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arXiv:1709.06046v1 [math.NT] 18 Sep 2017

BY THE MULTISET OF ITS 𝑠-SUMS?

DMITRI V. FOMIN

This paper is dedicated to the memory of Oleg Izhboldin (1963-2000)

ABSTRACT. This is a survey of all available information on a remarkable problem in number theory proposed by Leo Moser in 1957. In general form the question is: can a collection of𝑛numbers be uniquely restored given the collection of its𝑠-sums? We describe results and techniques from sixty years of research in this area. Some new finds and open questions are presented.

1. INTRODUCTION

Sixty years ago, in 1957, American Mathematical Monthly has published the following relatively simple problem in number theory proposed by Leo Moser (see [2]):

Problem 1.1. (a) The ten numbers 𝑠₁ ⩽ 𝑠₂ ⩽ ... ⩽ 𝑠₁₀ are the sums of the five unknown numbers

𝑥₁ ⩽𝑥₂⩽ ...⩽𝑥₅taken two at a time. Determine the𝑥’s in terms of the𝑠’s.

(b) Show that if 𝑠₁ < 𝑠₂ < ... < 𝑠₆ are six distinct numbers formed by taking the sums of four numbers two at a time, then there exist four other numbers which give the same sums when added in pairs.

Naturally – after it was quickly solved – the problem was immediately generalized and reformu-lated. And then it turned out to be quite an interesting little question...

Problem 1.2. Let 𝐴be a collection (multiset) of 𝑛 numbers 𝑎₁ ⩽ 𝑎₂ ⩽ ... ⩽ 𝑎_𝑛. Consider the multiset𝐴(2) of (𝑛

2

)

2-sums of multiset𝐴, i.e., collection of all sums of the kind 𝑎_𝑖

1 +𝑎𝑖2, where

1⩽𝑖₁< 𝑖₂ ⩽𝑛. Is it possible to restore𝐴given𝐴(2)?

The numbers here could be complex or even belong to an arbitrary field of characteristic zero. That really doesn’t matter as we will learn shortly.

In this generalization the problem asks whether a multiset is uniquely determined by (or can be recovered from) the multiset of its2-sums.

It was later presented in the literature (e.g., see [8]) using somewhat different terminology.

Problem 1.3. A malicious farmer’s apprentice was asked to provide the list of weights of𝑛bags of grain. Instead he weighed them two at a time and recorded all𝑛(𝑛− 1)∕2combined weights written down in some random order. Is it possible to find the weights of bags (up to permutation of bags)? By the way, it seems that the apprentice was not only malicious but also somewhat dense – instead of performing only𝑛weighings he did a whole lot more of them.

Again the problem is posed as a “recovery” question – whether an unknown multiset can be uniquely restored from the multiset of its pairwise sums.

Date: September 19, 2017.

2010Mathematics Subject Classification. Primary: 11B75, 11P70; Secondary: 05A15.

Key words and phrases. integer multisets, multiset recovery, sumsets, symmetric polynomials. 1

Now, each interesting question, theorem or conjecture deserves a nice name that easily rolls of the tongue in lectures and discussions. “Fermat’s Last Theorem”, “Riemann Hypothesis”, “P=NP”, “Collatz 3k+1 Conjecture” – all these names are short and to the point. I submit that “Multiset Recovery Problem” sounds just as neat while describing the issue with decent precision.

The original problem1.1 was indeed quite easy. However, its generalization1.2 was not. Still, it did not present a serious obstacle; the answer was quickly discovered and so the problem was generalized even further.

Notation. For any pair of positive integers 𝑛 and 𝑠 such that 𝑛 ⩾ 𝑠 we will denote by 𝐴(𝑠) _the multiset of𝑠-sums of𝐴, i.e. collection of all sums of the kind

𝑎_𝑖

1 +𝑎𝑖2 + … +𝑎𝑖𝑠,

where1⩽𝑖₁< 𝑖₂ <…< 𝑖_𝑠 ⩽𝑛.

Problem 1.4. Consider positive integers𝑛and𝑠with𝑛 > 𝑠. Do there exist two distinct𝑛-multisets

𝐴and𝐵such that𝐴(𝑠)_{= 𝐵}(𝑠)_?

Such a pair of multisets would represent a "recovery failure". Indeed, in this case, given multiset

𝑀 =𝐴(𝑠) =𝐵(𝑠) it is impossible to determine the original multiset.

Definition. If two multisets𝐴and 𝐵 have the same collections of𝑠-sums – that is, 𝐴(𝑠) _{= 𝐵}(𝑠) _– then we will call these multisets 𝑠-equivalent (or when this will not cause any confusion, simply equivalent) and we will denote that relation as𝐴∼𝑠 𝐵 (or simply𝐴∼ 𝐵).

Definition. We will call a pair of natural numbers (𝑛, 𝑠) singular if it represents a non-trivial “multiset recovery failure” – i.e.,𝑛 > 𝑠and there exist two different𝑠-equivalent𝑛-multisets𝐴and

𝐵 (𝐴≠𝐵 & 𝐴∼𝑠 𝐵).

So all the above problems can be reworded as questions about singular pairs. The ultimate goal is to describe those pairs in some easily “computable” way.

Notation. For any natural number 𝑠by_𝑠 we will denote the set of all natural numbers𝑛 > 𝑠

such that pair(𝑛, 𝑠)is singular.

For instance,₁is obviously empty. AlsoQuestion 1.1could be reformulated as follows: does ₂contain numbers 10 and 4?

This article will present all currently known results on the Multiset Recovery Problem and the methods involved. We will also discuss some new facts and conjectures.

2. HISTORICAL TIMELINE

(1957). Just a few words about Moser’s possible motivation for this problem. A few years before that, in 1954, Leo Moser and Jim Lambek published article [1] about pairs of complementary subsets ofℕwhere they have provedLambek-Moser Theoremabout partitions ofℕand how they are related to sequences of numbers of the form{𝑓(𝑛) +𝑛}where𝑓 ∶ℕ_→ℤ_⩾0is some arbitrary non-decreasing unbounded function.

This investigation seems to be quite close to questions about how finite or infinite sets of natural numbers overlap or complement each other when being translated.

I suspect that this was how Leo Moser stumbled upon questions about multiset recovery for the case of 𝑠 = 2 – but of course, this is pure speculation. However, in their later short article [4] Lambek and Moser mention both Multiset Recovery Problem and complementary sequences of integers literally on the same page.

(1958). In 1958, almost immediately after Moser has posed his original question, Selfridge and Straus in their article [3] provided solution for the “real” problem1.2as well as some other ques-tions. First, they have proved the following

Theorem 2.1. Multiset𝐴of𝑛numbers is uniquely determined by multiset of its2-sums𝐴(2)_{if and} only if𝑛is not a power of 2.

In other words, pair(𝑛,2)is singular if and only if𝑛is a power of 2. Or, using our notation, ₂ = {4,8,16,32, ...} = {2𝑘 _{∶ 𝑘 >1}.}

Second, they have explored case of𝑠 = 3of the more general problem 1.4. They have shown that for𝑛 = 6 there are easily constructed examples of different multisets of𝑛 numbers𝐴and𝐵

such that𝐴∼3 𝐵. For instance:

𝐴= {15,−5}, 𝐵 = {(−1)5,5}⟹ 𝐴≠𝐵 𝐴(3) = {310,(−3)10}, 𝐵(3) = {(−3)10,310}⟹ 𝐴(3)=𝐵(3),

where15_{is not1as you might have thought. In standard multiset notation𝑎}𝑏_{means element𝑎with}

multiplicity𝑏(i.e.,𝑏-multiple entry of number𝑎), so we simply mean that

𝐴= {1,1,1,1,1,−5}, 𝐵 = {−1,−1,−1,−1,−1,5}.

As for other values of𝑛, the authors of [3] have proved that such example for case𝑠= 3can exist only if𝑛2_{− (2}𝑘_{+ 1)𝑛+ 2⋅3}𝑘−1_{vanishes for some natural𝑘 < 𝑛. It is not very hard to prove that the} only other non-trivial values of𝑛for which that is possible are𝑛= 27(with𝑘= 5,9) and𝑛= 486

(𝑘= 9).

Thus, they have showed that

{6}⊂₃ ⊂ {6,27,486}.

Third, using the same technique for the case of𝑠= 4it was shown that the only non-trivial values of𝑛when recovery might not be always possible are𝑛= 8,12.

Presenting an example for𝑛 = 8is quite easy. Generally, one can always construct an example of “recovery failure” inProblem 1.4if𝑛= 2𝑠(we will do that later inSection 3). Again, this can be written as

{8}⊂ ₄⊂ {8,12}.

Naturally, that suggested a few additional questions.

Question 2.2. Do pairs(𝑛, 𝑠) = (27,3)and(486,3)represent actual “recovery failures”? In other words, do there exist for𝑛 = 27and𝑛= 486examples of different𝑛-multisets𝐴and𝐵 such that

𝐴(3) =𝐵(3)?

Question 2.3. Same question about pair(𝑛, 𝑠) = (12,4). That is, do there exist two different12 -multisets𝐴and𝐵such that𝐴(4) =𝐵(4)?

At that time both questions were left unsolved. Yet another important question from the same article:

Question 2.4. In cases when recovery is impossible, could there exist more than two 𝑛-multisets that generate identical multisets of𝑠-sums?

(1959). Soon after the paper by Selfridge and Straus, Leo Moser and his coauthor Joachim (Jim) Lambek wrote a small article [4]. It started by acknowledging results of their colleagues from UCLA, and then they proceeded to develop the problem in a slightly different direction.

Namely, they asked a question whether the set of non-negative integers can be split in two subsets

𝐴= {𝑎₁, 𝑎₂, ...}and𝐵 = {𝑏₁, 𝑏₂, ...}such that𝐴(2)and𝐵(2)coincide as multisets. They proved that the answer was positive and that there exists only one such decomposition ofℤ_⩾0.

They did that by using multiset generating functions.

Definition. For any finite multiset𝐴of positive integers of the form{𝑎𝑘1

1 , 𝑎

𝑘2

2 , ..., 𝑎

𝑘_𝑚

𝑚 }we define its

generating function (polynomial)𝑓_𝐴(𝑥)by formula

𝑓_𝐴(𝑥) = 𝑚 ∑

𝑖=1

𝑘_𝑖𝑥𝑎𝑖_.

Similarly this generating function can be defined for an infinite multiset𝐴 = {𝑎𝑘1

1 , 𝑎

𝑘₂

2 , ...}as long as sequence{𝑘1∕𝑎𝑖

𝑖 }is bounded.

Authors proved that in their particular case generating functions satisfied system of equations:

{

𝑓_𝐴(𝑥) +𝑓_𝐵(𝑥) = 1∕(1 −𝑥) 𝑓2

𝐴(𝑥) −𝑓𝐴(𝑥

2_{) =𝑓}2

𝐵(𝑥) −𝑓𝐵(𝑥

2_).

It was also proved that similar split of𝑍_𝑛 = {0,1,2, ..., 𝑛−1}is possible if and only if𝑛is a power of 2. That split is unique and is determined by the so-calledThue-Morse sequence{𝛼𝑛}defined as 𝛼_𝑛 =𝑠₂(𝑛) (mod 2)where𝑠₂(𝑛)is thebinaryweight of𝑛, i.e., sum of digits (or simply, the number of ones) in the binary representation of𝑛. So if 𝑛 = 2𝑝 _{and we define sets𝐴 = {𝑎}

1, ..., 𝑎𝑚}and 𝐵 = {𝑏₁, ..., 𝑏_𝑚}as follows

𝐴= {𝑘∈𝑍_𝑛∶ 𝛼_𝑘 = 0} 𝐵 = {𝑘∈𝑍_𝑛∶ 𝛼_𝑘 = 1}

then𝐴(2) =𝐵(2). Indeed, if you set𝑓_𝐴(𝑥) = ∑𝑚_𝑖=1𝑥𝑎𝑖 and_𝑓

𝐵(𝑥) = ∑𝑚

𝑖=1𝑥

𝑏_{𝑖then we have}

𝑓_𝐴(𝑥) +𝑓_𝐵(𝑥) = 𝑢(𝑥) = 1 −𝑥

2𝑝

1 −𝑥 , 𝑓𝐴(𝑥) −𝑓𝐵(𝑥) =𝑣(𝑥) = 𝑝−1

∏ 𝑖=0

(1 −𝑥2𝑖).

Thus𝑓_𝐴= (𝑢+𝑣)∕2and𝑓_𝐵 = (𝑢−𝑣)∕2and from that it follows quite easily that𝑓2

𝐴(𝑥) −𝑓𝐴(𝑥2) = 𝑓_𝐵2(𝑥) −𝑓_𝐵(𝑥2_{). It is left to notice that the sides in the the last equality are generating functions for} multisets𝐴(2)and𝐵(2)respectively.

Somewhat similar “generating functions” approach was used later in [5], [8] and [9] in conjunc-tion with some other ideas.

(1962). Next paper on the subject appeared in 1962, when Fraenkel, Gordon and Straus published [5] proving that answer toQuestion 2.4was negative. This was not the last time when Multiset Recovery Problem defied the expectations.

Authors have found numerous multi-singularity examples for the simplest case of𝑠 = 2. More precisely, they have showed how to construct examples of three different 8-multisets𝐴, 𝐵 and𝐶

such that𝐴(2) =𝐵(2)=𝐶(2). Here is one of these examples:

After this, naturally, the original question was adjusted into asking how many different𝑛-multisets could generate the same multiset of𝑠-sums. The maximum possible number of such multisets was denoted by _𝑠(𝑛). Of course, pair (𝑛, 𝑠)must be singular to begin with – which is equivalent to inequality_𝑠(𝑛)>1.

Then more inequalities for_𝑠(𝑛)were proved. For instance,

₂(16)⩽3, 2⩽ ₃(6)⩽6, ₄(12)⩽2.

However, no other examples of triple (or greater) “multiplicity” were found, which led to another open question:

Question 2.5. a) For𝑠 = 2does there exist𝑛 = 2𝑝 _{> 8such that some three distinct𝑛-multisets}

generate the same multisets of𝑠-sums (i.e.,₂(𝑛)> 2)?

b) Generally, does there exist any singular pair (𝑛, 𝑠)different from (8,2)with three pairwise distinct𝑠-equivalent𝑛-multisets (i.e.,_𝑠(𝑛)>2)?

Two more results in the same article deserve mention. One was to prove that when dealing with any question about multiset recovery it was enough to work with ring of integers ℤ, instead of arbitrary fields of characteristic zero or torsion-free Abelian groups. The other result resolved one of the questions about number of elements in_𝑠. Namely, the authors proved that_𝑠 was finite for all𝑠 >2.

(1962). The very same year the first part of the originalProblem 1.1was used in one of the top math contests in the Soviet Union – namely, Moscow City Mathematical Olympiad. It is quite possible that some Soviet mathematician had seen article [5] and liked the original question well enough to submit it to the olympiad committee. It was given to high school juniors and proved to be one of the more difficult problems of that year. An unpublished compilation of problems from that competition (translated into English) can be found in [13].

(1968). Among two questions about suspect pairs (Questions 2.2, 2.3) the latter –𝑛= 12, 𝑠= 4– seemed easier. So it was not surprising that “only” ten years after original article [3], John Ewell published his paper [6] claiming that pair(12,4)was not singular and recovery was always possible for this case (it became a part of his Ph.D thesis). He also found a purely combinatorial and more direct proof of the very important formula⟨3⟩(see below, inSection 4).

Many years later further investigation uncovered an error in calculations regarding pair(12,4). However, another result in the same article was clearly correct – namely, Ewell demonstrated that answer toQuestion 2.5 bwas positive. He has proved that₃(6) = 4and then went on to provide complete characterization of all possible quartets of pairwise different 3-equivalent 6-multisets.

We will give you one of these examples as a demonstration of Ewell’s discovery

𝐴= {0,5,9,10,11,13};𝐵= {1,5,8,9,10,15};𝐶= {1,6,7,8,11,15};𝐷= {3,5,6,7,11,16},

leaving the actual verification as an easy exercise for the reader.

(1981). Richard Guy mentioned multiset recovery in his compendium of unsolved problems in number theory, see [7], Problem C5. He explained that it had been solved for 𝑠 = 2 while the question for values of₂ still had not been answered in full. Case𝑠 = 3for𝑛 = 27and𝑛 = 486

(1991). As far as we know, after Ewell’s thesis Multiset Recovery Problem slipped into relative obscurity until 1991, when Boman, Bolker and O’Neil have explored a slightly different approach in their article [8].

More precisely, for point 𝑥 = (𝑥₁, 𝑥₂, ..., 𝑥_𝑛)in 𝑛-dimensional euclidean spaceℝ𝑛 and for any

𝑠-subset𝐴 = {𝑎₁, ..., 𝑎_𝑠}of𝐼_𝑛 = {1,2, ..., 𝑛}let us define𝑥_𝐴 as the sum𝑥_𝑎

1 +...+𝑥𝑎𝑠. Then we can define linear operator

𝑅_𝑛,𝑠 ∶ℝ𝑛 _→ℝ(𝑛𝑠), 𝑅

𝑛,𝑠(𝑥1, 𝑥2, ..., 𝑥𝑛) = (𝑥𝐴1, 𝑥𝐴2, ..., 𝑥𝐴₍𝑛

𝑠)

),

where𝐴₁, ...,𝐴₍𝑛 𝑠)

is the sequence of all𝑠-subsets in𝐼_𝑛. This is a “sort” of discrete (combinatorial) version of Radon integral transform. Mapping𝑅_𝑛,𝑠 can obviously be transferred from euclidean spaces to their reductions modulo standard actions (permutations of coordinates) of symmetric groups 𝑆_𝑛 and 𝑆₍𝑛

𝑠)

respectively so that we have 𝑅_𝑛,𝑠 ∶ 𝕃_{𝑛 →} 𝕃₍𝑛 𝑠)

where 𝕃_𝑛 = ℝ𝑛∕𝑆_𝑛. Then Multiset Recovery Problem can be posed as a question on whether𝑅_𝑛,𝑠 is an injection. Or more generally, as a question on the size of𝑅−1_𝑛,𝑠(𝑥), with𝑥∈𝕃₍𝑛

𝑠) .

Using this notation and terminology they proved – among other things – that when recovering

𝑛-multiset from the collection of its 2-sums one can not obtain more than𝑛− 2different multisets. Improving on that result they have also showed that for any𝑛≠ 8this upper bound could actually be lowered to 2 and almost always to 1. Thus they solvedQuestion 2.5 a.

It is worth noting that judging by the list of the open questions, at that time the authors did not know about Ewell’s paper [6].

(1992). By some happy “accident” in 1991 Question 1.2 was used in a students’ mathematical contest in St.Petersburg, USSR. Author of this survey was – as surely many other mathematicians before him – lured in by this seemingly simple problem, and started his own investigation. That resulted in article [9] by Fomin and Izhboldin submitted for Russian publication in 1992 (English translation was published in 1995).

Most of that article was about rediscovering the very same results already achieved in [3] and [5] – unfortunately, due to a rather poor access to international scientific magazines authors could not properly search for the papers already written on this issue. However, their article still contained one completely new result: singularity examples which positively answeredQuestion 2.2for both cases “under suspicion”. Pairs(27,3)and(486,3)were proved singular.

Thus, investigation of generalized Multiset Recovery Problem 1.4 for the case of 𝑠 = 3 was closed.

(1996). Just a few short years later, Boman and Linusson have independently come up with sin-gularity examples for pairs (27,3) and (486,3)in [11]. Alas, they thought that case (12,4) was already resolved by Ewell – at the end of their paper they mentioned that they were told (apparently at the very last moment) about article [6]. They also made some inroads into finding all possible singularity examples for𝑠= 3.

(1997). Ross Honsberger dedicated a chapter called “A Gem from Combinatorics” of his book [12] to the case𝑠 = 2 of Multiset Recovery Problem. It is curious that he never mentions Leo Moser. Instead Honsberger stated that the results he had reproduced came from Paul Erdős and John Selfridge. This is the only time when Erdős’s name appears in this story. It is not clear whether he really has done something there or possibly it was just a mistake in attribution.

(2003). In chapter 46 of their engaging book [14], Savchev and Andreescu explained the solution for Question 1.2and also went over the results from Lambek and Moser’s article [4] concerning Thue-Morse sequence.

(2008). A slightly expanded version ofQuestion 1.3with extra items repeating parts of [5] and [8] was published as another problem in American Mathematical Monthly – submitted by Chen and Lagarias, [15]. Some of the results were subsequently posted and discussed on the Cut-The-Knot website, see [16].

(2016). Nothing significant happened for quite some time until Isomurodov and Kokhas ([17]) have discovered that Ewell made a mistake in his lengthy polynomial computations for pair(12,4). We will never know how that happened, but nowadays mathematicians no longer have to do all these ex-hausting computations by hand – for instance, authors of [17] made use of symbolic computational package MAPLE™. After that the authors proved the existence of the “recovery failure” example and actually produced it, thus solvingQuestion 2.3and finalizing case𝑠= 4ofProblem 1.4(see below inSection 4).

3. SOME SIMPLE EXAMPLES

This short section explains how construct some simple examples of singular pairs.

Case 𝑠 = 2, 𝑛 = 2𝑘_{. The most obvious and trivial of all examples of “recovery failure” is pair} (2,2): one cannot hope to restore a set of two numbers knowing only their sum. This is not a “real” singular pair (because𝑛= 𝑠) but we can use it as a basis from which less trivial examples are built. Namely, if we have two𝑛-multisets𝐴and𝐵which are 2-equivalent, then for any number𝑑we have

𝐴∪ (𝐵+𝑑)∼2 𝐵∪ (𝐴+𝑑) ⟨1_⟩

where𝑋 +𝑑is multiset obtained from𝑋 by adding𝑑to all of its elements. So if we start with

𝐴= {1,1}, 𝐵 = {0,2}, 𝐴∼2 𝐵

then choosing𝑑 = 1we get

𝐴′= {1,1,1,3}, 𝐵′ = {0,2,2,2}, 𝐴′ 2∼𝐵′.

Proceeding in this manner, we can easily build examples of 2-equivalent𝑛-multisets for any𝑛which is a power of 2. Again, we will leave the proof of_⟨1_⟩as an exercise for the reader.

Case𝑛= 2𝑠. Remember that singularity example for𝑛= 6,𝑠= 3fromSection 2? It can be easily generalized for any pair(𝑛, 𝑠)where𝑛= 2𝑠.

Namely, you can take some2𝑠-multiset𝐴, find its arithmetic mean𝑎and reflect𝐴with respect to𝑎to obtain what we will call itsmirrormultiset𝐴̃ = 2𝑎−𝐴. As long as𝐴is not symmetric,𝐴̃

will be different from𝐴and𝐴 ∼𝑠 𝐴̃. To prove that it is sufficient to notice that for each𝑀 ⊂ 𝐴

with_|𝑀_|=𝑠the mirror image of𝐴∖𝑀 (which is a sub-multiset of𝐴̃consisting of𝑠numbers) has the same sum of elements.

For demonstration purposes we only need one example – let us consider𝐴= {12𝑠−1_{,1 − 2𝑠}and} its mirror𝐴̃ = {(−1)2𝑠−1_{,2𝑠− 1}.}

All the𝑠-sums of numbers in𝐴– there are(2𝑠

𝑠 )

of them – fall into two groups. One consists of the sums that include element(1 − 2𝑠)– there are(2𝑠−1

𝑠−1

)

of those, and each one of these sums is equal to(𝑠− 1)⋅1 + (1 − 2𝑠) = −𝑠. The other one has(2𝑠−1

𝑠 )

them, each one of them equal to𝑠. Thus we have𝐴(𝑠) _{= {(−𝑠)}𝑚_{, 𝑠}𝑚_{}where𝑚 =} (2𝑠−1 𝑠−1

)

= (2𝑠−1 𝑠

)

. You can see that𝐴(𝑠) _{is symmetric (with respect to zero) and thus𝐴̃}(𝑠) _=𝐴(𝑠)_.

Duality (𝑛, 𝑠) ↔ (𝑛, 𝑛− 𝑠). If we have two 𝑠-equivalent 𝑛-multisets 𝐴 and 𝐵, then these same

multisets are(𝑛−𝑠)-equivalent as well. To prove that, it is sufficient to demonstrate that sum of all elements of𝐴equals to that of𝐵. Quick computation shows that

∑

1⩽𝑖1<𝑖2...<𝑖𝑠⩽𝑛

(𝑎𝑖₁ +𝑎𝑖₂ +...+𝑎𝑖_𝑠) = (

𝑛− 1 𝑠− 1

) ∑

1⩽𝑖⩽𝑛 𝑎_𝑖.

Thus sum of numbers in𝐴equals the sum of numbers in𝐵 and denoting that number by𝑆we have

𝐴(𝑛−𝑠) =𝑆−𝐴(𝑠), 𝐵(𝑛−𝑠) =𝑆−𝐵(𝑠),

which proves the duality. This means we can always assume that𝑛⩾₂𝑠; if𝑠 < 𝑛 < 2𝑠then we can switch to pair(𝑛, 𝑠′_)where𝑠′ _{= 𝑛−𝑠and𝑛 > 2𝑠}′_.

This duality allows us to generate more examples of singular pairs. For instance, since (8,2)

is singular then(8,6)is singular, too. As we will see soon, pairs (27,3), (486,3)and(12,4)are singular – therefore, pairs(27,24),(486,483)and(12,8)are singular as well.

Later in this article (seeSection 5) we will talk more about this duality and its partial expansion.

Linear transformations. Finally, one obvious but useful fact. If𝐴∼𝑠 𝐵and𝑓(𝑥) =𝑝𝑥+𝑞is some arbitrary linear function then multisets𝑓(𝐴) and𝑓(𝐵)are also 𝑠-equivalent. That simply means we can translate and stretch/shrink singularity examples to obtain new ones.

For instance, if you consider 𝐴 = {0,5,9,10,11,13}, 𝐵 = {1,5,8,9,10,15}then 𝐴 ∼3 𝐵. Applying𝑓(𝑥) = 2𝑥− 13we obtain new pair of 3-equivalent multisets𝐴₁= {−13,−3,5,7,9,13},

𝐵₁ = {−11,−3,3,5,7,17}.

Of course, all the singularity examples that can be obtained from each other by such operations will be considered identical for the purposes of this investigation.

4. SYMMETRIC POLYNOMIALS. COMPLETE SOLUTION FOR CASES𝑠= 2,3,4

Now let us delve into specific techniques used in multiset recovery. The main one is based on the following approach that utilizes symmetric polynomials.

Given𝑛-multiset𝐴 = {𝑎₁, ..., 𝑎_𝑛}we can produce sequence of sums of its𝑘-th powers for𝑘 = 1, ..., 𝑛. That is, we can apply power-sum symmetric polynomials of𝑛variables

𝜎_𝑘(𝑥₁, ..., 𝑥_𝑛) = 𝑛 ∑

𝑖=1

𝑥𝑘_𝑖

to multiset𝐴to obtain sequence𝜎₁(𝐴), ..., 𝜎𝑛(𝐴). It is well known that𝐴can be restored from this

sequence – values of 𝜎_𝑘(𝐴) determine coefficients of polynomial(𝑥− 𝑎₁)(𝑥−𝑎₂)...(𝑥− 𝑎_𝑛) and therefore determine the multiset of its roots.

Thus if values of𝜎_𝑘(𝐴)for𝑘= 1, ..., 𝑛can be deduced from values of𝜎_𝑘(𝐴(𝑠)_{)then multiset𝐴is} uniquely determined by multiset𝐴(𝑠).

Let us start from small values of𝑘. For𝑘= 1we have already computed

𝜎₁(𝐴(𝑠)_{) =}

( 𝑛− 1 𝑠− 1

) 𝜎₁(𝐴)

and therefore, if𝐴(𝑠) _=𝐵(𝑠) _then𝜎

1(𝐴) =𝜎1(𝐵). This means that𝜎1(𝐴)can always be found from

Now, if𝑘= 2then

𝜎₂(𝐴(𝑠)_{) =} ∑ 1⩽𝑖1<...<𝑖𝑠⩽𝑛

(𝑎_𝑖

1 +...+𝑎𝑖𝑠)

2 ₌

( 𝑛− 1 𝑠− 1

) ∑

1⩽𝑖⩽𝑛 𝑎2_𝑖 +

( 𝑛− 2 𝑠− 2

) ∑

1⩽𝑖<𝑗⩽𝑛

2𝑎_𝑖𝑎_𝑗 =

= (

𝑛− 1 𝑠− 1

) ∑

1⩽𝑖⩽𝑛 𝑎2_𝑖 +

( 𝑛− 2 𝑠− 2

) ∑

1⩽𝑖≠𝑗⩽𝑛 𝑎_𝑖𝑎_𝑗 =

= ((

𝑛− 1 𝑠− 1

) −

( 𝑛− 2 𝑠− 2

)) ∑

1⩽𝑖⩽𝑛 𝑎2_𝑖 +

( 𝑛− 2 𝑠− 2

) ∑

1⩽𝑖,𝑗⩽𝑛 𝑎_𝑖𝑎_𝑗 =

= (

𝑛− 2 𝑠− 1

)

𝜎₂(𝐴) + (

𝑛− 2 𝑠− 2

)

𝜎₁(𝐴)2.

We already know𝜎₁(𝐴)and thus we can find𝜎₂(𝐴)as long as coefficient(𝑛−2

𝑠−1

)

is not zero. But

𝑛 > 𝑠 >0and therefore𝜎₂(𝐴)is also always “recoverable”. Since𝜎_𝑘(𝐴(𝑠)_{)is a symmetric polynomial of𝑎}

1, ... 𝑎𝑛 then it can be expressed in the following

way:

𝜎_𝑘(𝐴(𝑠)) =𝛼𝜎_𝑘(𝐴) +(𝜎1(𝐴), 𝜎2(𝐴), ..., 𝜎𝑘−1(𝐴)), ⟨2⟩ where 𝛼 is a constant (in terms of variables𝑎_𝑖) defined by three numbers𝑛, 𝑠, 𝑘, and  is some polynomial of 𝑘− 1 variables whose coefficients are fully defined by that triplet as well. For instance, as we have just shown, for𝑘= 2we have𝛼 =(𝑛−2

𝑠−1

)

and(𝑥) = (𝑛−2 𝑠−2

) 𝑥2_.

It follows that if we could show that coefficient𝛼 does not vanish then we will have proved that

𝜎_𝑘(𝐴)is determined by𝜎₁(𝐴(𝑠)_),𝜎 2(𝐴(

𝑠)_{), ...,𝜎}

𝑘(𝐴( 𝑠)_).

Let us denote that coefficient𝛼as𝐹_𝑠,𝑘(𝑛). Then the following equality is true:

Theorem 4.1.

𝐹_𝑠,𝑘(𝑛) = 𝑠 ∑

𝑝=1

(−1)𝑝−1_𝑝𝑘−1

( 𝑛 𝑠−𝑝

)

. _⟨3⟩

It was proved in [5] by some neat manipulation of a formula from [3]. Later a purely combinato-rial and more direct proof of_⟨3_⟩was given in [6]. And then it was again “rediscovered” and proved (in a somewhat different manner, by making use of exponential generating functions) in [9].

We will leave it to the reader as a simple exercise to prove that for𝑘 = 2formula_⟨3_⟩is indeed equivalent to𝐹_𝑠,2(𝑛) =(𝑛−2

𝑠−1

)

.

Incidentally, even without this formula case𝑠= 2(Problem 1.2) can now be resolved in a very straightforward manner. Computing𝜎_𝑘(𝐴(2)_{)we obtain (using notation from⟨2⟩)}

𝛼 = 𝑛− 2𝑘−1_,

which means that𝑛-multiset is always recoverable from the multiset of its 2-sums if𝑛is not a power of 2. As we already know, if𝑛is a power of 2 then𝑛-multiset𝐴cannot always be recovered from

𝐴(2)_.

Case𝑠= 3. Equation⟨3⟩gives us

𝐹_3,𝑘(𝑛) = (

𝑛 2

) − 2𝑘−1

( 𝑛 1

)

+ 3𝑘−1, 2𝐹_3,𝑘(𝑛) = 𝑛2−𝑛(2𝑘+ 1) + 2⋅3𝑘−1.

Investigation here is again relatively straightforward. First, we can prove that𝐹_3,𝑘(𝑛)cannot be zero for positive integer𝑛 if𝑘 > 12. Second, we check all the cases with𝑘 ⩽ 12and verify that polynomials𝐹_3,𝑘have integer roots if and only if𝑘∈ {1,2,3,5,9}. For these five special cases we have

𝐹_3,1(𝑛) = 1

2(𝑛− 1)(𝑛− 2), 𝐹_3,2(𝑛) = 1

2(𝑛− 2)(𝑛− 3), 𝐹_3,3(𝑛) = 1

2(𝑛− 3)(𝑛− 6), 𝐹_3,5(𝑛) = 1

2(𝑛− 6)(𝑛− 27), 𝐹_3,9(𝑛) = 1

2(𝑛− 27)(𝑛− 486).

FromSection 3 we already know that pair(6,3)is singular. To prove the same for pairs(27,3)

and(486,3)we present the following examples:

𝑛= 27 𝐴′₂₇ = {0,116,210}, 𝐴′′₂₇= {05,110,210,32}, 𝐴′′′₂₇ = {0,15,210,36,45}, 𝑛= 486 𝐴₄₈₆ = {022_,1176_,2231_,356_,4}.

We will leave it to the reader to verify that all these multisets are 3-equivalent to their mirrors. Nowadays, this can be done in minutes, using just a few lines of code in some decent computational package.

Summary: ₃= {6,27,486}.

Case𝑠= 4. From_⟨3_⟩we obtain

𝐹_4,𝑘(𝑛) = (

𝑛 3

) − 2𝑘−1

( 𝑛 2

) + 3𝑘−1

( 𝑛 1

)

− 4𝑘−1_,

6𝐹₄,𝑘(𝑛) =𝑛

3_−𝑛2_{(3 + 3⋅2}𝑘−1_{+ 1) +𝑛(2 + 3⋅2}𝑘−1_{+ 2⋅3}𝑘_{) − 6⋅4}𝑘−1_.

Then using divisibility and other pretty routine number theory ideas we can prove that for𝑘 >7

polynomials𝐹_3,𝑘do not have positive integer roots. And, finally,

𝐹_4,1(𝑛) = 1

6(𝑛− 1)(𝑛− 2)(𝑛− 3), 𝐹_4,2(𝑛) = 1

6(𝑛− 2)(𝑛− 3)(𝑛− 4), 𝐹_4,3(𝑛) = 1

6(𝑛− 3)(𝑛− 4)(𝑛− 8), 𝐹_4,4(𝑛) = 1

6(𝑛− 4)(𝑛

2_{− 23𝑛+ 96),}

𝐹_4,5(𝑛) = 1

6(𝑛− 8)(𝑛

2_{− 43𝑛+ 192),}

𝐹_4,6(𝑛) = 1

6(𝑛− 12)(𝑛

2_{− 87𝑛+ 512),}

𝐹_4,7(𝑛) = 1

6(𝑛− 8)(𝑛

Case𝑛= 8is “trivial” – this is the situation𝑛= 2𝑠which is well known to us by now. The only other non-trivial root of𝐹_4,𝑘polynomials is 12. As we mentioned before, this case turned out to be tougher than the others – a pair of 4-equivalent 12-multisets was found only in 2016 by Isomurodov and Kokhas. Namely, if we consider the two following different 12-multisets

𝐴= {12,4,6,7,82,9,10,12,152}, 𝐵 = {0,3,4,5,6,7,9,10,11,12,13,16},

then𝐴∼4 𝐵.

In [17] the authors have actually proved that this is the only possible singularity example for case

(12,4)(considering pairs of multisets that differ only by linear transformation as identical). Summary: ₄= {8,12}.

5. DIGGING FOR ROOTS(OF𝐹_𝑠,𝑘)

Since we know that pair (𝑛, 𝑠) can be singular only if 𝑛 is a root of 𝐹_𝑠,𝑘 then let us turn our attention to finding out more about those roots. (The rest of this section was inspired by the proof ofTheorem 7from [3]).

For this let us take another, closer, look at the polynomials𝐹_𝑠,𝑘for the first few values of𝑘.

Case𝑘= 1. We already know that

𝐹_𝑠,1(𝑛) = (

𝑛− 1 𝑠− 1

)

= 1

(𝑠− 1)!(𝑛− 1)(𝑛− 2)...(𝑛−𝑠+ 1) = 1 (𝑠− 1)!

𝑠−1

∏ 𝑝=1

(𝑛−𝑝).

Thus the roots are 1 through𝑠− 1and of no interest to us –𝑛has to be greater than𝑠to provide us with a possibly singular pair(𝑛, 𝑠).

Case𝑘= 2. This one we have also computed before.

𝐹_𝑠,2(𝑛) = (

𝑛− 2 𝑠− 1

)

= 1 (𝑠− 1)!

𝑠 ∏

𝑝=2

(𝑛−𝑝).

Again, no roots of interest. Let us go on.

Case𝑘= 3. It is still fairly easy to compute

𝐹_𝑠,3(𝑛) = 1

(𝑠− 1)!(𝑛− 2𝑠) 𝑠 ∏

𝑝=3

(𝑛−𝑝),

which (finally!) has a non-trivial root𝑛 = 2𝑠. However, we already know about it – pair(2𝑠, 𝑠)is always singular.

Case𝑘= 4. This computation might take a little longer but eventually you will get the following formula (for𝑠> 2)

𝐹_𝑠,4(𝑛) = 1 (𝑠− 1)!(𝑛

2_{− (6𝑠− 1)𝑛+ 6𝑠}2₎

𝑠 ∏

𝑝=4

(𝑛−𝑝). _⟨4_⟩

A-ha! We now have quadratic Diophantine equation for non-trivial roots𝑛

𝑛2− (6𝑠− 1)𝑛+ 6𝑠2 = 0, _⟨5_⟩

which can be also rewritten as a quadratic equation for𝑠

Sum of the roots of equation⟨6_⟩is𝑛– hence, if pair (𝑛, 𝑠)with𝑛 > 𝑠is a root of equation ⟨4_⟩

then so is pair(𝑛, 𝑛−𝑠). This is clearly a direct analog of singular pairs’ duality we have described inSection 3. Let us call such pairs_⟨𝑛,4⟩-conjugatedor simply𝑛-conjugated.

In the same manner from equation_⟨5_⟩we can conclude that if pair(𝑛, 𝑠)is a root of equation_⟨4_⟩

then its other root is pair(6𝑠− 1 −𝑛, 𝑠). These two pairs will be called_⟨𝑠,4⟩-conjugated.

Obviously, both types of conjugation are symmetric. It is also clear from equations_⟨5_⟩and_⟨6_⟩

that for any positive solution(𝑛, 𝑠)we have6𝑠− 1 > 𝑛and𝑛 > 𝑠 – otherwise left sides of these equations are positive. Thus, conjugate pair always consists of two positive integers as well.

Let us consider the smallest possible root of _⟨5_⟩, namely 𝑟 = (2,1) (since we are solving the equation in positive integers, we are allowed to talk about “smallest” solution). That pair is not something we can directly use because𝑠must be at least 3 for the formula⟨4⟩to make sense. But it is still a root of our quadratic equation⟨5⟩and we will use it to produce others.

It is important to mention that pair𝑟is self-𝑛-conjugated (2 − 1 = 1) so the only way to produce a different solution is via𝑠-conjugation. So we jump to pair(3,1), then through𝑛-conjugation to pair(3,2), then to(8,2), then to(8,6)and so on.

Proceeding like that we will obtain one infinite chain of pairs-solutions of equation_⟨5_⟩:

(2,1)↔𝑠 (3,1)

𝑛

↔(3,2)

𝑠

↔(8,2)

𝑛

↔(8,6)

𝑠

↔(27,6)

𝑛

↔

(27,21)↔𝑠 (98,21)

𝑛

↔(98,77)

𝑠

↔(363,77)

𝑛

↔(363,286)

𝑠

↔... ⟨7⟩

We have marked the arrows with small letters "𝑛" and "𝑠" to show which type of conjugation was used in each case; also we have underlined pairs which are not “fully compliant” – they are roots of equation_⟨5_⟩ but they are not roots of corresponding polynomial𝐹_𝑠,𝑘 with𝑠 > 2(pair(8,2)is singular but we have discounted it because of 𝑠 > 2requirement). So, starting from (8,6), pairs in the chain represent valid roots of polynomials 𝐹_𝑠,4. Thus, they are all “suspect” as possible singularities for Multiset Recovery Problem.

Case𝑘= 5. In this case computation is also not terribly complicated. For𝑠 > 3we obtain

𝐹_𝑠,5(𝑛) = 1 (𝑠− 1)!(𝑛

2_{− (12𝑠− 5)𝑛+ 12𝑠}2_{)(𝑛− 2𝑠)}

𝑠 ∏

𝑝=5

(𝑛−𝑝).

Again we have a quadratic Diophantine equation which can be written like this

𝑛2− (12𝑠− 5)𝑛+ 12𝑠2 = 0. _⟨8_⟩

or like this

𝑠2−𝑠𝑛+𝑛(𝑛+ 5)∕12 = 0. _⟨9⟩

As before, equation_⟨9_⟩gives us 𝑛-conjugation “duality”(𝑛, 𝑠) ↔ (𝑛, 𝑛−𝑠). And equation⟨8⟩

provides us with_⟨𝑠,5⟩-conjugation – namely,(𝑛, 𝑠)↔(12𝑠− 5 −𝑛, 𝑠).

Similarly to the previous subsection we obtain infinite chain of pairs-solutions:

(3,1)↔𝑠 (4,1)

𝑛

↔(4,3)

𝑠

↔(27,3)

𝑛

↔(27,24)

𝑠

↔(256,24)

𝑛

↔(256,232)

𝑠

↔(2523,232)

𝑛

↔...

However this case differs somewhat from the previous one. The “minimum solution” pair(3,1)

has𝑛-conjugate(3,2)that does not coincide with it – thus the chain can be extended in the other direction as well. Therefore we obtain more solutions:

(3,2)↔𝑠 (16,2)

𝑛

↔(16,14)

𝑠

↔(147,14)

𝑛

↔(147,133)

𝑠

↔(1444,133)

𝑛

↔(1444,1311)

𝑠

↔...

...(1444,1311)↔𝑠 (1444,133)

𝑛

↔(147,133)

𝑠

↔(147,14)

𝑛

↔(16,14)

𝑠

↔

𝑠

↔(16,2)

𝑛

↔(3,2)

𝑠

↔(3,1)

𝑠

↔(4,1)

𝑛

↔(4,3)

𝑠

↔(27,3)

𝑛

↔

𝑛

↔(27,24)

𝑠

↔(256,24)

𝑛

↔(256,232)

𝑠

↔(2523,232)... ⟨10⟩

In both cases 𝑠= 4 and𝑠= 5 it is easy to prove that all positive integer solutions of equations

⟨5_⟩and_⟨8_⟩belong to the chains 7 and 10 respectively. We will leave that to the reader.

Case𝑘= 6. It would be great if same ideas could be applied for this and subsequent cases as well. However, computation of𝐹_𝑠,6shows that for𝑠 >4we have the following formula:

𝐹_𝑠,6(𝑛) = 1

(𝑠− 1)!𝑔𝑠,6(𝑛) 𝑠 ∏

𝑝=6

(𝑛−𝑝),

where

𝑔_𝑠,6(𝑛) =𝑛4− (30𝑠− 16)𝑛3+ (150𝑠2− 90𝑠+ 11)𝑛2− (240𝑠3− 90𝑠2+ 4)𝑛+ 120𝑠4.

Some of the polynomials𝑔_𝑠,6have integer roots. For instance,

𝑔_8,6(𝑛) = (𝑛− 12)(𝑛3− 212𝑛2+ 6347𝑛− 40960), 𝑔_10,6(𝑛) = (𝑛− 32)(𝑛3− 252𝑛2+ 6047𝑛− 37500), 𝑔_22,6(𝑛) = (𝑛− 32)(𝑛3− 612𝑛2+ 51047𝑛− 878460), 𝑔_30,6(𝑛) = (𝑛− 32)(𝑛3− 852𝑛2+ 105047𝑛− 3037500).

But that is basically all we can get for𝑠 = 6. Alas, no more quadratic Diophantine equations, no chains of conjugation. Also, it seems likely that polynomials𝑔_𝑠,6do not have integer roots other than the ones shown above.

And, of course, same happens with cases of even greater values of𝑠– and so this line of inves-tigation ends here.

6. COMPUTER TO THE RESCUE

Roots of𝐹_𝑠,𝑘. Trying to find more roots for cases 𝑠 > 4in hope of some insight, I have written a short program in SAGE which was then run through SageMath web interface at CoCalc.com for 𝑠= 3,4,5,6,7, etc. until the server started to stumble (that happened somewhere around𝑠= 40). After that I have switched to local install of SAGEand proceeded until 𝑠 = 200 when every new value of𝑠started to require almost a day to process (and then my computer ran out of operational memory).

The program did the following. For every fixed value of𝑠it ran the loop for𝑘from 1 to 1000, where at each step it computed polynomial 𝐹_𝑠,𝑘, factorized it over ℤ and in case of non-trivial factorization printed out the roots of the polynomial. At the end it also produced𝑘_max(𝑠)– the last value of𝑘for which a non-trivial factorization of𝐹_𝑠,𝑘occurred.

Below (in Table 1) you can see the summary of all non-trivial roots (with pairs(2𝑠, 𝑠)excluded) obtained from this experiment.

We have marked each entry𝑛with the first value of 𝑘for which𝐹_𝑠,𝑘(𝑛) = 0. So, for instance, mark[4]corresponds to chain⟨7⟩, and[5]– to chain⟨10⟩.

For all other values of𝑠between 3 and200the only roots found were either𝑛= 2𝑠(which would have been marked with[3]) or trivial (1 through𝑠) and therefore of no interest for us.

𝑠 𝑛

3 27[5], 486[9] 4 12[6]

6 8[4],27[4] 8 12[6] 10 32[6]

14 16[5]_,147[5] 21 27[4],98[4]

22 32[6]

24 27[5]_,256[5] 30 32[6]

62 64[7]

77 98[4],363[4] 126 128[8]

133 147[5], 1444[5]

TABLE1. Non-trivial roots of𝐹_𝑠,𝑘for3⩽𝑠⩽200

Roots of𝐹_𝑠,𝑘which have not been yet verified as multiset recovery singularities are emphasized inbold. They represent the current “suspect” cases.

In addition the experiment showed that for all 3 < 𝑠 ⩽ 200the value of𝑘_max(𝑠) was equal to

2𝑠− 1. Claiming that to be always true is what we will call𝑘_max-Conjecture– seeConjecture 7.6

below, inSection 7.

The following proposition can be considered as a very easy “half” of this conjecture.

Proposition 6.1. For any𝑠 >2,𝑛= 2𝑠and any odd𝑘such that1< 𝑘 < 𝑛we have 𝐹_𝑠,𝑘(𝑛) = 0. Proof. We can rewrite this statement by using_⟨3_⟩, adding summand with𝑝= 0, and reversing the summation index𝑝. As a result we obtain

𝑠 ∑

𝑝=0

(−1)𝑝_{(𝑠−𝑝)}𝑟 (

𝑛 𝑝

) = 0

for any even number0< 𝑟 < 𝑛. Now, since

( 𝑛 𝑝

) =

( 𝑛 𝑛−𝑝

)

, (𝑠−𝑝)𝑟 = (𝑠− (𝑛−𝑝))𝑟,

the equation above is equivalent to

𝑛 ∑

𝑝=0

(−1)𝑝_{(𝑠−𝑝)}𝑟 (

𝑛 𝑝

) = 0.

Any polynomial of 𝑝 with degree less than 𝑛 (such as (𝑠 − 𝑝)𝑟_{) can be expressed as a linear}

proposition then follows from well-known formula

𝑛 ∑

𝑝=0

(−1)𝑝_𝑝[𝑟]

( 𝑛 𝑝

) = 0,

which can be easily proved using generating function𝜆(𝑥) = (1 +𝑥)𝑛 ₌ ∑𝑛 𝑝=0

(𝑛 𝑝 )

𝑥𝑝_{. Polynomial} 𝜆(𝑥)has(−1)as a root of order𝑛, thus its𝑟-th derivative𝜆(𝑟)_{(𝑥) =}∑𝑛

𝑝=0

(𝑛 𝑝 )

𝑝[𝑟]_𝑥𝑝−𝑟 _{also must have} (−1)as a root for any0< 𝑟 < 𝑛.

■ And then from𝐹_{𝑠,2𝑠−1}(2𝑠) = 0follows

Corollary 6.2. For any𝑠 >2we have𝑘_max(𝑠)⩾2𝑠− 1.

Singularity search. Well, since we already started using computer assistance, let us continue down this slippery slope. The next idea in automating our investigation is to hunt not for the roots of polynomials𝐹_𝑠,𝑘but for the singular multisets themselves.

The objective is to try and find singularity examples for the smallest “suspect” pairs(27,6)and

(32,10). The other suspects, not 𝑛-conjugated to these two, are too large to hope for any “brute force” computer search to succeed.

The main idea of this approach is to restrict the realm of the𝑛-multisets that we deal with. Con-sider all (𝑛+𝑚−1

𝑚−1

)

weak compositionsof𝑛 into𝑚parts, that is, representations of 𝑛as a sum of𝑚

non-negative integers𝑘_𝑖,(𝑖= 1,2, ..., 𝑚):

 ∶ 𝑛=𝑘₁+𝑘₂+...+𝑘_𝑚, 𝑘_𝑖∈ℤ_⩾0.

Each weak composition of this form can be treated as sequence of multiplicities – that is, from each composition we will construct𝑛-multiset

𝐴_ = {1𝑘1_,2𝑘2_{, ..., 𝑚}𝑘𝑚}_. ⟨₁₁⟩ Alas, in both cases(27,6)and(32,10)we cannot hope to find𝑛-multiset𝑋which is𝑠-equivalent to its own mirror𝑋̃. Indeed, if𝑋 ∼𝑠 𝑋̃ then without loss of generality we can assume that𝜎₁(𝑋) = 0. Thus𝑋̃ = −𝑋 and for any even𝑘 >0the sum of𝑘-th powers of numbers in𝑋will be equal to that of𝑋̃.

From the experiment we already know (and can easily verify this formally) that 𝐹_6,𝑘(27) = 0

iff𝑘 = 4and𝐹_10,𝑘(32) = 0 iff𝑘 = 6. Therefore, for any𝑋 such that 𝑋(𝑠) _{= 𝑋̃}(𝑠) _{we will have}

𝜎_𝑘(𝑋) = 𝜎_𝑘(𝑋)̃ for all values of1⩽ 𝑘 ⩽ 𝑛– the only exceptions we could have hoped for were 4 and 6 (for𝑛= 27and𝑛= 32respectively) and we have just eliminated them.

So, how can we proceed and what are the challenges?

First of all, we cannot afford to generate an array of all weak compositions (or multisets of type

⟨11_⟩) and then analyze the result – the computer would soon run out of memory. For example, if

𝑛= 27and𝑚= 10then we get almost 100 million of such compositions (94,143,280 to be exact). Thus, the algorithms here have to be iterative. It is fairly easy to write an iterator function which generates next weak composition based on the previous one. Hint: find the last non-zero part, increase the previous part by one and make all the following parts zero except for the last one. If necessary, same can be done when going through all𝑠-subsets in an𝑛-multiset.

Second, the check function that verifies whether the two given multisets𝐴and𝐵are𝑠-equivalent to its mirror 𝐴̃ must be written very carefully and very efficiently because it will be called quite a few times. To make it work as fast as possible the function needs to implement some “quick

rejection” checks. For instance, the sum of the first𝑠numbers from𝐴(let us assume it is sorted) is always equal to the minimum number in𝐴(𝑠)_{; hence these sums for𝐴and𝐵 must coincide. The} more simple checks of this sort are employed the better.

Third, calling this check function for every pair𝐴_ and𝐴_′of constructed multisets is absolutely

out of the question. So, some sort of simplified “signature” has to be computed for each multiset

𝐴(_𝑠) (alas, no quick rejections there) so we can compare these numbers instead of comparing very large multisets𝐴(_𝑠). But even with that we cannot go much farther beyond𝑚= 10for the reason I already mentioned above – such huge arrays of data will exhaust the computer memory.

My own implementation of this approach did not find any examples of 6-equivalent 27-multisets of type_⟨11_⟩ for𝑚 < 10. As a sanity check I ran the same code for pair(12,4)with𝑚 = 17and after a few hours of number crunching it resulted in the same unique example of two 4-equivalent 12-multisets already found in [17].

Clearly, absence of positive results in this computational experiment doesn’t mean much as there could be 6-equivalent 27-multisets that span longer stretches of integers. And it is always possible that singularity example for(27,6)simply doesn’t exist. For now, this remains an open question.

If some of the readers become interested in this line of investigation I will gladly send them my code – I am sure it can be made more effective while consuming less memory. And then, who knows, perhaps the next value of𝑚will finally yield the desired singularity.

7. OPEN QUESTIONS

Here is a list of a few open questions which have come up during this survey’s investigations.

Question 7.1. Is it true that₅ = {10}?

Question 7.2. Which of the new “suspect” pairs(𝑛, 𝑘)found inSection 6are singular?

Of course, we are talking here about pairs highlighted in bold in Table 1. Perhaps some cleverly written computer program could answer this question at least for the smallest “suspect” pairs(27,6)

and(32,10).

Question 7.3. Does a non-trivial root of 𝐹_𝑠,𝑘 guarantee an existence of corresponding singular pair? That is, is it true that for any positive integers𝑠,𝑘,𝑛such that𝑛 > 𝑠,𝑛 ⩾𝑘and𝐹_𝑠,𝑘(𝑛) = 0

there exists a pair of different𝑛-multisets𝐴and𝐵such that𝐴(𝑠) _=𝐵(𝑠)_?

All the results accumulated over the last sixty years so far confirm this hypothesis; however, it looks like an extremely difficult nut to crack. The following question could perhaps serve as a small step in that direction.

Question 7.4. A singular pair𝑃 = (𝑛, 𝑠)is such that𝐹_𝑠,𝑘(𝑛) = 0for𝑘= 4or5. Is pair𝑃′ = (𝑚, 𝑠)

obtained from𝑃 by⟨𝑠, 𝑘_⟩-conjugation also singular?

We know that 𝑛-conjugation between roots of 𝐹_𝑠,𝑘 has its direct analog in (𝑛, 𝑠) ↔ (𝑛, 𝑛− 𝑠)

duality between singular pairs. However, the similar question about𝑠-conjugation does not seem to be even remotely as simple.

Question 7.5. Is there a less “accidental” explanation for at least one singular pair with𝑛 ≠ 2𝑠,

𝑠 >2?

So far, all such examples were constructed in a ratherad hocmanner by grinding through solu-tions for simultaneous equasolu-tions𝜎_𝑘(𝐴(𝑠)_{) =𝜎}

𝑘(𝐵

overwhelmingly complex. Perhaps for at least some singular pairs there exists a less “accidental” construction, combinatorial or algebraic.

I think that the following hypothesis is the most important among reasonably hard open questions on multiset recovery.

Conjecture 7.6. (𝑘_max-Conjecture)Is it true that𝑘_max(𝑠) = 2𝑠− 1for all𝑠 > 3? In other words, prove or disprove that polynomial𝐹_𝑠,𝑘can have integer roots only if𝑘⩽ 2𝑠− 1.

Positive answer to this question would mean a considerable breakthrough in any – computer-aided or not – search for the roots of polynomials𝐹_𝑠,𝑘.

For starters, we could immediately claim that_𝑠 = {2𝑠}for many small values of𝑠. Proof of that for𝑠= 5(that is, a solution toQuestion 7.1) would go like this:

Proof. If𝑛 >5and𝑛≠10then𝐹_5,𝑘(𝑛)≠0for any𝑘 >0. Indeed,𝑘_max-Conjectureimplies there is no need to check values𝑘⩾10. FromSection 5we know the same is true for𝑘⩽ 5. So we only need to examine𝐹_5,6,𝐹_5,7,𝐹_5,8and𝐹_5,9:

𝐹_5,6(𝑛) =𝑛4− 134𝑛3+ 3311𝑛2− 27754𝑛+ 75000, 𝐹_5,7(𝑛) =𝑛4− 262𝑛3+ 9527𝑛2− 107570𝑛+ 375000, 𝐹_5,8(𝑛) =𝑛4− 518𝑛3+ 27791𝑛2− 420490𝑛+ 1875000, 𝐹_5,9(𝑛) =𝑛4− 1030𝑛3+ 81815𝑛2− 1653650𝑛+ 9375000.

How do we do that? Again, we can use computer to help us produce a verifiable computer-independent proof. As an example, let us prove (quite formally) that 𝐹_5,6 has no integer roots in[5; ∞[⧵{10}. A couple of lines of code in MATLAB™ or in SAGE will get us real roots of this polynomial

𝑥₁= 6.014875..., 𝑥₂ = 7.745287..., 𝑥₃= 15.348149..., 𝑥₄ = 104.891687... ,

We cannot use that as a proof, but we can computeby handvalues𝐹_5,6(𝑥)for 𝑥= 6, 7, 8, 15, 16, 104 and 105. The results – 24, (-600), 360, 2040, (-4776), (-745560) and 93480 – prove that there is a non-integer root inside each one of segments[6; 7],[7; 8],[15; 16]and[104; 105]. Those four non-integer numbers obviously constitute the set of all roots of𝐹_5,6.

In exactly the same way (but with longer computations) one can prove the same for polynomials

𝐹_5,7,𝐹_5,8and𝐹_5,9(polynomials𝐹_5,7and𝐹_5,9both have one integer root but it is equal to 10). Finally, since𝐹_5,𝑘(𝑛)≠0, then fromTheorem 4.1it follows that𝑛-multiset𝐴is always

recover-able from𝐴(5). ■

The following hypothesis proposes an update toQuestion 2.5.

Conjecture 7.7. (Magical Triplet Conjecture) If 𝑠 > 2 and 𝑛 > 2𝑠 then for any three distinct

𝑛-multisets some two of them are not𝑠-equivalent to each other.

If we agree to exclude fully investigated cases of𝑠= 2and𝑛= 2𝑠then let us call three different

𝑛-multisets such that they are all𝑠-equivalent to each other, amagical triplet. I submit thatmagical triplets do not exist– in other words, when one tries to recover a multiset from its collection of

...(1444,1311)↔𝑠 (1444,133) 𝑛

↔(147,133) 𝑠

↔(147,14) 𝑛

↔(16,14) 𝑠 ↔ 𝑠

↔(16,2) 𝑛 ↔(3,2)

𝑠 ↔(3,1)

𝑠 ↔(4,1)

𝑛 ↔(4,3)

𝑠

↔(27,3) 𝑛 ↔ 𝑛

↔(27,24) 𝑠

↔(256,24) 𝑛

↔(256,232) 𝑠

↔(2523,232)... ⟨10⟩

In both cases 𝑠= 4 and𝑠= 5 it is easy to prove that all positive integer solutions of equations

⟨5_⟩and_⟨8_⟩belong to the chains 7 and 10 respectively. We will leave that to the reader.

Case𝑘= 6. It would be great if same ideas could be applied for this and subsequent cases as well. However, computation of𝐹_𝑠,6shows that for𝑠 >4we have the following formula:

𝐹_𝑠,6(𝑛) = 1

(𝑠− 1)!𝑔𝑠,6(𝑛)

𝑠 ∏

𝑝=6

(𝑛−𝑝),

where

𝑔_𝑠,6(𝑛) =𝑛4− (30𝑠− 16)𝑛3+ (150𝑠2− 90𝑠+ 11)𝑛2− (240𝑠3− 90𝑠2+ 4)𝑛+ 120𝑠4.

Some of the polynomials𝑔_𝑠,6have integer roots. For instance,

𝑔_8,6(𝑛) = (𝑛− 12)(𝑛3− 212𝑛2+ 6347𝑛− 40960), 𝑔_10,6(𝑛) = (𝑛− 32)(𝑛3− 252𝑛2+ 6047𝑛− 37500), 𝑔_22,6(𝑛) = (𝑛− 32)(𝑛3− 612𝑛2+ 51047𝑛− 878460), 𝑔_30,6(𝑛) = (𝑛− 32)(𝑛3− 852𝑛2+ 105047𝑛− 3037500).

But that is basically all we can get for𝑠 = 6. Alas, no more quadratic Diophantine equations, no chains of conjugation. Also, it seems likely that polynomials𝑔_𝑠,6do not have integer roots other than the ones shown above.

And, of course, same happens with cases of even greater values of𝑠– and so this line of inves-tigation ends here.

6. COMPUTER TO THE RESCUE

Roots of𝐹_𝑠,𝑘. Trying to find more roots for cases 𝑠 > 4in hope of some insight, I have written a short program in SAGE which was then run through SageMath web interface at CoCalc.com for 𝑠= 3,4,5,6,7, etc. until the server started to stumble (that happened somewhere around𝑠= 40). After that I have switched to local install of SAGEand proceeded until 𝑠 = 200 when every new value of𝑠started to require almost a day to process (and then my computer ran out of operational memory).

The program did the following. For every fixed value of𝑠it ran the loop for𝑘from 1 to 1000, where at each step it computed polynomial 𝐹_𝑠,𝑘, factorized it over ℤ and in case of non-trivial factorization printed out the roots of the polynomial. At the end it also produced𝑘_max(𝑠)– the last value of𝑘for which a non-trivial factorization of𝐹_𝑠,𝑘occurred.

Below (in Table 1) you can see the summary of all non-trivial roots (with pairs(2𝑠, 𝑠)excluded) obtained from this experiment.

We have marked each entry𝑛with the first value of 𝑘for which𝐹_𝑠,𝑘(𝑛) = 0. So, for instance, mark[4]corresponds to chain⟨7⟩, and[5]– to chain⟨10⟩.

For all other values of𝑠between 3 and200the only roots found were either𝑛= 2𝑠(which would have been marked with[3]) or trivial (1 through𝑠) and therefore of no interest for us.

𝑠 𝑛

3 27[5], 486[9]

4 12[6]

6 8[4],27[4]

8 12[6]

10 32[6]

14 16[5]_,147[5] 21 27[4],98[4] 22 32[6]

24 27[5]_,256[5] 30 32[6]

62 64[7]

77 98[4],363[4] 126 128[8]

133 147[5], 1444[5]

TABLE1. Non-trivial roots of𝐹_𝑠,𝑘for3⩽𝑠⩽200

Roots of𝐹_𝑠,𝑘which have not been yet verified as multiset recovery singularities are emphasized inbold. They represent the current “suspect” cases.

In addition the experiment showed that for all 3 < 𝑠 ⩽ 200the value of𝑘_max(𝑠) was equal to

2𝑠− 1. Claiming that to be always true is what we will call𝑘_max-Conjecture– seeConjecture 7.6 below, inSection 7.

The following proposition can be considered as a very easy “half” of this conjecture.

Proposition 6.1. For any𝑠 >2,𝑛= 2𝑠and any odd𝑘such that1< 𝑘 < 𝑛we have 𝐹_𝑠,𝑘(𝑛) = 0. Proof. We can rewrite this statement by using_⟨3_⟩, adding summand with𝑝= 0, and reversing the summation index𝑝. As a result we obtain

𝑠 ∑

𝑝=0

(−1)𝑝_{(𝑠−𝑝)}𝑟 (

𝑛 𝑝

)

= 0

for any even number0< 𝑟 < 𝑛. Now, since

( 𝑛 𝑝

)

=

( 𝑛 𝑛−𝑝

)

, (𝑠−𝑝)𝑟 = (𝑠− (𝑛−𝑝))𝑟,

the equation above is equivalent to

𝑛 ∑

𝑝=0

(−1)𝑝_{(𝑠−𝑝)}𝑟 (

𝑛 𝑝

)

= 0.

Any polynomial of 𝑝 with degree less than 𝑛 (such as (𝑠 − 𝑝)𝑟_{) can be expressed as a linear}

proposition then follows from well-known formula

𝑛 ∑

𝑝=0

(−1)𝑝_𝑝[𝑟]

( 𝑛 𝑝

)

= 0,

which can be easily proved using generating function𝜆(𝑥) = (1 +𝑥)𝑛 ₌ ∑𝑛 𝑝=0

(𝑛 𝑝 )

𝑥𝑝_{. Polynomial} 𝜆(𝑥)has(−1)as a root of order𝑛, thus its𝑟-th derivative𝜆(𝑟)_{(𝑥) =}∑𝑛

𝑝=0

(𝑛 𝑝 )

𝑝[𝑟]_𝑥𝑝−𝑟 _{also must have}

(−1)as a root for any0< 𝑟 < 𝑛.

■

And then from𝐹_{𝑠,2𝑠−1}(2𝑠) = 0follows

Corollary 6.2. For any𝑠 >2we have𝑘_max(𝑠)⩾2𝑠− 1.

Singularity search. Well, since we already started using computer assistance, let us continue down this slippery slope. The next idea in automating our investigation is to hunt not for the roots of polynomials𝐹_𝑠,𝑘but for the singular multisets themselves.

The objective is to try and find singularity examples for the smallest “suspect” pairs(27,6)and

(32,10). The other suspects, not 𝑛-conjugated to these two, are too large to hope for any “brute force” computer search to succeed.

The main idea of this approach is to restrict the realm of the𝑛-multisets that we deal with. Con-sider all (𝑛+𝑚−1

𝑚−1

)

weak compositionsof𝑛 into𝑚parts, that is, representations of 𝑛as a sum of𝑚

non-negative integers𝑘_𝑖,(𝑖= 1,2, ..., 𝑚):

 ∶ 𝑛=𝑘₁+𝑘₂+...+𝑘_𝑚, 𝑘_𝑖∈ℤ_⩾0.

Each weak composition of this form can be treated as sequence of multiplicities – that is, from each composition we will construct𝑛-multiset

𝐴_ = {1𝑘1_,2𝑘2_{, ..., 𝑚}𝑘𝑚}_. ⟨₁₁⟩

Alas, in both cases(27,6)and(32,10)we cannot hope to find𝑛-multiset𝑋which is𝑠-equivalent to its own mirror𝑋̃. Indeed, if𝑋 ∼𝑠 𝑋̃ then without loss of generality we can assume that𝜎₁(𝑋) = 0. Thus𝑋̃ = −𝑋 and for any even𝑘 >0the sum of𝑘-th powers of numbers in𝑋will be equal to that of𝑋̃.

From the experiment we already know (and can easily verify this formally) that 𝐹_6,𝑘(27) = 0

iff𝑘 = 4and𝐹_10,𝑘(32) = 0 iff𝑘 = 6. Therefore, for any𝑋 such that 𝑋(𝑠) _{= 𝑋̃}(𝑠) _{we will have}

𝜎_𝑘(𝑋) = 𝜎_𝑘(𝑋̃)for all values of1⩽ 𝑘 ⩽ 𝑛– the only exceptions we could have hoped for were 4 and 6 (for𝑛= 27and𝑛= 32respectively) and we have just eliminated them.

So, how can we proceed and what are the challenges?

First of all, we cannot afford to generate an array of all weak compositions (or multisets of type

⟨11_⟩) and then analyze the result – the computer would soon run out of memory. For example, if

𝑛= 27and𝑚= 10then we get almost 100 million of such compositions (94,143,280 to be exact). Thus, the algorithms here have to be iterative. It is fairly easy to write an iterator function which generates next weak composition based on the previous one. Hint: find the last non-zero part, increase the previous part by one and make all the following parts zero except for the last one. If necessary, same can be done when going through all𝑠-subsets in an𝑛-multiset.

Second, the check function that verifies whether the two given multisets𝐴and𝐵are𝑠-equivalent to its mirror 𝐴̃ must be written very carefully and very efficiently because it will be called quite a few times. To make it work as fast as possible the function needs to implement some “quick

rejection” checks. For instance, the sum of the first𝑠numbers from𝐴(let us assume it is sorted) is always equal to the minimum number in𝐴(𝑠)_{; hence these sums for𝐴and𝐵 must coincide. The} more simple checks of this sort are employed the better.

Third, calling this check function for every pair𝐴_ and𝐴_′of constructed multisets is absolutely out of the question. So, some sort of simplified “signature” has to be computed for each multiset

𝐴(_𝑠) (alas, no quick rejections there) so we can compare these numbers instead of comparing very large multisets𝐴(_𝑠). But even with that we cannot go much farther beyond𝑚= 10for the reason I already mentioned above – such huge arrays of data will exhaust the computer memory.

My own implementation of this approach did not find any examples of 6-equivalent 27-multisets of type_⟨11_⟩ for𝑚 < 10. As a sanity check I ran the same code for pair(12,4)with𝑚 = 17and after a few hours of number crunching it resulted in the same unique example of two 4-equivalent 12-multisets already found in [17].

Clearly, absence of positive results in this computational experiment doesn’t mean much as there could be 6-equivalent 27-multisets that span longer stretches of integers. And it is always possible that singularity example for(27,6)simply doesn’t exist. For now, this remains an open question.

If some of the readers become interested in this line of investigation I will gladly send them my code – I am sure it can be made more effective while consuming less memory. And then, who knows, perhaps the next value of𝑚will finally yield the desired singularity.

7. OPEN QUESTIONS

Here is a list of a few open questions which have come up during this survey’s investigations. Question 7.1. Is it true that₅ = {10}?

Question 7.2. Which of the new “suspect” pairs(𝑛, 𝑘)found inSection 6are singular?

Of course, we are talking here about pairs highlighted in bold in Table 1. Perhaps some cleverly written computer program could answer this question at least for the smallest “suspect” pairs(27,6)

and(32,10).

Question 7.3. Does a non-trivial root of 𝐹_𝑠,𝑘 guarantee an existence of corresponding singular pair? That is, is it true that for any positive integers𝑠,𝑘,𝑛such that𝑛 > 𝑠,𝑛 ⩾𝑘and𝐹_𝑠,𝑘(𝑛) = 0 there exists a pair of different𝑛-multisets𝐴and𝐵such that𝐴(𝑠) _=𝐵(𝑠)_?

All the results accumulated over the last sixty years so far confirm this hypothesis; however, it looks like an extremely difficult nut to crack. The following question could perhaps serve as a small step in that direction.

Question 7.4. A singular pair𝑃 = (𝑛, 𝑠)is such that𝐹_𝑠,𝑘(𝑛) = 0for𝑘= 4or5. Is pair𝑃′ = (𝑚, 𝑠) obtained from𝑃 by⟨𝑠, 𝑘_⟩-conjugation also singular?

We know that 𝑛-conjugation between roots of 𝐹_𝑠,𝑘 has its direct analog in (𝑛, 𝑠) ↔ (𝑛, 𝑛− 𝑠)

duality between singular pairs. However, the similar question about𝑠-conjugation does not seem to be even remotely as simple.

Question 7.5. Is there a less “accidental” explanation for at least one singular pair with𝑛 ≠ 2𝑠,

𝑠 >2?

So far, all such examples were constructed in a ratherad hocmanner by grinding through solu-tions for simultaneous equasolu-tions𝜎_𝑘(𝐴(𝑠)_{) =𝜎}

𝑘(𝐵

overwhelmingly complex. Perhaps for at least some singular pairs there exists a less “accidental” construction, combinatorial or algebraic.

I think that the following hypothesis is the most important among reasonably hard open questions on multiset recovery.

Conjecture 7.6. (𝑘_max-Conjecture)Is it true that𝑘_max(𝑠) = 2𝑠− 1for all𝑠 > 3? In other words, prove or disprove that polynomial𝐹_𝑠,𝑘can have integer roots only if𝑘⩽ 2𝑠− 1.

Positive answer to this question would mean a considerable breakthrough in any – computer-aided or not – search for the roots of polynomials𝐹_𝑠,𝑘.

For starters, we could immediately claim that_𝑠 = {2𝑠}for many small values of𝑠. Proof of that for𝑠= 5(that is, a solution toQuestion 7.1) would go like this:

Proof. If𝑛 >5and𝑛≠10then𝐹_5,𝑘(𝑛)≠0for any𝑘 >0. Indeed,𝑘_max-Conjectureimplies there is no need to check values𝑘⩾10. FromSection 5we know the same is true for𝑘⩽ 5. So we only need to examine𝐹_5,6,𝐹_5,7,𝐹_5,8and𝐹_5,9:

𝐹_5,6(𝑛) =𝑛4− 134𝑛3+ 3311𝑛2− 27754𝑛+ 75000, 𝐹_5,7(𝑛) =𝑛4− 262𝑛3+ 9527𝑛2− 107570𝑛+ 375000, 𝐹_5,8(𝑛) =𝑛4− 518𝑛3+ 27791𝑛2− 420490𝑛+ 1875000, 𝐹_5,9(𝑛) =𝑛4− 1030𝑛3+ 81815𝑛2− 1653650𝑛+ 9375000.

How do we do that? Again, we can use computer to help us produce a verifiable computer-independent proof. As an example, let us prove (quite formally) that 𝐹_5,6 has no integer roots in[5; ∞[⧵{10}. A couple of lines of code in MATLAB™ or in SAGE will get us real roots of this polynomial

𝑥₁= 6.014875..., 𝑥₂ = 7.745287..., 𝑥₃= 15.348149..., 𝑥₄ = 104.891687... ,

We cannot use that as a proof, but we can computeby handvalues𝐹_5,6(𝑥)for 𝑥= 6, 7, 8, 15, 16, 104 and 105. The results – 24, (-600), 360, 2040, (-4776), (-745560) and 93480 – prove that there is a non-integer root inside each one of segments[6; 7],[7; 8],[15; 16]and[104; 105]. Those four non-integer numbers obviously constitute the set of all roots of𝐹_5,6.

In exactly the same way (but with longer computations) one can prove the same for polynomials

𝐹_5,7,𝐹_5,8and𝐹_5,9(polynomials𝐹_5,7and𝐹_5,9both have one integer root but it is equal to 10). Finally, since𝐹_5,𝑘(𝑛)≠0, then fromTheorem 4.1it follows that𝑛-multiset𝐴is always

recover-able from𝐴(5). ■

The following hypothesis proposes an update toQuestion 2.5.

Conjecture 7.7. (Magical Triplet Conjecture) If 𝑠 > 2 and 𝑛 > 2𝑠 then for any three distinct

𝑛-multisets some two of them are not𝑠-equivalent to each other.

If we agree to exclude fully investigated cases of𝑠= 2and𝑛= 2𝑠then let us call three different

𝑛-multisets such that they are all𝑠-equivalent to each other, amagical triplet. I submit thatmagical triplets do not exist– in other words, when one tries to recover a multiset from its collection of

REFERENCES

[1] J. Lambek, L. Moser, (1954),Inverse and Complementary Sequences of Natural Numbers, Amer. Math. Monthly.,

33, pp.454–458

[2] L. Moser, (1957)Problem E1248, Amer. Math. Monthly,64, p.507

[3] J.L. Selfridge, E.G. Straus, (1958),On the determination of numbers by their sums of a fixed order, Pacific J. Math.,

8, pp.847–856

[4] J. Lambek, L. Moser, (1959),On some two way classifications of integers, Canad.J. Math.,2, pp.85–89

[5] A.S. Fraenkel, B. Gordon and E.G. Straus, (1962),On the determination of sets by the sets of sums of a certain

order, Pacific J. Math.,12, pp.187–196

[6] J.A. Ewell, (1968),On the determination of sets by sets of sums of fixed order, Canad. J. Math.,20, pp.596–611 [7] R.K. Guy, (1981),Unsolved problems in Number Theory, Springer-Verlag New York Inc.

[8] J. Boman, E. Bolker, P. O’Neil, (1991),The Combinatorial Radon Transform modulo the Symmetric Group, Ad-vances in Applied Mathematics,12, pp.400-411

[9] D.V. Fomin, O.T. Izhboldin (1994), "Nabory Kratnyh Summ" (in Russian), Proceedings of the St. Petersburg

Mathematical Society, Volume III, St.Petersburg University Publishing, St.Petersburg, Russia

[10] D.V. Fomin, O.T. Izhboldin (1995), "Collections of Multiple Sums",Proceedings of the St. Petersburg

Mathe-matical Society Volume III, AMS Translations, Series 2, vol.166, American Mathematical Society, Providence,

R.I.

[11] J. Boman, S. Linusson, (1996),Examples of Non-Uniqueness for the Combinatorial Radon Transform modulo

the Symmetric Group, Mathematica Scandinavica,78, pp.207-212

[12] R. Honsberger (1997),In Pólya’s Footsteps. Miscellaneous Problems and Essays, Dolciani Mathematical Expo-sitions,19, The Mathematical Association of America

[13] G. Galperin, A. Tolpygo, D. Leites (transl.) (1997), 60-odd Years of Moscow Mathematical Olympiads, MMO 1935-1997

[14] S. Savchev, T.Andreescu (2003),Mathematical Miniatures, New Mathematical Library,43, The Mathematical Association of America

[15] E.R. Chen, J.C. Lagarias (2008),Problem E11389, Amer. Math. Monthly,115(8), p.758 [16] A. Bogomolny (2011),A Property of the Powers of 2, Cut-The-Knot

[17] J.E. Isomurodov, K.P. Kokhas (2017),A set of 12 numbers is not determined by its set of 4-sums, Journal of Math. Sciences,224, No.2, pp. 258-262

BOSTON, USA