, then
(36)supp⁡A⁡x,Zm+D⊂0,mk+1∩Zm+D,supp⁡B⁡x,Zm+D⊂0,2mk∩Zm+D,supp⁡C⁡x,Zm+D⊂0,mk∩Zm+D.Proof.
Fix the nonnegative integerx∈[0,mk) with k≥1. (For simplicity we do not explicitly require x∉Zm+D, although Theorem 8 does imply that members of Zm+D have empty support sets.)
Suppose there existb,c∈Zm+D such that x,b,c∈ΛZ+, so x+c=2b. Base m arithmetic for this sum takes the form
(37)xm,i+cm,i+δi=2bm,i+mδi+1
for every i≥0, with carry-overs δi∈0,1 satisfying δ0=0 and
(38)xm,i+cm,i+δi<m⟹δi+1=0,xm,i+cm,i+δi≥m⟹δi+1=1
for i>0. But x<mk so ⟦x⟧m,i=0 for i≥k. In particular,
(39)xm,k+cm,k+δk<m2+1<m
so δk+1=0. For all i≥k+1, it follows that δi=0 and
(40)xm,i+cm,i+δi=cm,i=2bm,i=2bm,i.
But D is midpoint-free, so bm,i=cm,i=0 for all i≥k+1, since otherwise 0,bm,i,cm,i∈ΛD yields the contradiction ΛD≠∅. Hence b<c<mk+1.
The other two cases follow simply by noting that ifa,x,c∈ΛZ+ then a<c≤a+c=2x<2mk, and if a,b,x∈ΛZ+ then a<b<x<mk.Corollary 13.
IfD is a midpoint-free base m≥3 digit subset, then relative to Zm+D the three multiplicities of any x∈Z+ are finite.Corollary 14.
IfD is a midpoint-free base m≥3 digit subset and x∈[0,mk) with k≥1, then Mk≔[0,mk+1)∩Zm+D contains all three support sets forx relative to Zm+D. Moreover #Mk=dk+1, where d≔#D.Corollary 15.
Ifm≥5 then m+3 is eccentric for Zm+0,1.Corollary 16.
Ifm≥8 then 2m+4 is eccentric for Zm+0,1,3.Corollary 17.
Ifm≥14 then 4m+9 is eccentric for Zm+0,1,3,4.The last three corollaries leave open the possibility that their subject setsZm+D might be maximal midpoint-free subsets of Z+ when m is small enough. In the next section we shall consider Z3+0,1 and pursue other cases later.


## 5. Greedy Midpoint-Free Subset ofZ+

Thegreedy midpoint-free subset of Z+ is the set
(41)S0≔{si∈Z+∣i∈Z+}
in which s0=0 and each si with i>0 is the smallest integer satisfying si>si-1 such that {s0,s1,…,si} is midpoint-free. It has long been known [4] that
(42)S0=Z3+0,1=0,1,3,4,9,10,12,13,27,…,
corresponding to sequence A00536 of OEIS [5]. For brevity, let
(43)A0+≔AS0,Z+,B0+≔BS0,Z+,C0+≔CS0,Z+.
We shall attach the adjectivegreedy when referring to these sets. Before we prove that each of these greedy balance point sets contains all positive integers not in Z3+0,1, let us check the example 32=10123∈Z+∖Z3+0,1. The following base3computations are transparent:
(44)10123+10103=2·10113,10123+11113=2·11003,2·10123=21013=10003+11013.
Hence we have the midpoint triples 30,31,32, 32,36,40, 27,32,37∈ΛZ+, showing that 32∈A0+∩B0+∩C0+. Now consider the general case.Theorem 18.
The greedy midpoint-free subsetZ3+0,1⊂Z+ has greedy balance point sets satisfying
(45)A0+=B0+=C0+=Z+∖Z3+0,1.Proof.
Letx∈Z+∖Z3+0,1. First we show that x is a midpoint for Z3+0,1. Note that there is at least one j≥0 such that 2x3,j=1. Specify a,c∈Z3+0,1 by their base3digits; thus
(46)2x3,i=1⟹a3,i=0,c3,i=1;2x3,i≠1⟹a3,i=c3,i=122x3,i.
We have a3,i≤c3,i≤2x3,i for all i≥0. Also a3,j=0<c3,j=1, because 2x3,j=1. Hence 0≤a<c≤2x. It is easily checked that
(47)a3,i+c3,i=2x3,i
for all i≥0, so a+c=2x. Hence x∈B0+.
Next we show thatx is an upper endpoint for Z3+0,1. Let a,b∈Z3+0,1 be specified by their base3digits; thus
(48)x3,i=2⟹a3,i=0,b3,i=1;x3,i≠2⟹a3,i=b3,i=x3,i.
There is at least one k≥0 such that x3,k=2, so 0≤a<b<x. Also
(49)a3,i+x3,i=2b3,i=2b3,i
for all i≥0, so a+x=2b. Hence x∈C0+.
Finally we show thatx is a lower endpoint for Z3+0,1. We need to find b,c∈Z3+0,1 such that 0<x<b<c and x+c=2b. Base3arithmetic for this sum takes the form
(50)x3,i+c3,i+δi=2b3,i+3δi+1
for every i≥0, with carry-overs δi∈0,1 satisfying δ0=0 and
(51)x3,i+c3,i+δi≤2⟹δi+1=0,x3,i+c3,i+δi>2⟹δi+1=1.
Specify b,c by their base3digits; thus
(52)x3,i+δi∈0,3⟹c3,i=b3,i=0,x3,i+δi=1⟹c3,i=b3,i=1,x3,i+δi=2⟹c3,i=1,b3,i=0.
Ifx3,k=2 and x3,i<2 for 0≤i<k then δi=0 for 0≤i≤k. It follows that c3,k=1>b3,k=0. Since c3,i≥b3,i, for every i≥0, we have c>b.
There are integersh≥g≥k such that x3,h>0 and x3,i=0 for i>h and x3,g=2 and x3,i≤1 for i>g. If δg=0 then c3,g=1, b3,g=0; if δg=1 then c3,g=b3,g=0. In either case it follows that δg+1=1. If δh+1=1 then x3,h+1<c3,h+1=b3,h+1=1 so x<b since x3,i=0 for all i>h. If δh+1=0 then there is an integer f such that h≥f>g and δi=1 for f≥i>g while δf+1=0. If x3,i≤1 and δi=0 then x3,i=c3,i=b3,i∈0,1 so δi+1=0. Hence δi=0 for all i>f. Also δf=1, δf+1=0 and c3,f=1, so x3,f=0. Then x3,f=0<b3,f=1 and x3,i=b3,i for all i>f, so x<b.
Finally, in all three casesx3,i+c3,i+δi=2b3,i+3δi+1 for each i≥0, so x+c=2b. Hence x∈A0+.Corollary 19.
The greedy midpoint-free subsetZ3+0,1 is maximal in Z+.In [3] it was shown that S0=Z3+0,1 is actually a maximal midpoint-free subset of Z. The proof is not repeated here, but let us note that the single example -32∈AS0,Z follows from computing -32+c=2b with b,c∈Z3+0,1 in the form 32+2b=c. Base3considerations yield
(53)10123+223=11113,
corresponding to -32,4,40∈ΛZ, so -32∈AS0,Z as claimed.For brevity, letA0≔AS0,Z, B0≔BS0,Z, C0≔CS0,Z. With Theorem 18, this yields the following result.Corollary 20.
The greedy midpoint-free subsetZ3+0,1⊂Z has greedy balance point sets satisfying A0=Z∖Z3+0,1, B0=C0=Z+∖Z3+0,1.Reversing implications, this yields the following result.Corollary 21.
The greedy midpoint-free subsetZ3+0,1 is maximal in Z.Now the multiplicities forx∈Z+∖Z3+0,1 can be examined. First note that the sum of base3digits of any even integer is 0mod2, so even integers have an even number of base3digits equal to1. If x∈Z+∖Z3+0,1, at least one digit in 2x3 must be1, so the total number of such digits is 2k, and k>0.Theorem 22.
Ifx∈Z+∖Z3+0,1, then the midpoint multiplicity of x is
(54)βx,Z3+0,1=22k-1,
where 2k is the number of digits equal to1in 2x3 and k>0.Proof.
Supposea,c∈Z3+0,1 satisfy a<c and a+c=2x. There is no carry-over in the base3arithmetic for the sum, so a3,i+c3,i=2x3,i, for all i≥0. If 2x3,i∈0,2 then we must have
(55)a3,i=c3,i=122x3,i.
There is an integer j>0 such that 2x3,j=1 and 2x3,i∈0,2 for all i>j, so a<c forces a3,j=0<c3,j=1. However, when j>i≥0 and 2x3,i=1, the requirement a3,i+c3,i=1 is satisfied if a3,i,c3,i=0,1. Both possibilities are consistent with a<c, so there are 22k-1 solutions in total.To illustrate, whenx=50 then 2x=(10201)3 so ΛZ3+0,1,Z+ has just two triples with50as midpoint: (9,50,91), (10,50,90). In contrast, when x=20 then 2x=(1111)3 so ΛZ3+0,1,Z+ has eight triples with20as midpoint, ranging from (0,20,40) to (13,20,27), namely, a,20,40-a for all a∈Z3+0,1∩0,13.Fixx∈Z+∖Z3+0,1. The digits of the base3representation x3 include at least one2. An ordered pair of integers j,k with 0≤j<k iscritical for x if x3,j=2 and x3,k<2, with x3,i>0 when j≤i<k. Any two ordered pairs (j,k) and (j′,k′) critical for x areindependent if j,k∩j′,k′=∅. Any set of ordered pairs critical for x is independent if every two members are independent. Let Critx denote the family of all sets of independent ordered pairs critical for x, including the empty set.Theorem 23.
Ifx∈Z+∖Z3+0,1, then x has total endpoint multiplicity
(56)αx,Z3+0,1+γx,Z3+0,1=#Critx.Proof.
Givenx∈Z+∖Z3+0,1, we seek y,b∈Z3+0,1 such that x+y=2b. Any solution necessarily satisfies x≠y. Base3arithmetic for this sum takes the form
(57)x3,i+y3,i+δi=2b3,i+3δi+1
for every i≥0, with carry-overs δi∈0,1 satisfying δ0=0 and
(58)x3,i+y3,i+δi≤2⟹δi+1=0,x3,i+y3,i+δi>2⟹δi+1=1.
Specifying y,b by their base3digits, the equation x+y=2b requires
(59)x3,i+δi∈0,3⟹y3,i=b3,i=0,x3,i+δi=1⟹y3,i=b3,i=1,x3,i+δi=2⟹y3,i+b3,i=1.
As is easily verified, these digit specifications satisfy the requirements of base3arithmetic for x+y=2b.
Any integeri≥0 is a base3transition point for the sum x+y if δi+1≠δi. Note that the digits y3,i and b3,i are forced unless x3,i+δi=2. In the latter case there are two options:(1)
y3,i=0, b3,i=1, δi+1=0,(2)
y3,i=1, b3,i=0, δi+1=1.If x3,i=2, δi=0 then choosing option (1) preserves the carry-over state (“off”) with δi+1=0 whereas choosing option (2) switches the carry-over state (to “on”) with δi+1=1, so option (2) causes i to be a transition point. If x3,i=1, δi=1 then choosing option (1) switches the carry-over state (to “off”) with δi+1=0 and causes i to be a transition point, whereas choosing option (2) preserves the carry-over state (“on”) with δi+1=1.
LetX be a maximal block of nonzero digits in the base3representation x3, say X≔(x3,i∣g≤i≤h), and assume that X contains the digit2at least once. There must be at least one such maximal block X in x3. Then δg=0, and the “off” carry-over state δi=0 can only switch to δj+1=1 for some j such that g≤j≤h when x3,j=2 and y3,j=1, corresponding to an option (2) choice in the construction of y. The “on” carry-over state δi=1 can only switch back to δk+1=0 for some k such that j<k≤h if x3,k=1 and y3,k=0, corresponding to an option (1) choice in constructing y. If no such option is exercised, then maximality of X ensures the carry-over state δi=1 inevitably switches back at i=h+1, for in this case x3,h>0, x3,h+1=0, δh=1 and an option (2) choice for y3,h results in δh+1=1, y3,h+1=1, b3,h+1=1, δh+2=0. Note that the ordered pair j,k is critical for x, as is the ordered pair j,h+1.
Now consider any setPX of independent ordered pairs j,k critical for x, with g≤j<k≤h+1. The set PX determines a unique sequence of carry-over digits Δ≔(δi∣g≤i≤h+1), with δi=1 if j≤i≤k and j,k∈PX, and δi=0 for every other i in the interval [g,h+1]. Then X and Δ determine blocks
(60)Y≔y3,i∣g≤i≤h+1,B≔b3,i∣g≤i≤h+1
such that X+Y=2B. Each member of Critx is of the form Px≔⋃XPX, where X runs through all maximal blocks of nonzero digits containing2in x3. Any such Px uniquely determines a collection of suitable blocks of base3digits for y,b while all other digits are forced by base3arithmetic for x+y=2b, so the number of solutions for y,b is precisely #Critx.To illustrate, whenx=50=(1212)3 the possible ordered pairs critical for x are 0,1, 0,3, 0,4, 2,3, 2,4. There are #Crit50=8 sets of independent critical pairs, namely, ∅, the five singletons, 0,1,2,3 and 0,1,2,4. The corresponding triples in ΛZ+ with50as an endpoint are
(61)4,27,50,10,30,50,12,31,50,28,39,50,30,40,50,soγ50,Z3+0,1=5;50,81,112,50,84,118,50,85,120,soα50,Z3+0,1=3.Givenx∈Z+∖Z3+0,1, let X≔(x3,i∣g≤i≤h) be the leading maximal block of nonzero digits containing2in x3, so x3,i<2 when i>h, and in particular x3,h+1=0. Then any ordered pair j,k critical for x is constrained by 0≤j<k≤h+1. In particular, theleading ordered pairs critical for x are all those of the form j,h+1. In this case g≤j≤h, and there is at least one such ordered pair since x3,j=2 for at least one j∈[g,h]. Let LeadCritx denote the family of just those sets of independent ordered pairs critical for x that include a leading pair.Corollary 24.
Ifx∈Z+∖Z3+0,1, then x has lower endpoint multiplicity
(62)αx,Z3+0,1=#LeadCritx.Proof.
Continuing with the definitions and notation in the proof of Theorem23, if Px∈LeadCritx contains the leading ordered pair j,h+1 critical for x then the corresponding y,b for x+y=2b satisfy x3,j=2, y3,j=1, b3,j=0 with y3,i=b3,i when i>j; moreover, x3,h+1=0, y3,h+1=1, and x3,i=y3,i when i≥h+2. Thus x<b<y and b,y∈supp⁡Ax,Z3+0,1.
On the other hand, ifP*x∈Critx∖LeadCritx then the corresponding y,b for x+y=2b satisfy x3,k+δk=2, y3,k=0, b3,k=1 for some k∈g,h, with x3,i=y3,k=b3,i when i>k. Thus b,y∈supp⁡Cx,Z3+0,1.The multiplicities in Theorem23 and Corollary 24 are implicit, so it is of some interest to identify a class of positive integers with endpoint multiplicities that can be specified simply and explicitly. Let u3,n≔3n-1/2 be the integer with base3representation comprising a block of n digits all equal to1, so u3,n is a base3rep-unit, following terminology of Yates [6]. (See A003462 in OEIS [5].)Corollary 25.
For any integersn≥m>0 and s≥0, t>n+s, let
(63)x≔3su3,n+u3,m+3tv+w,
where v∈Z3+0,1 and w∈Z3+0,1∩0,3s. Then
(64)αx,Z3+0,1=m,γx,Z3+0,1=mn-m+1.Proof.
The only maximal block of nonzero digits containing2in x3 is X=(x3,i∣s≤i<n+s), with x3,i=2 when s≤i<m+s, and x3,i=1 when m+s≤i<n+s, with x3,n+s=0. Therefore, the only ordered pairs j,k critical for x have s≤j<m+s≤k≤n+s. No two of these critical pairs are disjoint, so the only sets in Critx are m(n-m+1) singletons and the empty set. The sets comprising LeadCritx are the m singletons {j,n+s}.For instance,x=862=10112213 arises by taking m=2, n=4, s=1, t=6, and v=w=1 in Corollary 25. The leading critical pairs for x are (j,5) with j∈1,2, yielding two midpoint triples (x,b,y) with b,y∈Z3+0,1:
(65)b=11000B13,y=1111Y13
for blocks B,Y∈0,01,1,10, so x,b,y=(862,973+r,1084+2r) with r∈0,3. Thus α(862,Z3+0,1)=2. The non-leading critical pairs for x are (j,k) with j∈1,2, k∈3,4, yielding five midpoint triples (y,b,x) with b,y∈Z3+0,1:
(66)b=101B13,y=10Y13
for B,Y∈{(000,0101), (001,0110), (100,1001), (101,1010), (111,1100)}, so y,b,x=(760+2r,811+r,862) for r∈0,3,27,30,39. Thus γ(862,Z3+0,1)=5.


## 6. Doubling the Greedy Midpoint-Free Subset ofZ+

DoublingS0≔Z3+0,1 as in Theorem 4 shows that S0(2)≔2S0∪-2S0+1 is a maximal midpoint-free subset of Z. (This is an alternative demonstration to the proof given in [3].) Here 2S0=Z3+0,2 and 2S0+1=Z3+0,2;1, where Z3+0,2;1 comprises those positive integers with base3representation in which the trailing digit (the last nonzero digit) is1and every other digit is in 0,2. The relevant balance point sets and multiplicities will now be examined briefly.For brevity, let us write(67)A0,2+≔A2S0,Z+,B0,2+≔B2S0,Z+,C0,2+≔C2S0,Z+;A0,2≔A2S0,Z,B0,2≔B2S0,Z,C0,2≔C2S0,Z;A0,2(2)≔AS0(2),Z,B0,2(2)≔BS0(2),Z,C0,2(2)≔CS02,Z.
Note that A2S0+1,Z+=A0,2++1, A2S0+1,Z=A0,2+1. Similar identities hold for other balance point sets of 2S0+1.It can be shown that every odd integerx∈Z+ is the sum of two distinct members of Z3+0,1. Indeed, there are precisely 4k such sums, where 2k+1 is the number of digits equal to1in x3 for some integer k≥0. The balance point sets and multiplicities for 2S0, 2S0+1, and S0(2) can now be specified.Theorem 26.
The midpoint-free subsetZ3+0,2⊂Z+ has balance point sets satisfying A0,2+=C0,2+=2Z+∖Z3+0,2 and B0,2+=Z+∖Z3+0,2.Corollary 27.
Ifx∈Z+∖Z3+0,1, the endpoint multiplicities satisfy
(68)α2x+1,Z3+0,2;1=α2x,Z3+0,2=αx,Z3+0,1,γ2x+1,Z3+0,2;1=γ2x,Z3+0,2=γx,Z3+0,1.Corollary 28.
Ifx∈Z+∖Z3+0,2, the midpoint multiplicities satisfy
(69)βx+1,Z3+0,2;1=βx,Z3+0,2=2k-1,
where k>0 is the number of digits equal to1in x3.Corollary 29.
The midpoint-free subsetZ3+0,2⊂Z has balance point sets satisfying A0,2=A0,2+∪2Z-, B0,2=B0,2+, C0,2=C0,2+.Corollary 30.
The midpoint-free subsetS0(2)≔Z3+0,2∪-Z3+0,2;1⊂Z has balance point sets satisfying
(70)A0,2(2)=A0,2∪-C0,2+1=Z∖S02∖2Z++1,C0,2(2)=C0,2∪-A0,2+1=Z∖S02∖2Z-,B0,2(2)=B0,2+∪-B0,2,1+=Z∖S0(2).Corollary 31.
The setS0(2) is a maximal midpoint-free subset of Z, with balance point sets satisfying A0,2(2)∪C0,2(2)=B0,2(2)=Z∖S0(2).Note that treblingS0≔Z3+0,1 as in Theorem 6 yields S0(3)≔3S0∪3S0+1, and earlier we saw that this is “trivial” in this case, because S0(3)=S0. In the notation of Corollary 7, for any integer n≥1 this implies
(71)S0(3,n)=S0,S0(2,3,n)=S0(2).
If S is any set satisfying the hypotheses of Theorem 6, and s0≔min⁡S, then thenormalized set S′=S-s0 is an affine transform which satisfies the hypotheses of Theorem 6. Without loss of generality, assume S is any normalized compliant set; then Corollary 7 shows that iterated trebling yields a sequence of maximal midpoint-free subsets of Z which asymptotically approach the greedy subset S0, because
(72)S(3,n)∩0,3n=S0∩0,3n.
In this sense we may write the asymptotic equivalences
(73)S(3,n)~S0,S(2,3,n)~S0(2).


## 7. The Midpoint-Free SetZ4+0,1

Let us now study the set(74)S1≔Z4+0,1=0,1,4,5,16,17,20,21,64,…,
corresponding to sequence A000695 in OEIS [5]. Corollaries 9 and 15 leave open the possibility that S1 is a maximal midpoint-free subset of Z+. We now settle that matter. Let A1+≔AS1,Z+, B1+≔BS1,Z+, C1+≔CS1,Z+.Theorem 32.
The midpoint-free subsetZ4+0,1⊂Z+ has endpoint sets satisfying A1+∪C1+=Z+∖Z4+0,1.Proof.
Letx∈Z+∖Z4+0,1. We seek y,b∈Z4+0,1 such that x+y=2b. Clearly any such solution has x≠y. Base4computation requires
(75)x4,i+y4,i+δi=2b4,i+4δi+1
for every i≥0, with carry-overs δi∈0,1 satisfying δ0=0 and
(76)x4,i+y4,i+δi≤3⟹δi+1=0,x4,i+y4,i+δi>3⟹δi+1=1.
Specifying y,b by their base4digits, the equation x+y=2b requires
(77)x4,i+δi∈0,4⟹y4,i=b4,i=0,x4,i+δi=1⟹y4,i=b4,i=1,x4,i+δi=2⟹y4,i=0,b4,i=1,x4,i+δi=3⟹y4,i=1,b4,i=0.
These digit specifications are easily seen to satisfy the requirements of base4arithmetic for x+y=2b. Therefore, every x∈Z+∖Z4+0,1 is in A1+∪C1+. Since Z4+0,1 is midpoint-free, it follows that A1+∪C1+=Z+∖Z4+0,1.Corollary 33.
The setZ4+0,1 is a maximal midpoint-free subset of Z+.Corollary 34.
For anyx∈Z+∖Z4+0,1 the total endpoint multiplicity is
(78)αx,Z4+0,1+γx,Z4+0,1=1.Anyx∈Z+∖Z4+0,1 has at least one base4digit in 2,3. Ahigh block in x4 is a maximal block of base4digits all in 2,3. If X≔(x4,i∣j≤i≤k) is a high block in x4, it isclear if x4,k+1=0 andleading if x4,i≤1 when i>k.Corollary 35.
The lower endpoint setA1+ for Z4+0,1 in Z+ comprises all positive integers in which the leading base4high block is clear and contains3. The upper endpoint set C1+ is the complement of A1+ in Z+∖Z4+0,1.Proof.
With the definitions and notation in the proof of Theorem32, let x4,h>0 for some h>0 and x4,i=0 when i>h. Let X≔(x4,i∣j≤i≤k) be the leading base4high block for x, so 0≤j≤k≤h and x4,i≤1 when i>k.(1)
Ifx4,h=1 then δh+1=0, so y4,i=b4,i=0 when i>h. Also x4,k≥2, while x4,i≤1 when i>k. If x4,k+δk≥3 then δk+1=1. Therefore, x4,k+1+y4,k+1=1 and δi=0 when i>k+1. This case arises just when x4,t=3 for some t such that j≤t≤k, and x4,i=2 if t<i≤k. Also x4,i=y4,i when i>k+1, so x>y if x4,k+1=1 and x<y if x4,k+1=0, the latter condition holding precisely when X is clear. On the other hand, if x4,k+δk=2 then x4,k=2, y4,k=0, while δi=0 and x4,i=y4,i when i>k, so x>y.(2)
Ifx4,h=2 and δh=0 then y4,h=0, so δi=0 and x4,i=y4,i=0 when i>h. Then k=h but the high block X does not contain3. In this case x>y.(3)
Ifx4,h+δh=3 then y4,h=1, so δh+1=1, x4,h+1=0, y4,h+1=1, and δi=x4,i=y4,i=0 when i>h+1. Once again k=h, but now the high block X must contain3. In this case x<y.(4)
Ifx4,h=3, δh=1 then δh+1=1, x4,h+1=0, y4,h+1=1 and the same conclusions as in (3) apply.Corollary 36.
The midpoint setB1+ for Z4+0,1 in Z+ comprises those positive integers x with 2x∈Z4+0,1,2∖Z4+0,2. Any such x has midpoint multiplicity βx,Z4+0,1=2k-1, where k>0 is the number of digits in 2x4 equal to1.Proof.
Letx∈Z+∖Z4+0,1. We seek a,c∈Z4+0,1 such that a+c=2x. Clearly any such solution has a≠x, so a≠c holds. Base4arithmetic requires
(79)a4,i+c4,i=2x4,i
for every i≥0. This forces 2x4,i∈0,1,2. Since x∉Z4+0,1 it follows that 2x4,j=1 for some j≥0, with 2x4,i≠1 for all i>j. Specifying a,c by their base4digits, the equation a+c=2x requires
(80)2x4,i∈0,2⟹a4,i=c4,i=122x4,i,2x4,i=1⟹a4,i+c4,i=1.
With a4,j=0, c4,j=1, clearly all solutions meet base4arithmetic requirements for a+c=2x and a<c.An alternative characterization of the midpoint setB1+ is that it comprises every x∈Z+ with x4,i∈2,3 for at least one i≥0, and in each such instance x4,i+1∈0,2.Now consider the midpoint-free setS1≔Z4+0,1 as a subset of Z. For brevity, let A1≔AS1,Z, B1≔BS1,Z, C1≔CS1,Z.Corollary 37.
The midpoint-free subsetZ4+0,1⊂Z has balance point sets satisfying A1=A1+∪Z-, B1=B1+, C1=C1+.Proof.
Let-x∈Z-. We seek b,c∈Z4+0,1 such that x+2b=c. Clearly any such solution has -x<b<c. Base4computation requires
(81)x4,i+2b4,i+δi=c4,i+4δi+1
for every i≥0, with carry-overs δi∈0,1 satisfying δ0=0 and
(82)x4,i+2b4,i+δi≤3⟹δi+1=0,x4,i+2b4,i+δi>3⟹δi+1=1.
Specifying b,c by their base4digits, we require
(83)x4,i+δi∈0,4⟹b4,i=c4,i=0,x4,i+δi=1⟹b4,i=0,c4,i=1,x4,i+δi=2⟹b4,i=1,c4,i=0,x4,i+δi=3⟹b4,i=c4,i=1.
It is straightforward to verify thatb,c are uniquely determined by these specifications, and -x,b,c∈ΛZ, so -x∈A1.Corollary 38.
The setZ4+0,1 is a maximal midpoint-free subset of Z.Corollary 39.
Anyx∈Z- has lower endpoint multiplicity αx,Z4+0,1=1.


## 8. Doubling and Trebling the SetZ4+0,1

By Theorem32 and Corollary 38, the set S1≔Z4+0,1 is compliant with the requirements for the doubling and trebling constructions of Theorems 4 and 6, so both sets
(84)S1(2)≔2S1∪-2S1+1,S1(3)≔3S1∪3S1+1
are maximal midpoint-free subsets of Z.Let us now briefly examine the balance point sets associated withS1(2):
(85)A1,2+≔A2S1,Z+,B1,2+≔B2S1,Z+,C1,2+≔C2S1,Z+;A1,2≔A2S1,Z,B1,2≔B2S1,Z,C1,2≔C2S1,Z;A1,2(2)≔AS1(2),Z,B1,2(2)≔BS12,Z,C1,22≔CS1*,Z.
Easy digit and parity considerations in combination with Theorem 32 yield the following corollaries.Theorem 40.
The midpoint-free subset2S1=Z4+0,2⊂Z+ has balance point sets satisfying
(86)A1,2+=2A1+,C1,2+=2C1+,B1,2+=Z4+0,1,2∖Z4+0,2.Corollary 41.
The midpoint-free subset2S1⊂Z has balance point sets satisfying
(87)A1,2=A1,2+∪2Z-,B1,2=B1,2+,C1,2=C1,2+.Corollary 42.
The subsetS1(2)≔2S1∪-2S1+1⊂Z is midpoint-free and has balance point sets satisfying
(88)A1,2(2)=A1,2∪-C1,2+1,C1,2(2)=C1,2∪-A1,2+1,B1,2(2)=B1,2+∪-B1,2++1.Corollary 43.
The sets2S1 and S1(2)≔2S1∪-2S1+1 are midpoint-free subsets of Z with endpoint sets satisfying A1,2∪C1,2=2Z∖2S1 and A1,2(2)∪C1,2(2)=Z∖S1(2).The balance point sets ofS1(3)≔3S1∪3S1+1 can now be considered. Let
(89)A1,3+≔AS13,Z+,B1,3+≔BS13,Z+,C1,3+≔CS13,Z+,A1,3≔AS13,Z,B1,3≔BS13,Z,C1,3≔CS13,Z.
If a+x=2b and a,b∈S1≔Z4+0,1, the conditions
(90)3a+r,3b+s∈S1(3)=3S1∪3S1+1
and (3a+r)+(3x+t)=2(3b+s) are satisfied by the triples r,s,t=0,0,0,1,1,1,1,0,-1,0,1,2. Hence, if x∈A1+∪C1+ then 3x+-1,0,1,2⊂A1,3+∪C1,3+. Such observations yield the following results.Theorem 44.
The subsetS1(3)≔3S1∪3S1+1⊂Z+ is midpoint-free and has balance point sets satisfying
(91)A1,3+∪C1,3+=Z+∖S13,B1,3+=3B1+∪3B1++1∪3B1~-1,
where B1~≔(B1+∪{1})∖2Z+.Corollary 45.
The subsetS1(3)≔3S1∪3S1+1⊂Z is midpoint-free and has balance point sets satisfying
(92)A1,3=A1,3+∪Z-,B1,3=B1,3+,C1,3=C1,3+.


## 9. The Midpoint-Free SetZ7+0,1,3

Next consider the set(93)S2≔Z7+0,1,3=0,1,3,7,8,10,21,22,24,49,….
At the time of writing, no corresponding sequence appears in OEIS [5]. However, Corollary 10 asserts that it is midpoint-free, and it will be shown that S2 is in fact a maximal midpoint-free subset of Z. It will follow from Theorems 4 and 6 that
(94)S2(2)≔2S1∪-2S1+1,S2(3)≔3S2∪3S2+1
are maximal midpoint-free subsets of Z.The methods used in earlier sections are again applicable, so fewer details are now required. Following earlier practice, let(95)A2+≔AS2,Z+,B2+≔BS2,Z+,C2+≔CS2,Z+,A2≔AS2,Z,B2≔BS2,Z,C2≔CS2,Z.Theorem 46.
The midpoint-free subsetS2≔Z7+0,1,3⊂Z+ has endpoint sets satisfying A2+∪C2+=Z+∖S2.Proof.
Givenx∈Z+∖S2, we seek y,b∈S2 such that x+y=2b. Any solution has x≠y. Base7computation requires
(96)x7,i+y7,i+δi=2b7,i+7δi+1
for i≥0, with appropriate carry-overs δi∈0,1 beginning with δ0=0. All base7digits of y,b are determined by
(97)x7,i+δi∈0,7⟹y7,i=b7,i=0,x7,i+δi=1⟹y7,i=b7,i=1,x7,i+δi=2⟹y7,i=0,b7,i=1,x7,i+δi=3⟹y7,i=b7,i=3,x7,i+δi=4⟹y7,i=3,b7,i=0,x7,i+δi=5⟹y7,i=1,b7,i=3,x7,i+δi=6⟹y7,i=0,b7,i=3.
It is straightforward to verify that all requirements are satisfied.Corollary 47.
The setZ7+0,1,3 is a maximal midpoint-free subset of Z+.Corollary 48.
ForS2≔Z7+0,1,3, any x∈Z+∖S2 has endpoint multiplicity
(98)αx,S2+γx,S2=1.Corollary 49.
The midpoint setB2+ for S2≔Z7+0,1,3 in Z+ comprises those integers x∈Z+∖S2 with 2x∈Z7+0,6∖{5}. Any such x has midpoint multiplicity βx,S2=2k-1, where k>0 is the number of digits in 2x7 belonging to 1,3,4.Proof.
Givenx∈Z+∖S2, we seek a,c∈S2 such that a+c=2x. Necessarily, any solution has a≠x, so a≠c holds. Base7arithmetic requires
(99)a7,i+c7,i=2x7,i
for every i≥0. This forces 2x7,i∈0,6∖{5}. Since x∉S2 there is an integer j≥0 such that 2x7,j∉0,2,6 but 2x7,i∈0,2,6 for all i>j. Specify a,c by
(100)2x7,i∈0,2,6⟹a7,i=c7,i=122x7,i2x7,i=1⟹a7,i,c7,i=0,1,2x7,i=3⟹a7,i,c7,i=0,3,2x7,i=4⟹a7,i,c7,i=1,3.
With a4,j<c4,j, evidently all solutions satisfy a+c=2x and a<c.The base7representation of the integer u7,n≔7n-1/6 is a block of n digits, all equal to1. Thus u7,n is a base7rep-unit [6]. (See A023000 in OEIS [5].) For any x∈Z+ let x7,k denote the integer y resulting from x7 by deleting all but the last k base7digits:
(101)y7,i=0fori≥k,y7,i=x7,ifork>i≥0.
An alternative characterization of the midpoint set B2+ is that it comprises every x∈Z+ with x7,i∈2,4,5,6 for at least one i≥0 and
(102)x7,k=2⟹x7,k≤3u7,k,x7,k=6⟹x7,k>3u7,k.Corollary 50.
The midpoint-free subsetS2≔Z7+0,1,3⊂Z has endpoint sets satisfying A2=A2+∪Z-, B2=B2+, C2=C2+.Proof.
Given-x∈Z-, we seek b,c∈S2 such that x+2b=c. Clearly -x≠c. Base7computation requires
(103)x7,i+2b7,i+δi=c7,i+7δi+1
for i≥0, with appropriate carry-overs δi∈0,1 beginning with δ0=0. All base7digits of b,c are determined by
(104)x7,i+δi∈0,7⟹b7,i=c7,i=0,x7,i+δi=1⟹b7,i,c7,i∈0,1,1,3,3,0,x7,i+δi=2⟹b7,i=3,c7,i=1,x7,i+δi=3⟹b7,i=0,c7,i=3,x7,i+δi=4⟹b7,i=c7,i=3,x7,i+δi=5⟹b7,i=1,c7,i=0,x7,i+δi=6⟹b7,i=c7,i=1.
Letx have leading digit in position h≥0. If b7,h<c7,h then δh+1=0 so b<c, since b,c have leading digits in position h. All requirements are satisfied.
Now suppose thatb7,h≥c7,h. Then in every case δh+1=1, and
(105)b7,h+1,c7,h+1∈0,1,1,3,3,0.
The first two options here ensure that δh+2=0 and b,c have leading digits in position h+1, and b<c. However, b7,h+1>c7,h+1 holds if the third option is chosen, and then δh+2=1, so
(106)b7,h+2,c7,h+2∈0,1,1,3,3,0.
This behaviour can be iterated any finite number of times but must terminate at some stage in order to determine integers b,c. We may choose any integer k>0 and assign
(107)b7,h+k,c7,h+k∈0,1,1,3,
with b7,h+i=3, c7,h+i=0 if 0<i<k. The leading digits of b,c are in position h+k, so b<c. Once again, all requirements are satisfied.Corollary 51.
The setZ7+0,1,3 is a maximal midpoint-free subset of Z.The proof of Corollary50 implies a surprising result.Corollary 52.
Letx7,h∈4,5,6 be the leading base7digit of x∈Z+. Then -x is a lower endpoint for the midpoint-free subset S2≔Z7+0,1,3⊂Z with multiplicity α-x,S2=ℵ0.Proof.
The conditionx7,h∈4,5,6 necessitates b7,h≥c7,h. Let b,c∈S2 be a solution to x+2b=c with leading digits b7,h+1=1, c7,h+1=3. For g>h, replace the leading1of b7 by a block comprising a leading digit1followed by g-h digits all equal to3. This yields a new solution with the leading3of c7 replaced by a block comprising a leading digit3followed by g-h digits all equal to0. Using base7rep-units, this yields
(108)x+2b-7h+1+3u7,g+1-u7,h+1+7g+1=c-3·7h+1+3·7g+1,
since 6u7,g+1-u7,h+1=7g+1-7h+1. Thus,
(109)-x,b+9u7,g+1-u7,h+1,c+18u7,g+1-u7,h+1∈ΛZ
for every g>h, so there are infinitely many triples in ΛZ having -x as lower endpoint, with midpoint and upper endpoint in S2.


## 10. Closing Remarks

Remark 1.
We have seen that the midpoint-free setS5≔Z5+0,1 is not maximal in Z+. The smallest member of E(S5,Z+) is8, confirming Corollary 15 when m=5. If S5⊂T⊂Z+ and T is a maximal midpoint-free subset of Z+ then
(110)T∖S5⊂ES5,Z+.
This raises some intriguing open questions. For which subsets X⊂E(S5,Z+) is S5∪X a maximal midpoint-free subset of Z+? What is the greedy subset S5*⊂E(S5,Z+) which makes S5∪S5* a maximal midpoint-free subset of Z+?Note thatS5∪x is midpoint-free for any x∈E(S5,Z+), but S5∪x,y is not always midpoint-free if x,y∈E(S5,Z+). For instance, the midpoint triple 8,25,42 comprises 25∈S5 and 8,42∈E(S5,Z+).Remark 2.
Consider the balance points ofS5≔Z5+0,1. Using notation defined after the proof of Corollary 49, along with base5rep-units u5,k (see A003463 in OEIS [5]), for x∈Z+∖S5 we have x∈A(S5,Z+)∪C(S5,Z+) precisely when
(111)x5,k=2⟹x5,k≤3u5,k,x5,k=3⟹x5,k>3u5,k.
Hence, in particular, Z5+0,1,4∖S5⊂A(S5,Z+)∪C(S5,Z+). Similarly, if x∈Z+∖S5 then x∈B(S5,Z+) precisely when x5,i≠4 for all i≥0, and
(112)x5,k=1⟹x5,k≤2u5,k,x5,k=2⟹x5,k>2u5,k.
In particular, Z5+0,3∖{0}⊂B(S5,Z+).
Ifx∈Z+∖S5 is eccentric for S5, the digit configuration (x)5 violates each of these conditions. If x<5h then (x+5hy)5 contains the same digit configuration for every y∈Z+, so x+5hy∈E(S5,Z+). Every eccentric point x is the lower endpoint of infinitely many midpoint triples
(113)x,x+5hy,x+2·5hy∈ΛES5,Z+,
so E(S5,Z+) is densely packed with midpoint triples.Remark 3.
The setT5≔Z5+0,1,3 is not midpoint-free, since 0,3,6∈ΛT5. The digit3is responsible for “most” members of T5 being midpoints, since it can be shown that BT5,Z+=Z+∖S5. Moreover, for any x∈Z+∖S5 there is at least one triple a,x,c∈ΛZ+ with a∈S5 and c∈T5.
It can be shown thatA(T5,Z+)∪C(T5,Z+)=Z+ and Z-⊂A(T5,Z), so no integer is eccentric for T5. Hence there are maximal midpoint-free subsets T⊂Z such that S5⊂T⊂T5. What is the greedy subset T5*⊂T5∖S5 which makes S5∪T5* a maximal midpoint-free subset of Z?
Note also that ifx,b,c∈ΛZ and b,c∈T5, then
(114)x,b+3·5h,c+6·5h∈ΛZ
and b+3·5h, c+6·5h∈T5 for every sufficiently large h∈Z+. It follows that every x∈Z has lower endpoint multiplicity αx,T5=ℵ0.Remark 4.
The setT6≔Z6+0,1,3 is not midpoint-free. Once again, the digit3is responsible for “most” members of T6 being midpoints, since it can be shown that T6∖S6⊂BT6,Z+, where S6≔Z6+0,1. However, ET6,Z+≠∅. For example, it is easily verified that {x∈Z+∣x6,0=4,x6,1=2}⊂ET6,Z+. Thus there is no maximal midpoint-free subset T⊂Z that satisfies S6⊂T⊆T6. What is the greedy subset S6*⊂ES6,Z+ which makes S6∪S6* a maximal midpoint-free subset of Z+?Remark 5.
The midpoint-free setS8≔Z8+0,1,3 is not maximal in Z+. Indeed, confirming Corollary 16 when m=8, the smallest member of E(S8,Z+) is20. In fact,
(115)x∈Z+x8,0=4,x8,1=2⊂ES8,Z+.
Forx∈Z+∖S8 it can be shown that x∈A(S8,Z+)∪C(S8,Z+) precisely when
(116)x8,k=3⟹x8,k≤4u8,k,x8,k=4⟹x8,k>4u8,k
so, in particular, Z8+[0,3]∖S8⊂A(S8,Z+)∪C(S8,Z+). (For base8rep-units u8,k, see A023001 in OEIS [5].) Again, if x∈Z+∖S8 then x∈B(S8,Z+) precisely when
(117)x8,k∈2,3,6,7⟹x8,k≤3u8,k
so, in particular, Z8+[0,3]∖S8⊂B(S8,Z+). What is the greedy subset S8*⊂E(S8,Z+) which makes S8∪S8* a maximal midpoint-free subset of Z+?Remark 6.
The setS9≔Z9+0,1,3,4 is a maximal midpoint-free subset of Z+ since Z9+0,1,3,4=Z3+0,1. This follows immediately from the observation that
(118)3r+s9n=r32n+1+s32nwhenr,s∈0,1,n≥0.
More generally, Zm+D=Z3+0,1 when m=3n and D=S0∩[0,u3,n] for any positive integer n.Remark 7.
The midpoint-free setsSm≔Zm+0,1,3,4 are maximal in Z+ for each m∈[9,13], and the endpoint sets satisfy A(Sm,Z+)∪C(Sm,Z+)=Z+∖Sm. Thus Sm(2)≔2Sm∪-2Sm+1 and Sm(3)≔3Sm∪3Sm+1 are maximal midpoint-free subsets of Z when m∈[9,13], by Theorems 4 and 6. However, the midpoint-free set S14 is not maximal in Z+ since, confirming Corollary 17 when m=14, the smallest member of E(S14,Z+) is65. What is the greedy subset S14*⊂E(S14,Z+) which makes S14∪S14* a maximal midpoint-free subset of Z+?



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