#### Abstract

A set is midpoint-free if no ordered triple satisfies and . Midpoint-free subsets of and are studied, with emphasis on those sets characterized by restrictions on the base digits of their elements when , and with particular attention to maximal midpoint-free subsets with .

#### 1. Introduction

An ordered triple of integers is a* midpoint triple* of if and . The* midpoint* of this triple is , its* lower endpoint* is and its* upper endpoint* is . For any subset , let denote the set of all midpoint triples . When , let
Generically, the members of are the* balance points* for in . The balance points comprise the* lower endpoint set *, the* midpoint set *, and the* upper endpoint set *, for in . The members of are the* eccentric points* for in . Attention to these sets appears to be a new focus, suggested by the underlying geometrical viewpoint. There is an extensive literature associated with treating as specifying an arithmetic progression of length. A compact discussion and rich bibliography are given in Guy’s survey work [1, Section ]. For an example of recent work in this field, see Dybizbański [2].

If it will be convenient to say that is* midpoint-free*; moreover, is a* maximal* midpoint-free subset of if and , whenever . Hence we have for any . This implies the following characterization.

Theorem 1. *If , then is a maximal midpoint-free subset of if and only if , or equivalently, if and only if and .*

Note that if is infinite, any maximal midpoint-free subset must also be infinite: any pair of elements of has one midpoint and two endpoints, so precludes at most three elements of from membership of ; thus would be infinite if were finite, but then would be midpoint-free for any , contradicting maximality of .

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 to and subsets, especially and . Here, the main focus will be on , , and when is a maximal midpoint-free subset of , or .

Initially has been defined to be midpoint-free if contains no midpoint triple. There are several semantic equivalents for this condition.

Theorem 2. *If , then any one of the sets
**
is empty if and only if all three are empty.*

*Proof. *Consider the contrary. If there is a triple ; this triple ensures that and . The other two cases follow in the same way.

Take in Theorem 2. Then recovers the “natural” terminology that is midpoint-free if and only if* it does not contain the midpoint of any two of its members*. Equally, 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 subset is midpoint-free if and only if
*

This yields another semantic equivalent: is midpoint-free exactly when it is balance point-free.

For any define the* multiplicity* of as a lower endpoint, midpoint, or upper endpoint for , respectively, as
The multiplicities for in particular cases of will be of interest.

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

Let us begin with an explicit example to illustrate these notions and sample some of the typical features encountered.

For any integer , let be the set of integers with regular base representation in which all digits are restricted to the set . When are distinct members of , all digits of the base representation of lie in the set , but base computation of involves no carry over, so contains the digitin each place where the base digits of and differ. Hence is impossible. It follows that is midpoint-free.

Let denote the regular base representation of . An easy base computation shows that This corresponds to the identities Note that is an even positive integer, so is a midpoint for when this set is treated as a subset of . Digit considerations make it clear that has no other solutions with , so the midpoint multiplicity of is. In summary, Again, the general base computation corresponds to the identity which implies when . Additionally, in the special case we also have Hence . There are no other solutions to with , so the multiplicities of as a lower and upper endpoint are , .

When , it can be seen that has no solutions with , so is not an endpoint for , and Now seek such that and . As , it follows that . Then it is routine to verify that when , but note that digit arguments must be sensitive to the magnitude of ; for instance, when , while for smaller we have It follows that and is not a midpoint for when . This completes the demonstration that is an eccentric point, so the midpoint-free subset is not maximal if .

We will later return to a more systematic study of and .

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

If and with , then is an* affine transform* of . Clearly if and only if when or when , so any affine transform is midpoint-free if and only if is midpoint-free.

For example, let be the set of all integers with regular base representation in which all digits are restricted to the set . Then is midpoint-free because is midpoint-free when , and is an affine transform of . The last identity shows that . Similarly, is an affine transform of so is a midpoint-free subset of when . Then is a midpoint-free subset of , since the positive and negative components of this set are midpoint-free, and has no solution with , , and , because these conditions require to be an odd integer and to be an even integer.

Suppose is known to be a maximal midpoint-free subset of . It turns out that the principle illustrated by the example in the previous paragraph holds strongly for .

Theorem 4. *If is a maximal midpoint-free subset of , then the set is a maximal midpoint-free subset of .*

*Proof. *As is a maximal midpoint-free subset of , Theorem 1 implies
But , so and ; hence and . Therefore, .

Given , there exist such that and . Let . Then , so
Therefore,
Conversely, suppose , so there exist such that and . But , where , so implies , whence with . But is midpoint-free, so is midpoint-free, and therefore, . Thus . Now and imply that and , so . Therefore, the reverse containments also hold:
Hence

For , similar reasoning shows that , so
However, this containment can in fact be proper. For instance, if , then only contains even integers, whereas .

Since is a maximal midpoint-free subset of , no integers are eccentric for , so no members of are eccentric for . Then the positive integers eccentric for satisfy
where . Since it follows that
Specifically, all integers eccentric for are odd, and all integers eccentric for are even. Therefore, no integer is eccentric for both and , so . Thus is a maximal midpoint-free subset of .

Corollary 5. *If is a maximal midpoint-free subset of , the balance point sets for and with satisfy
*

It is convenient to refer to the construction in Theorem 4 as “doubling” the given set . Other constructions involving affine transforms of a set are also of interest. For example, since is midpoint-free when , it follows that the disjoint sets and are midpoint-free when . In fact, their union is midpoint-free. This turns out to be “trivial.” Multiplying a member of by simply shifts its base digits one place, and a terminalemerges to occupy the zeroth place; then addingreplaces the terminalby. Hence .

When , the disjoint midpoint-free sets and are more interesting. Let us verify that is also midpoint-free. The two component sets are midpoint-free, so any midpoint triple must have at least one member in each set. Thus in each case with . If , then for . Then and . But , so . If then , while if then . In each case and , so because . Thus , as claimed.

Generalising the latter example, a “trebling” construction which produces new maximal midpoint-free subsets of will now be studied.

Theorem 6. *If is a maximal midpoint-free subset of , and all members of are endpoints for , then is a maximal midpoint-free subset of , and all members of are endpoints of .*

*Proof. *Because is midpoint-free, each of the affine transforms and is midpoint-free. Assume . Since , there is a such that . Then , so . Thus , so no such triple exists. Hence is midpoint-free.

By hypothesis, . Also by Corollary 5, so every is an endpoint for . Suppose and . Two integers from complementary sets cannot be equal, so must be different. Also
holds when . It follows that
If are consecutive members of , all members of the interval
are endpoints for . Then are consecutive members of and all members of are endpoints for . Also each is an endpoint for , so all members of are endpoints for . Thus , and .

Let for each , and let . Iterating the construction in Theorem 6 and combining with Theorem 4 yields the following result.

Corollary 7. *If is a maximal midpoint-free subset of , and all members of are endpoints for , then the set
**
is a maximal midpoint-free subset of , for any integer , and all members of are endpoints of . Moreover, the set
**
is a maximal midpoint-free subset of .*

#### 4. Subsets of with Base Digit Restrictions

Fix an integer . Let . Then is a* digit subset* for base representations of the integers or, briefly, a* base ** digit subset*. Let be the set of nonnegative integers with base representation using only digits in , and let denote the digit in position of the regular base representation of , so

Let us say that is midpoint-free as a base digit subset if and there is no ordered triple such that and .

Theorem 8. *If is a midpoint-free base digit subset with and , then the set is midpoint-free.*

*Proof. *Suppose . Then . There is no carry-over in computing this sum in base arithmetic since all its digits are less than , so
for every . Since is midpoint-free, it follows that
for every , so , contradicting the initial choice of . Thus , so is midpoint-free.

Three early instances of Theorem 8, the first of which was independently demonstrated earlier, are of considerable interest.

Corollary 9. *Each is midpoint-free when .*

Corollary 10. *Each is midpoint-free when .*

Corollary 11. *Each is midpoint-free when .*

If is a midpoint-free base digit subset, it is of interest to decide whether has any members that are eccentric for , since this is equivalent to deciding whether is a maximal midpoint-free subset of . For , let the* support* for , as a lower endpoint, midpoint, or upper endpoint for , be defined by
The following result is useful for settling whether the eccentric set is empty in specific cases.

Theorem 12. *Suppose and is a midpoint-free base digit subset. If with , then
*

*Proof. *Fix the nonnegative integer with . (For simplicity we do not explicitly require , although Theorem 8 does imply that members of have empty support sets.)

Suppose there exist such that , so . Base arithmetic for this sum takes the form
for every , with carry-overs satisfying and
for . But so for . In particular,
so . For all , it follows that and
But is midpoint-free, so for all , since otherwise yields the contradiction . Hence .

The other two cases follow simply by noting that if then , and if then .

Corollary 13. *If is a midpoint-free base digit subset, then relative to the three multiplicities of any are finite.*

Corollary 14. *If is a midpoint-free base digit subset and with , then contains all three support sets for relative to . Moreover , where .*

Corollary 15. *If then is eccentric for .*

Corollary 16. *If then is eccentric for .*

Corollary 17. *If then is eccentric for .*

The last three corollaries leave open the possibility that their subject sets might be maximal midpoint-free subsets of when is small enough. In the next section we shall consider and pursue other cases later.

#### 5. Greedy Midpoint-Free Subset of

The* greedy midpoint-free subset* of is the set
in which and each with is the smallest integer satisfying such that is midpoint-free. It has long been known [4] that
corresponding to sequence A00536 of OEIS [5]. For brevity, let
We shall attach the adjective* greedy* when referring to these sets. Before we prove that each of these greedy balance point sets contains all positive integers not in , let us check the example . The following basecomputations are transparent:
Hence we have the midpoint triples , , , showing that . Now consider the general case.

Theorem 18. *The greedy midpoint-free subset has greedy balance point sets satisfying
*

*Proof. *Let . First we show that is a midpoint for . Note that there is at least one such that . Specify by their basedigits; thus
We have for all . Also , because . Hence . It is easily checked that
for all , so . Hence .

Next we show that is an upper endpoint for . Let be specified by their basedigits; thus
There is at least one such that , so . Also
for all , so . Hence .

Finally we show that is a lower endpoint for . We need to find such that and . Basearithmetic for this sum takes the form
for every , with carry-overs satisfying and
Specify by their basedigits; thus

If and for then for . It follows that . Since , for every , we have .

There are integers such that and for and and for . If then , ; if then . In either case it follows that . If then so since for all . If then there is an integer such that and for while . If and then so . Hence for all . Also , and , so . Then and for all , so .

Finally, in all three cases for each , so . Hence .

Corollary 19. *The greedy midpoint-free subset is maximal in .*

In [3] it was shown that is actually a maximal midpoint-free subset of . The proof is not repeated here, but let us note that the single example follows from computing with in the form . Baseconsiderations yield corresponding to , so as claimed.

For brevity, let , , . With Theorem 18, this yields the following result.

Corollary 20. *The greedy midpoint-free subset has greedy balance point sets satisfying , .*

Reversing implications, this yields the following result.

Corollary 21. *The greedy midpoint-free subset is maximal in .*

Now the multiplicities for can be examined. First note that the sum of basedigits of any even integer is , so even integers have an even number of basedigits equal to. If , at least one digit in must be, so the total number of such digits is , and .

Theorem 22. *If , then the midpoint multiplicity of is
**
where is the number of digits equal toin and .*

*Proof. *Suppose satisfy and . There is no carry-over in the basearithmetic for the sum, so , for all . If then we must have
There is an integer such that and for all , so forces . However, when and , the requirement is satisfied if . Both possibilities are consistent with , so there are solutions in total.

To illustrate, when then so has just two triples withas midpoint: , . In contrast, when then so has eight triples withas midpoint, ranging from to , namely, for all .

Fix . The digits of the baserepresentation include at least one. An ordered pair of integers with is* critical* for if and , with when . Any two ordered pairs and critical for are* independent* if . Any set of ordered pairs critical for is independent if every two members are independent. Let denote the family of all sets of independent ordered pairs critical for , including the empty set.

Theorem 23. *If , then has total endpoint multiplicity
*

*Proof. *Given , we seek such that . Any solution necessarily satisfies . Basearithmetic for this sum takes the form
for every , with carry-overs satisfying and
Specifying by their basedigits, the equation requires
As is easily verified, these digit specifications satisfy the requirements of basearithmetic for .

Any integer is a base*transition point* for the sum if . Note that the digits and are forced unless . In the latter case there are two options:(1), , ,(2), , .If , then choosing option (1) preserves the carry-over state (“off”) with whereas choosing option (2) switches the carry-over state (to “on”) with , so option (2) causes to be a transition point. If , then choosing option (1) switches the carry-over state (to “off”) with and causes to be a transition point, whereas choosing option (2) preserves the carry-over state (“on”) with .

Let be a maximal block of nonzero digits in the baserepresentation , say , and assume that contains the digitat least once. There must be at least one such maximal block in . Then , and the “off” carry-over state can only switch to for some such that when and , corresponding to an option (2) choice in the construction of . The “on” carry-over state can only switch back to for some such that if and , corresponding to an option (1) choice in constructing . If no such option is exercised, then maximality of ensures the carry-over state inevitably switches back at , for in this case , , and an option (2) choice for results in , , , . Note that the ordered pair is critical for , as is the ordered pair .

Now consider any set of independent ordered pairs critical for , with . The set determines a unique sequence of carry-over digits , with if and , and for every other in the interval . Then and determine blocks
such that . Each member of