International Journal of Combinatorics
Volume 2015 (2015), Article ID 216475, 14 pages
http://dx.doi.org/10.1155/2015/216475

Research Article

Maximal Midpoint-Free Subsets of Integers

Roger B. Eggleton

School of Mathematical and Physical Sciences, University of Newcastle, Callaghan, NSW 2308, Australia

Received 3 September 2014; Accepted 9 December 2014

Academic Editor: Chris A. Rodger

Copyright © 2015 Roger B. Eggleton. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 basetransition 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 is of the form , where runs through all maximal blocks of nonzero digits containingin . Any such uniquely determines a collection of suitable blocks of basedigits for while all other digits are forced by basearithmetic for , so the number of solutions for is precisely .

To illustrate, when the possible ordered pairs critical for are , , , , . There are sets of independent critical pairs, namely, , the five singletons, and . The corresponding triples in withas an endpoint are

Given , let be the leading maximal block of nonzero digits containingin , so when , and in particular . Then any ordered pair critical for is constrained by . In particular, the leading ordered pairs critical for are all those of the form . In this case , and there is at least one such ordered pair since for at least one . Let denote the family of just those sets of independent ordered pairs critical for that include a leading pair.

Corollary 24. If, then has lower endpoint multiplicity

Proof. Continuing with the definitions and notation in the proof of Theorem 23, if contains the leading ordered pair critical for then the corresponding for satisfy , , with when ; moreover, , , and when . Thus and .
On the other hand, if then the corresponding for satisfy , , for some , with when . Thus .

The multiplicities in Theorem 23 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 be the integer with baserepresentation comprising a block of digits all equal to, so is a baserep-unit, following terminology of Yates [6]. (See A003462 in OEIS [5].)

Corollary 25. For any integers and , , let

where and . Then

Proof. The only maximal block of nonzero digits containingin is , with when , and when , with . Therefore, the only ordered pairs critical for have . No two of these critical pairs are disjoint, so the only sets in are singletons and the empty set. The sets comprising are the singletons .

For instance, arises by taking , , , , and in Corollary 25. The leading critical pairs for are with , yielding two midpoint triples with :

for blocks , so with . Thus . The non-leading critical pairs for are with , , yielding five midpoint triples with :

for , , , , , so for . Thus .

6. Doubling the Greedy Midpoint-Free Subset of

Doubling as in Theorem 4 shows that is a maximal midpoint-free subset of . (This is an alternative demonstration to the proof given in [3].) Here and , where comprises those positive integers with baserepresentation in which the trailing digit (the last nonzero digit) isand every other digit is in . The relevant balance point sets and multiplicities will now be examined briefly.

For brevity, let us write

Note that , . Similar identities hold for other balance point sets of .

It can be shown that every odd integer is the sum of two distinct members of . Indeed, there are precisely such sums, where is the number of digits equal toin for some integer . The balance point sets and multiplicities for , , and can now be specified.

Theorem 26. The midpoint-free subset has balance point sets satisfying and .

Corollary 27. If, the endpoint multiplicities satisfy

Corollary 28. If, the midpoint multiplicities satisfy

where is the number of digits equal toin .

Corollary 29. The midpoint-free subset has balance point sets satisfying , , .

Corollary 30. The midpoint-free subset has balance point sets satisfying

Corollary 31. The set is a maximal midpoint-free subset of , with balance point sets satisfying .

Note that trebling as in Theorem 6 yields , and earlier we saw that this is “trivial” in this case, because . In the notation of Corollary 7, for any integer this implies

If is any set satisfying the hypotheses of Theorem 6, and , then the normalized set is an affine transform which satisfies the hypotheses of Theorem 6. Without loss of generality, assume is any normalized compliant set; then Corollary 7 shows that iterated trebling yields a sequence of maximal midpoint-free subsets of which asymptotically approach the greedy subset , because

In this sense we may write the asymptotic equivalences

7. The Midpoint-Free Set

Let us now study the set

corresponding to sequence A000695 in OEIS [5]. Corollaries 9 and 15 leave open the possibility that is a maximal midpoint-free subset of . We now settle that matter. Let , , .

Theorem 32. The midpoint-free subset has endpoint sets satisfying .

Proof. Let . We seek such that . Clearly any such solution has . Basecomputation requires

for every , with carry-overs satisfying and

Specifying by their basedigits, the equation requires

These digit specifications are easily seen to satisfy the requirements of basearithmetic for . Therefore, every is in . Since is midpoint-free, it follows that .

Corollary 33. The set is a maximal midpoint-free subset of .

Corollary 34. For any the total endpoint multiplicity is

Any has at least one basedigit in . A high block in is a maximal block of basedigits all in . If is a high block in , it is clear if and leading if when .

Corollary 35. The lower endpoint set for in comprises all positive integers in which the leading basehigh block is clear and contains. The upper endpoint set is the complement of in .

Proof. With the definitions and notation in the proof of Theorem 32, let for some and when . Let be the leading basehigh block for , so and when .

(1)If then , so when . Also , while when . If then . Therefore, and when . This case arises just when for some such that , and if . Also when , so if and if , the latter condition holding precisely when is clear. On the other hand, if then , , while and when , so .(2)If and then , so and when . Then but the high block does not contain. In this case .(3)If then , so , , , and when . Once again , but now the high block must contain. In this case .(4)If , then , , and the same conclusions as in (3) apply.

Corollary 36. The midpoint set for in comprises those positive integers with . Any such has midpoint multiplicity , where is the number of digits in equal to.

Proof. Let . We seek such that . Clearly any such solution has , so holds. Basearithmetic requires

for every . This forces . Since it follows that for some , with for all . Specifying by their basedigits, the equation requires

With , , clearly all solutions meet basearithmetic requirements for and .

An alternative characterization of the midpoint set is that it comprises every with for at least one , and in each such instance .

Now consider the midpoint-free set as a subset of . For brevity, let , , .

Corollary 37. The midpoint-free subset has balance point sets satisfying , , .

Proof. Let . We seek such that . Clearly any such solution has . Basecomputation requires

for every , with carry-overs satisfying and

Specifying by their basedigits, we require

It is straightforward to verify that are uniquely determined by these specifications, and , so .

Corollary 38. The set is a maximal midpoint-free subset of .

Corollary 39. Any has lower endpoint multiplicity .

8. Doubling and Trebling the Set

By Theorem 32 and Corollary 38, the set is compliant with the requirements for the doubling and trebling constructions of Theorems 4 and 6, so both sets

are maximal midpoint-free subsets of .

Let us now briefly examine the balance point sets associated with :

Easy digit and parity considerations in combination with Theorem 32 yield the following corollaries.

Theorem 40. The midpoint-free subset has balance point sets satisfying

Corollary 41. The midpoint-free subset has balance point sets satisfying

Corollary 42. The subset is midpoint-free and has balance point sets satisfying

Corollary 43. The sets and are midpoint-free subsets of with endpoint sets satisfying and .

The balance point sets of can now be considered. Let

If and , the conditions

and are satisfied by the triples . Hence, if then . Such observations yield the following results.

Theorem 44. The subset is midpoint-free and has balance point sets satisfying

where .

Corollary 45. The subset is midpoint-free and has balance point sets satisfying

9. The Midpoint-Free Set

Next consider the set

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 is in fact a maximal midpoint-free subset of . It will follow from Theorems 4 and 6 that

are maximal midpoint-free subsets of .

The methods used in earlier sections are again applicable, so fewer details are now required. Following earlier practice, let

Theorem 46. The midpoint-free subset has endpoint sets satisfying .

Proof. Given , we seek such that . Any solution has . Basecomputation requires

for , with appropriate carry-overs beginning with . All basedigits of are determined by

It is straightforward to verify that all requirements are satisfied.

Corollary 47. The set is a maximal midpoint-free subset of .

Corollary 48. For , any has endpoint multiplicity

Corollary 49. The midpoint set for in comprises those integers with . Any such has midpoint multiplicity , where is the number of digits in belonging to .

Proof. Given , we seek such that . Necessarily, any solution has , so holds. Basearithmetic requires

for every . This forces . Since there is an integer such that but for all . Specify by

With , evidently all solutions satisfy and .

The baserepresentation of the integer is a block of digits, all equal to. Thus is a baserep-unit [6]. (See A023000 in OEIS [5].) For any let denote the integer resulting from by deleting all but the last basedigits:

An alternative characterization of the midpoint set is that it comprises every with for at least one and

Corollary 50. The midpoint-free subset has endpoint sets satisfying , , .

Proof. Given , we seek such that . Clearly . Basecomputation requires

for , with appropriate carry-overs beginning with . All basedigits of are determined by

Let have leading digit in position . If then so , since have leading digits in position . All requirements are satisfied.
Now suppose that . Then in every case , and

The first two options here ensure that and have leading digits in position , and . However, holds if the third option is chosen, and then , so

This behaviour can be iterated any finite number of times but must terminate at some stage in order to determine integers . We may choose any integer and assign

with , if . The leading digits of are in position , so . Once again, all requirements are satisfied.

Corollary 51. The set is a maximal midpoint-free subset of .

The proof of Corollary 50 implies a surprising result.

Corollary 52. Let be the leading basedigit of . Then is a lower endpoint for the midpoint-free subset with multiplicity .

Proof. The condition necessitates . Let be a solution to with leading digits , . For , replace the leadingof by a block comprising a leading digitfollowed by digits all equal to. This yields a new solution with the leadingof replaced by a block comprising a leading digitfollowed by digits all equal to. Using baserep-units, this yields

since . Thus,

for every , so there are infinitely many triples in having as lower endpoint, with midpoint and upper endpoint in .

10. Closing Remarks

Remark 1. We have seen that the midpoint-free set is not maximal in . The smallest member of is, confirming Corollary 15 when . If and is a maximal midpoint-free subset of then

This raises some intriguing open questions. For which subsets is a maximal midpoint-free subset of ? What is the greedy subset which makes a maximal midpoint-free subset of ?

Note that is midpoint-free for any , but is not always midpoint-free if . For instance, the midpoint triple comprises and .

Remark 2. Consider the balance points of . Using notation defined after the proof of Corollary 49, along with baserep-units (see A003463 in OEIS [5]), for we have precisely when

Hence, in particular, . Similarly, if then precisely when for all , and

In particular, .
If is eccentric for , the digit configuration violates each of these conditions. If then contains the same digit configuration for every , so . Every eccentric point is the lower endpoint of infinitely many midpoint triples

so is densely packed with midpoint triples.

Remark 3. The set is not midpoint-free, since . The digitis responsible for “most” members of being midpoints, since it can be shown that . Moreover, for any there is at least one triple with and .
It can be shown that and , so no integer is eccentric for . Hence there are maximal midpoint-free subsets such that . What is the greedy subset which makes a maximal midpoint-free subset of ?
Note also that if and , then

and , for every sufficiently large . It follows that every has lower endpoint multiplicity .

Remark 4. The set is not midpoint-free. Once again, the digitis responsible for “most” members of being midpoints, since it can be shown that , where . However, . For example, it is easily verified that . Thus there is no maximal midpoint-free subset that satisfies . What is the greedy subset which makes a maximal midpoint-free subset of ?

Remark 5. The midpoint-free set is not maximal in . Indeed, confirming Corollary 16 when , the smallest member of is. In fact,

For it can be shown that precisely when

so, in particular, . (For baserep-units , see A023001 in OEIS [5].) Again, if then precisely when

so, in particular, . What is the greedy subset which makes a maximal midpoint-free subset of ?

Remark 6. The set is a maximal midpoint-free subset of since . This follows immediately from the observation that

More generally, when and for any positive integer .

Remark 7. The midpoint-free sets are maximal in for each , and the endpoint sets satisfy . Thus and are maximal midpoint-free subsets of when , by Theorems 4 and 6. However, the midpoint-free set is not maximal in since, confirming Corollary 17 when , the smallest member of is. What is the greedy subset which makes a maximal midpoint-free subset of ?

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

Thanks are due to James MacDougall for suggesting the problem of determining multiplicity of solutions for the greedy midpoint-free sequence .

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