Proof of a Conjecture of Nath and Sellers on Simultaneous Core Partitions

In 2016, Nath and Sellers proposed a conjecture regarding the precise largest size of ${(s,ms-1,ms+1)}$-core partitions. In this paper, we prove their conjecture. One of the key techniques in our proof is to introduce and study the properties of generalized-$\beta$-sets, which extend the concept of $\beta$-sets for core partitions. Our results can be interpreted as a generalization of the well-known result of Yang, Zhong, and Zhou concerning the largest size of $(s,s+1,s+2)$-core partitions.


Introduction
Recall that an integer partition, or simply a partition, is a finite non-increasing sequence of positive integers λ = (λ 1 , λ 2 , . . ., λ ℓ ) with λ 1 λ 2 . . .λ ℓ > 0 (see [22,30]).Here ℓ is called the length, λ i (1 i ℓ) are the parts and |λ| := 1 i ℓ λ i is the size of λ.Each partition λ can be visualized by its Young diagram, which is an array of boxes arranged in left-justified rows with λ i boxes in the i-th row.By flipping the Young diagram of the partition along its main diagonal, we obtain another partition corresponding to the new Young diagram.Such partitions are said to be conjugate to each other.A partition is called self-conjugate if it is equal to its conjugate partition.For each box = (i, j) in the i-th row and the j-th column of the Young diagram of λ, its hook length h = h ij is defined to be the number of boxes exactly below, exactly to the right, or the box itself.Let s > 0 be a positive integer.A partition λ is called an s-core partition if its hook length set doesn't contain any multiple of s.Furthermore, λ is called an (s 1 , s 2 , . . ., s m )-core partition if it is simultaneously an s 1 -core, an s 2 -core, . .., and an s m -core partition (see [1,19]).For instance, Figure 1 gives the Young diagram and hook lengths of the partition (6, 3, 2, 1) and Figure 2 gives the Young diagram and hook lengths of its conjugation.The partition (6, 3, 2, 1) is a (4,6,11)-core partition since its hook length set doesn't contain multiples of 4, 6 or 11.One can note the first column hook lengths are 9, 5, 3, 1 and they uniquely determine the partition.Core partitions arise naturally in the study of modular representation theory and combinatorics.For example, core partitions label the blocks of irreducible characters of symmetric groups (see [27]).Therefore, simultaneous core partitions play important roles in the study of relations between different blocks in the modular group representation theory.Simultaneous core partitions are connected with rational combinatorics (see [16]).Also, simultaneous core partitions are connected with Motzkin paths and Dyck paths (see [9,11,39,40]).Some statistics of simultaneous core partitions, such as numbers of partitions, numbers of corners, largest sizes and average sizes, have attracted much attention in the past twenty years (see [2,3,4,7,8,10,13,14,15,18,20,21,23,24,25,29,31,33,35,37,42,43,44]).For example, Anderson [3] showed that the number of (s 1 , s 2 )-core partitions is equal to 1 s 1 +s 2 s 1 +s 2 s 1 when s 1 and s 2 are coprime to each other.Armstrong [4] conjectured that the average size of (s 1 , s 2 )-core partitions equals (s 1 − 1)(s 2 − 1)(s 1 + s 2 + 1)/24 when s 1 and s 2 are coprime to each other, which was first proved by Johnson [18] and later by Wang [33].However, there are still a lot of unsolved problems in this field.
In this paper, we focus on the largest size of simultaneous core partitions.For (s 1 , s 2 )core partitions, Olsson and Stanton [27] showed that the largest size of such partitions is (s 1 2 −1)(s 2 2 −1) 24 when s 1 and s 2 are coprime to each other.Straub [32] studied the largest size of (s, s + 2)-core partitions with distinct parts, and conjectured that the largest size should be equal to 1  384 (s 2 − 1)(s + 3)(5s + 17) in 2016.Straub's conjecture was first proved by Yan, Qin, Jin and Zhou [38].Later, the second author [36] obtained the largest sizes of (t, mt + 1) and (t, mt − 1)-core partitions with distinct parts.Recently, Nam and Yu [23] derived formulas for the largest sizes of (s, s + 1)-core partitions whose all parts are odd or all parts are even.
In this paper, we prove the following result, which verifies Nath and Sellers' conjecture [26] on the largest size of (s, ms − 1, ms + 1)-core partitions.
Theorem 1 (see Conjecture 57 of [26]).The largest size of an (s, ms − 1, ms + 1)-core partition is There are two such maximal partitions; one's β-set is L m (s) (please see Section 2 for the definition of β-sets and Section 3 for the definition of L m (s)), and the other one is the conjugate of the first one.
Remark 2. Theorem 1 can be seen as a generalization of the well-known result of Yang, Zhong and Zhou [41] on the largest size of (s, s + 1, s + 2)-core partitions.Thus in the following discussion, we could assume that s, m 2.
Remark 3. Nath and Sellers [26] computed the size of the longest (s, ms − 1, ms + 1)-core partition, i.e. the partition with the largest β-set.Our result shows that the longest of (s, ms − 1, ms + 1)-core partitions is also the largest one.
Next, we provide two examples for Theorem 1.

Example 4.
(1) When s = 5 and m = 3, Figures 3 and 4 show the two β-sets of maximal (s, ms − 1, ms + 1)-core partitions (here the β-set is just the collection of first column hook lengths, visually represented as beads on the s-abacus, see Definition 5 for the formal definition of the β-set).(2) When s = 6 and m = 3, Figures 5 and 6 show the two β-sets of maximal (s, ms − 1, ms + 1)-core partitions.Figure 5 corresponds to the partition (22,17,12,12,9,9,9,6,6,6,3,3,3,3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1). Figure 6 corresponds to the partition (24,19,14,10,10,10,7,7,7,4,4,4, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1).Both of them have the size 135 and are conjugate to each other.The rest of the paper is structured as follows.In Section 2, we recall the definitions of β-sets and s-abacus diagrams, which serve as basic tools to study simultaneous core partitions.Following [26], we present the structure of ms-abacus diagrams of β-sets of (s, ms − 1, ms + 1)-core partitions in Section 3. Next, we define and study the properties of generalized-β-sets in Section 4. This will allow us to loosen the restriction of β-sets to make room for the adjustments we will need.In Section 5, we prove several lemmas that will be used repeatedly in the later sections.In the following sections, we will use adjustments step by step to find the necessary conditions of S for S being the generalizedβ-set that maximizes f (S) (the definition of f (S) is given in Definition 8).In Section 6, we determine the possible shape of each row of S. In Section 7, we determine the possible shape of the entire S. Finally, we prove the main theorem in Section 8.

β-sets and s-abacus diagrams
In this section, we recall the definitions of β-sets and s-abacus diagrams for partitions.
Definition 5 ([6, 27, 28]).The β-set β(λ) of a partition λ which contains hook lengths of boxes in the first column of the corresponding Young diagram of λ.Remark 6.It is easy to see that the β-set completely determines the partition.
We introduce the notion of the size-counting function.It can be used to compute the size of a partition with the knowledge of its β-set.
Definition 8 (size-counting function, see [17]).Let S be a finite set of integers.We define where |S| is the cardinality of the set S.
From now on we will focus on core partitions.The β-set of an s-core partition has the following property.
Lemma 10 (see [28,34]).Let s be a positive integer and S be the β-set of a partition λ.Then λ is an s-core partition, if and only if for any x ∈ S and x s, we have x − s ∈ S.
Next, we recall the definition of s-abacus.Definition 11 (s-abacus (see [17,26])).Let X ⊆ N be a set of positive integers and s be a positive integer.Then the s-abacus diagram S of X is defined to be the set Remark 12. Usually, we use a diagram like Figure 7 to represent an s-abacus.When (i, j) ∈ S, the number in the i-th row and j-th column is circled.We begin with Row 0 at the bottom and Column 0 at the far left.
Remark 14.The s-abacus is just a graphic representation of the β-set, thus we will use the terms β-set and s-abacus interchangeably.When we deal with the ms-abacus, we will write is + j instead of (i, j) for i 0 and 0 j s − 1.When we mention a certain row of a given set, we mean the corresponding row in the ms-abacus of the set.
Lemma 18.For any positive integer s and m, the partition with the beta set L m (s) is not self-conjugate.
the electronic journal of combinatorics 30 (2023), #P00 Proof.We prove it by contradiction.Otherwise, the partition λ On the other hand, max L m (s) equals

Generalized-β-sets
Now we generalize the concept of the β-set of an (s, ms − 1, ms + 1)-core partition to the generalized-β-set.In the following sections, we will use this more general concept instead of the original β-sets studied in previous papers (see [17,26,28,34]).To better formulate the concept of a generalized-β-set, we introduce the following notation.For For 1 i ⌈(s − 1)/2⌉m, define For 1 k ⌈(s − 1)/2⌉, define When there is only one set S, we write t, B i , a i , n i instead of t(S), B i (S), a i (S), n i (S) for simplicity.When we deal with two sets S, S ′ , the same is for S and we use t Example 19.Let m = 3, s = 5 and S = L 3 (5).Then t = 2.We have (see Figure 3) B 6 = {27}, a 6 = 1, n 6 = 2; Next, we give the definition of generalized-β-sets.
The following lemma is the reason why we use the term generalized-β-set.
Remark 23.A generalized-β-set is not necessarily the β-set of an (s, ms − 1, ms + 1)-core partition.Let s = 6 and m = 3. See Figure 8   The following proposition gives an upper bound of n i .
Proposition 24.Let S ⊆ L m (s) be a generalized-β-set.For 0 i t−1 and 1 k m, we have Proof.Since S ⊆ L m (s), we have Then by Condition (5) in Definition 20, we obtain the result.
The next proposition derives several inequalities for A i .
the electronic journal of combinatorics 30 (2023), #P00 (2 Proof.By the definition of t, we have a (t−1)m+1 > 0. By Condition (1) in Definition 20, we have a i > 0 for 1 i (t − 1)m + 1.Then Conclusion (1) can be directly derived by Condition (3) in Definition 20.Next, we prove Conclusion (2), assuming that (2) Let m = 3, s = 6, and S = L 3 (6).Then we have Let A m (s) be the set of all β-sets of (s, ms − 1, ms + 1)-core partitions and B m (s) be the set of all generalized-β-sets S ⊆ L m (s).By Lemma 22, we know that max We aim to prove Nath-Sellers' conjecture by showing that max and arg max f (S ′′ ), which will be proved in the following sections while we are searching for all generalized-βsets S that maximize f (S).Here and in the rest of the paper, for a function g : D → R, We use generalized-β-sets instead of β-sets since the original definition of the β-set is too tight for the adjustments in Sections 5, 6 and 7. We can easily make a generalized-βset remain a generalized-β-set after each adjustment in our following proofs, while a β-set may not remain a β-set after the adjustments.

Preliminaries for the proofs
In the next three sections, we will introduce several adjustments for a generalized-β-set S which doesn't maximize f (S).After the adjustments, we will get a new set S ′ , such that f (S) < f (S ′ ).We wish to prove that S ′ is a generalized-β-set.The following lemma will be repeatedly used to verify Condition (3) in Definition 20 for S ′ .
the electronic journal of combinatorics 30 (2023), #P00 Proof.We prove it by contradiction.Assume that there exists 1 i n, such that The following lemma shifts our focus from all generalized-β-sets to the ones with some specific structure.
Lemma 28.Let S be a generalized-β-set that maximizes f (S).Then consists of consecutive integers for 1 i tm.Furthermore, for 0 i t − 1 and 1 k m, we have Proof.First, we construct a generalized-β-set | consecutive numbers, and n ′ i = max {x mod s : x ∈ B i (S ′ )} satisfies the equality in Proposition 24.Therefore, Conditions (1)(2)(3)(6) in Definition 20 hold for S ′ since a ′ i = a i for 1 i tm.Conditions (4) (5) in Definition 20 are also true since the equality in Proposition 24 is achieved.Thus S ′ is also a generalized-β-set and If S = S ′ , the equality above cannot be achieved, which means that f (S) < f (S ′ ).However, at the beginning we assume that S is a generalized-β-set that maximizes f (S), which means that f (S) f (S ′ ), a contradiction.Therefore, S = S ′ and thus for 0 i t − 1 and 1 k m, B im+k consists of consecutive integers and the electronic journal of combinatorics 30 (2023), #P00 Definition 29.Recall that A m (s) is the set of all β-sets of (s, ms − 1, ms + 1)-core partitions and B m (s) is the set of all generalized-β-sets S ⊆ L m (s).Let C m (s) be the set of all generalized-β-sets S ⊆ L m (s) such that S satisfies Conditions (1) and ( 2) in Lemma 28.Let D m (s) be the set of all nonempty sets S ⊆ [⌈(s − 1)/2⌉ms − 1] such that the nonempty set B i consists of a i consecutive integers for 1 i ⌈(s − 1)/2⌉m.
The following lemma is obvious and the proof is omitted.
Lemma 30.For the sets C m (s) and D m (s) defined in Definition 29, we have We use Figure 9 to visualize that Remark 31.By Lemma 28 it is easy to see that arg max The following lemma gives a characterization of D m (s).
Then for 1 i N, by definition we have 0 s and a i (S) − 1 n i (S).Therefore, f (S) ∈ V and F are well-defined.
We first prove that Next, we prove that F is surjective.For any and F (S) = Z.From the discussion above, we obtain that F is bijective.
Next, we show how to calculate f (S) for S ∈ D m (s).
/2, we only need to prove the first equation.In fact, since Lemma 33 implies the following result.
Corollary 34.Let S, S ′ ⊆ D m (s) and t = t ′ .Assume that there exist positive integers i < j, such that the following three conditions hold: (1) (2) (n ′ j , a ′ j ) = (n j + 1, a j + 1) or (n i , a i + 1); (3) For all 1 k tm, k = i, j, we have For other cases of (n ′ i , a ′ i ) and (n ′ j , a ′ j ), the proof is similar.

Adjustments in one row
The aim of this section is to show that if S maximizes f (S), then a (i−1)m+1 − a im 2 for 1 i t, just as Example 1.2 shows.The following lemma gives some properties of a generalized-β-set S that maximizes f (S).
Lemma 35.Assume that a generalized-β-set S maximizes f (S).Then we have the following conclusions: Remark 36.Note that the inequality a i+m a i − 2 may be violated when a i+m = 0.This needs to be discussed in the following proof.
Proof.By Lemma 28, S ∈ C m (s).We prove Conclusion (2) here.The proof of Conclusion ( 1) is similar and omitted.Assume that Conclusion (2) is not true.Then either there exists some 0 i t − 2, such that a im+p+1 − a (i+1)m 2; or there exists some 0 j t − 2, such that a jm+1 − a jm+p 2. These two cases are similar.Without loss of generality, we consider the first case, i.e., there exists some 0 i t − 2, such that for some positive integers u and v. Then u + v < m − p.We will do adjustments to S ∩ [(im + p)s, (i + 1)ms − 1], which can be divided into the following three cases.
(1) v = 0 and a (i+1)m−1 a (i+1)m + 2. Let S ′ ∈ D m (s) satisfy the following conditions: the electronic journal of combinatorics 30 (2023), #P00 It's easy to check that S ′ is well-defined and |S| = |S ′ |.Then by Corollary 34, we have f (S ′ ) > f (S). ( It's easy to check that S ′ is well-defined and Thus f (S ′ ) > f (S).Notice that here S ′ may not be a generalized-β-set.
The adjustments when there exists some 0 j t − 2, such that a jm+1 − a jm+p 2 are similar.The process above can be repeated in S ′ if there exists some 0 2 or there exists some 0 thus the adjustments will stop after finite steps, when we get a final set S ′′ , such that a ′′ (i−1)m+1 − a ′′ and for 1 i t.The analog notations U ′′ i and V ′′ i are for S ′′ .By the discussion above, for 1 i t, we have Since S is a generalized-β-set, we obtain Thus S ′′ is also a generalized-β-set and f (S) < f (S ′′ ), which contradict the assumption that S is a generalized-β-set that maximizes f (S).
With this lemma, we can control the shape of a certain row of a generalized-β-set S that maximizes f (S).Example 39.Let m = 3 and s = 6. Figure 10 shows the generalized-β-set S before the adjustment.Figure 11 shows the generalized-β-set S ′ after the adjustment.

Adjustments in different rows
In this section, we aim to prove that if a generalized-β-set S ⊆ L m (s) maximizes f (S), then for nonzero a i and a i+m , we have a i = a i+m + 2, thus the equality of Condition (3) in Definition 20 holds.Lemma 40.Let S be a generalized-β-set that maximizes f (S).Then we have the following conclusions: (1 for 1 i (t − 2)m + p and a (t−2)m+p = 3 > a (t−2)m+p+1 .
Proof.By Lemma 35, we know that S ∈ E m (s).
Step 1. First, we prove the following weaker version of Lemma 40: (1)' If A t−1 − A t 2m, then 2m A i − A i+1 2m + 1 for 1 i t − 1.The equation A i − A i+1 = 2m + 1 only holds by at most one i. ( The equation A i − A i+1 = 2m + 1 only holds by at most one i. First, we deal with the case A t−1 − A t 2m.We prove Conclusion (1)' by contradiction.Assume that Conclusion (1)' is not true.There are two possible cases.
Adjustments in the i-th row involve Cases (a) and (b).Adjustments in the i + 1-th row involve Cases (c), (d) and (e).Obviously S ′ ∈ E m (s).If (c) doesn't hold, then f (S) < f (S ′ ).If (c) holds, there are two possibilities: (b) and (c) hold simultaneously or (a) and (c) hold simultaneously.Recall that D(i) denote Assume that (b) and (c) hold simultaneously.Notice that a (i+1)m s − 2 − i, thus Assume that (a) and (c) hold simultaneously.Similarly, the electronic journal of combinatorics 30 (2023), #P00 Therefore, f (S ′ ) > f (S).A contradiction.
(2) There exists i < j, such that A i − A i+1 2m + 1 and A j − A j+1 2m + 1.Then we construct S ′ ∈ D m (s) similar to Case (1).We first do the adjustments to the i-th row of S similar to the adjustments to the i-th row in Case (1), and then do the adjustments to the (j + 1)-th row of S similar to the adjustments to the i + 1-th row in Case (1).Then we obtain S ′ after the adjustment.It is obvious that S ′ ∈ E m (s).Similar to the discussion in Case (1), f (S ′ ) > f (S).A contradiction.
Next, we deal with the case A t−1 − A t < 2m.In this case, by Proposition 25, there exists 1 p < m, such that a (t−1)m = a (t−1)m+1 = • • • = a (t−1)m+p = 1 and a (t−1)m+p+1 = • • • = a tm = 0. We will prove Conclusion (2)' by contradiction.Assume that Conclusion (2)' isn't true.Then either there exists 1 i t − 2, such that A i − A i+1 2m + 2; or there exist 1 i < j t − 2, such that A i − A i+1 = 2m + 1 and A j − A j+1 = 2m + 1.We will deduce a contradiction for each case. Define for 1 i t.Then their cardinal numbers are Firstly, we deal with the case A i − A i+1 2m + 2. We have the following three cases to consider. (1) Consider S ′ ∈ D m (s) satisfying the following conditions: the electronic journal of combinatorics 30 (2023), #P00 im+u , a ′ im+u = (n im+u , a im+u + 1).Cases (b) and (c) are adjustments in R i .Cases (d),(e),(f) and (g) are adjustments in L i+1 .Then 3) in Definition 20 won't be violated after the adjustments.We can verify that S ′ ∈ E m (s).Similar to the discussion for the case A t−1 − A t 2m, we obtain |S| = |S ′ | and f (S) < f (S ′ ).Thus we reach a contradiction. ( Consider S ′ ∈ D m (s) satisfying the following conditions: the electronic journal of combinatorics 30 (2023), #P00 Case (b) and (c) are adjustments in R i .Case (d), (e) and (f) are adjustments in R i+1 .We claim that Case (d) is valid.Otherwise, S ′ violates Condition (2) in Definition 20, then Notice that L i − L i+1 = 2p and R i − R i+1 2(m − p) + 2, we have a (i−1)m+p < a (i−1)m+p+1 .A contradiction.Therefore, S ′ satisfies Condition (2) in Definition 20.We can easily prove that S ′ ∈ E m (s).When (f) doesn't happen, obviously f (S) < f (S ′ ).When (f) happens, similar to the discussions in Case (1) when

Therefore, we obtain 2m
A i − A i+1 2m + 1 for 1 i t − 2. Now we deal with the case that there exists 1 i < j t − 2, such that A i − A i+1 = 2m + 1 and A j − A j+1 = 2m + 1.There are four cases to be considered.
(1) R i − R i+1 = 2(m − p) + 1 and L j − L j+1 = 2p + 1.This case is similar to Case (1) when A i − A i+1 2m + 2. We can still reach a contradiction. ( We can still reach a contradiction. ( This case is similar to Case (3) when A i − A i+1 2m + 2. We can still reach a contradiction. ( Assume that a (i−1)m+1 = a (i−1)m+p+1 .Since Consider S ′ ∈ D m (s) satisfying the following conditions: Cases ) are adjustments in L i .Cases (f), (g), (h) are adjustments in R j+1 .By the discussion before, Cases (b) and (f) are valid.Similarly we can prove that S ′ ∈ E m (s) and f (S) < f (S ′ ).A contradiction.
Step 2. Now we provide the proofs of Conclusions (1)(2).First, we prove Conclusion (1) by contradiction.Assume that a i = a i+m + 2 for 1 i (t − 1)m doesn't hold.Then by Conclusion (1)', there exists a unique 1 l t − 1, such that A l − A l+1 = 2m + 1.Thus there exists a unique u, such that a u − a u+m = 3 and a i − a i+m = 2 for i = u.Assume that q m is the largest positive integer such that a 1 = • • • = a q .By Condition (2) in Definition 20 and Conclusion (2) in Lemma 35, we obtain u ≡ q (mod m).
Therefore, u = (l − 1)m + q.By Lemma 35, a (i−1)m+1 − a im 1 for 1 i t, so we have two cases: Firstly, consider the case a 1 = a m + 1.For 0 w t, let S w ∈ D m (s) satisfy the following conditions.Here we use (n w i , a w i ) to denote (n i (S w ) , a i (S w )). ( the electronic journal of combinatorics 30 (2023), #P00 Then S = S l .It's easy to check that S 0 , Secondly, consider the case a 1 = a m .For 0 w t, let S w ∈ D m (s) satisfy the following conditions.Here we still use (n w i , a w i ) to denote (n i (S w ) , a i (S w )). ( (2) Assume that 0 k l − 1.For km + 1 i lm, (n w i , a w i ) = (n i , a i − 1) if i ≡ q mod m and (n w i , a w i ) = (n i + 1, a i ) if i ≡ q mod m.Otherwise, (n w i , a w i ) = (n i , a i ).
(3) Assume that l k t.For lm Then S = S l .It's easy to check that S 0 , This also contradicts the assumption of S. Now we turn to the case where A t−1 − A t < 2m.We prove Conclusion (2) by contradiction.First, assume that the claim a i = a i+m + 2 for 1 i (t − 2)m is not true.Similarly, there exist unique 1 l t − 1 and 1 k m, such that a (l−1)m+k − a lm+k = 3 and a i − a i+m = 2 for i = (l − 1)m + k.We can make adjustments similar to the former case to the first p columns of S and the last m − p columns of S separately.Still, we can reach a contradiction.Thus a i − a i+m = 2 for 1 i (t − 2)m.
By Lemma 40, we obtain the following corollary.
Next, we give an example illustrating the adjustments in the proof of Lemma 40. the electronic journal of combinatorics 30 (2023), #P00 Example 43.Let s = 6 and m = 3. Figure 12 shows the generalized-β-set S before the adjustments.Figure 13 shows the generalized-β-set S ′ after the first adjustment.Figure 14 shows the same generalized-β-set S ′ with Figure 13. Figure 15 shows the generalized-β-set S ′′ after the second adjustment.The differences in the colors of some circles between Figure 13 and 14 are to highlight the changes of different adjustments.Here t = 2, A 1 = 13, A 2 = 3.Thus A t−1 − A t 2m.

Adjustments in columns
In this section, first we prove two useful results Lemma 44 and Lemma 46.These two lemmas deal with the cases A t−1 − A t 2m and A t−1 − A t > 2m respectively.Lemma 44.Let S be a generalized-β-set that maximizes f (S).Assume that A t−1 − A t 2m, then the electronic journal of combinatorics 30 (2023), #P00  (1) (2) a (t−1)m+1 ∈ {1, 2}.
Proof.By Lemma 40, we obtain that S ∈ F m (s) and a i − a i+m = 2 for 1 i (t − 1)m.We will prove Conclusion (1) by contradiction.Assume that Conclusion (1) is not true.
Then there exists 1 k m − 2 and a 0, such that For 0 n m − 1, let S n ∈ C m (s) denote the t-row generalized-β-set such that a and a i − a i+m = 2 for all i.Then S = S k and we have Therefore, the electronic journal of combinatorics 30 (2023), #P00 Then R or T is a generalized-β-set while f (S) < f (R) and f (S) < f (T ).A contradiction.
Next, we prove p = m − 1 by contradiction.Otherwise, we assume that p < m − 1.Since a (t−1)m+1 = • • • = a (t−1)m+p = 1 for 1 p < m − 1, s 2t.Let S ′ ∈ E m (s) be a t-row set satisfying the following conditions: for 1 i tm, A contradiction.Thus Conclusion (2) is proved.Now we prove Conclusion (1).By Lemma 40, a (t−2)m+1 = 3.Let S ′ ∈ E m (s) be a t-row set satisfying the following conditions: for 1 i tm, The equality may only be achieved for s = 2t.Then s is even and t = s 2 .Thus Conclusion (1) is proved.
There are three kinds of adjustments in Lemma 46.We give an example to help visualize those adjustments.Figure 23 and Figure 24 display the third kind of adjustment.Notice that S 32 = S 31 ∪ {15}.We color S 32 \ S 31 red.Indeed t = 2 and (n 3 (S 32 ), a 3 (S 32 )) = (n 3 (S 31 ), a 3 (S 31 ) + 1).We have f (S 31 ) = 72 < 78 = f (S 32 ).Lemmas 44 and 46 discuss about the necessary condition for the generalized-β-set S that maximizes f (S).The following theorem gives a detailed description of S.
the electronic journal of combinatorics 30 (2023), #P00  .From the discussion above, we know that S s−1 2 and Qs−1 are the only two generalized-β-sets that maximize f (S).Next, we discuss about the case where s is even.We first consider the case where A t−1 − A t 2m.Then by Lemma 40, a i − a i+m = 2 for 1 i (t − 1)m.Just like the former case, for 1 k s 2 − 1, we can define S k , T k , Q k ; for 1 k s 2 , we can define P k .With similar discussions, we only need to compare f S s   Thus we have f P s 2 > f T s 2 −1 .Similarly, we can prove that f P s 2 > f Q s 2 −1 .Therefore, S = P s 2 is the only generalized-β-set that maximizes f (S) when A t−1 −A t 2m.Let P ′ conjecture proposed by Nath and Sellers in [26].Currently, most research results on simultaneous core partitions are about (s 1 , s 2 , . . ., s m )-core partitions with m 3. When m 4, it still lacks proper tools for studying such general simultaneous core partitions.We believe that the concepts and techniques related to generalized-β-sets introduced in this paper offer some insights for exploring statistics of general simultaneous core partitions, of which we know very little at this moment.

Figure 9 :
Figure 9: Relations between A m (s), B m (s), C m (s), and D m (s).

Definition 37 .
Let E m (s) be the set of all S ∈ C m (s) such that S satisfies Conclusions (1) and (2) in Lemma 35.By Lemma 35, we obtain the following corollary.Corollary 38.We have E m (s) ⊆ C m (s) ⊆ B m (s) and arg max S ′ ∈Bm(s) f (S ′ ) = arg max S ′′ ∈Cm(s) f (S ′′ ) = arg max S ′′′ ∈Em(s) f (S ′′′ ).Next, we give an example of the adjustments inLemma 35.

Figure 13 :
Figure 13: After the first adjustment.

Figure 15 :
Figure 15: After the second adjustment.

Figure 22 :
Figure 22: S 22 : after the second kind of adjustment.
Figure 23: S 31 : before the third kind of adjustment.