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Complete classification of 3-multisets up to combinatorial equivalence
Accepted Manuscript Complete classification of 3-multisets up to combinatorial equivalence
Davi Lopes Alves de Medeiros, Lev Birbrair
PII: DOI: Reference:
S0022-314X(16)30282-7 http://dx.doi.org/10.1016/j.jnt.2016.09.035 YJNTH 5603
To appear in:
Journal of Number Theory
Received date: Revised date: Accepted date:
10 May 2016 27 September 2016 28 September 2016
Please cite this article in press as: D.L.A. de Medeiros, L. Birbrair, Complete classification of 3-multisets up to combinatorial equivalence, J. Number Theory (2017), http://dx.doi.org/10.1016/j.jnt.2016.09.035
This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
COMPLETE CLASSIFICATION OF 3-MULTISETS UP TO COMBINATORIAL EQUIVALENCE DAVI LOPES ALVES DE MEDEIROS* AND LEV BIRBRAIR** Abstract. Let A = {a1 , . . . , ak } be a finite multiset of positive real numbers. Consider the sequence of all positive integers multiples of all ai ’s, and note the multiplicity of each term in this sequence. This sequence of multiplicities is called the resonance sequence generated by {a1 , . . . , ak }. Two multisets are called combinatorially equivalent if they generate the same resonance sequence. The paper gives a complete criterion of classification of multisets with 3 elements up to combinatorial equivalence, by showing this is equivalent to a system of equations depending directly of the numbers in the multisets.
1. Introduction Resonance sequences appear in Geometric Theory of Ordinary Differential Equations as an invariant of the linear equations of the second order under the focal equivalence. In general, the resonance sequence is defined for a finite multiset of arbitrary real numbers, but a consideration of rational multisets leads to a discovery of some interesting arithmetical and combinatorial properties of resonance sequences. The main result of the paper is the classification theorem of resonance sequences generated by three elements (Theorem 5.1). We define 4 functions of the elements of the multisets. The multisets are combinatorially equivalent if and only if the values of these functions are equal. The main ingredient of the proof is the Alvarez Algorithm, defined in [3] and investigated in [4]. In order to obtain the result, we define a sort of the inverse Alvarez algorithm. The results of the paper cannot be directly generalized for multisets generated by more the three elements. The general case needs a deeper investigation. We would like to thank Daniel Berend for useful remarks. We also would like to thank the anonymous referee for extremely useful comments. 2010 Mathematics Subject Classification. 11B30, 05A17 . Key words and phrases. Resonance sequences, combinatorial equivalence, focal classification. *Research supported by CNPq grant 131709/2016-0. **Research supported under CNPq grant and by Capes-Cofecub 302655/2014-0. 1
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D.L.A. MEDEIROS AND L. BIRBRAIR
2. Definitions, Basic Properties and Examples Definition 2.1. Consider a finite multiset of positive real numbers A = {a1 , . . . , an }. Let Ai be the set of all positive integer multiples of ai : Ai = {ai , 2ai , 3ai , . . . }, 1 ≤ i ≤ n Let Q = ∪ni=1 Ai . Q is infinite countable and may be sorted by increasing order: Q = {mj : j ∈ N}, (m1 < m2 < · · · < mr < . . . ) Let ind(mj ) = {1 ≤ i ≤ n : mj ∈ Ai } and set rj = |ind(mj )|. The numbers rj are integers with 1 ≤ rj ≤ n. The sequence Ress(A) = (rj )j∈N is the resonance sequence generated by {a1 , . . . , an }. The number n is called the dimension of the multiset A. Example 2.2. Consider the set A = {3, 5, 6}. The set of positive integer multiples of the elements in A is: Q = {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, . . . } and its resonance sequence is: Ress(A) = 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 3, . . . Example 2.3. Consider the set A = {2,
√
5, 3}. The set of positive integer multiples of
the elements in A is: Q = {2,
√ √ √ √ √ √ 5, 3, 4, 2 5, 6, 3 5, 8, 4 5, 9, 10, 5 5, 12, 6 5 . . . }
and its resonance sequence is: Ress(A) = 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1 . . . √ √ √ Example 2.4. Consider the set A = { 2, 3, 5}. It is easy to see that Ai ∩ Aj = ∅ for all i = j, and then its resonance sequence is: Ress(A) = 1, 1, 1, 1, 1, 1, . . . Those examples motivates us the folowing definition: Definition 2.5. A multiset A = {a1 , . . . , an } is totally rational if αi /αj ∈ Q, for all i, j; / Q, for all i = j; and semi-rational if it is neither totally totally irrational if αi /αj ∈ rational nor totally irrational. The main interest in the paper is the combinatoral equivalence, defined below: Definition 2.6. Given two multisets A and A , they are combinatorially equivalent if Ress(A) = Ress(A ), i.e. they generate the same resonance sequence.
COMPLETE CLASSIFICATION OF 3-MULTISETS UP TO COMBINATORIAL EQUIVALENCE
3
It is easy to see that, for c > 0, A = {a1 , . . . , an } and cA = {ca1 , . . . , can } are combinatorially equivalent. Also, it’s easy to see that A = {a1 , . . . , an } and A = {a1 , . . . , an }, where {a1 , . . . , an } is a permutation of {a1 , . . . , an }, are combinatorially equivalent. The combinatorial equivalence of one of these two forms is considered as a trivial combinatorial equivalence. Our main interest is to study non-trivial combinatorial equivalence of multisets. Consider now multisets with 3 elements. Clearly, for every totally rational multiset A = {a1 , a2 , a3 }, there is a unique number c > 0 such that ni = cai (i = 1, 2, 3) are natural numbers such that gcd(n1 , n2 , n3 ) = 1. Note that {a1 , a2 , a3 } and {n1 , n2 , n3 } are trivially combinatorially equivalent. It is clear that, for every semi-rational mutiset A = {a1 , a2 , a3 }, there is a permutation A = {a1 , a2 , a3 } such that a1 /a2 ∈ Q, and a unique number c > 0 such that n1 = ca1 , n2 = ca2 and λ = ca3 , where n1 , n2 are coprime positive integers, and λ is an irrational number. Note that {a1 , a2 , a3 } and {n1 , n2 , λ} are trivially combinatorially equivalent. These observations motivates the next definiton. Definition 2.7. Let A be a totally rational multiset with 3 elements, trivially combinatorially equivalent to A0 = {n1 , n2 , n3 }, where n1 , n2 , n3 are positive integers with gcd(n1 , n2 , n3 ) = 1. We define the multiset A0 as the canonical form of A. In a similar way, let A be a semi-rational multiset with 3 elements, trivially combinatorially equivalent to A0 = {n1 , n2 , λ}, where n1 , n2 are coprime positive integers and λ ∈ / Q. We define the multiset A0 as the canonical form of A. The next two theorems were proved in [3], and they are a classification of totally irrational multisets and semi-rational multisets up to combinatorial equivalence. Theorem 2.8. [3], pg.272. All totally irrational multisets are combinatorially equivalent, and a totally irrational multiset is combinatorially equivalent only to another totally irrational one. Theorem 2.9. [3], pg.273, 276. Let A be a semi-rational multiset with 3 elements, with canonical form {n1 , n2 , λ}. A multiset with 3 element A is combinatorially equivalent to A if, and only if, A is semi-rational, with canonical form {n1 , n2 , λ }, and: n1 n 2 n n + n1 + n2 = 1 2 + n1 + n2 λ λ In order to complete the cassification of multisets with 3 elements up to combinatorial equivalence, it is necessary to know what happens with totally rational multisets. Since the canonical forms of them are multisets with positive integers, the main object of study in this paper will be the multisets with positive integers. Consider now A = {a1 , . . . , ak }, where all the ai are positive integers, and let Q as in Definition 2.1. It is clear that, for x ∈ N, M = lcm(a1 , . . . , ak ) and t ∈ Z such that
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D.L.A. MEDEIROS AND L. BIRBRAIR
x + tM ∈ N, we have x ∈ Q implies x + tM ∈ Q, and ind(x) = ind(x + tM ). This proves that Ress(A) is a periodic sequence. Also, M is the first number y ∈ Q such that |ind(y)| = n. Thus, Ress(A) is completely determined by the numbers until the first n (after that, the sequence Ress(A) will start to repeat). This motivates a new definition. Definition 2.10. Let A be a multiset with n positive integers. The Fundamental Word of A is the subsequence of Ress(A), starting from the first term and ending at the first occurence of n. Example 2.11. If A = {1, 2, 3, 4}, the fundamental word of A is 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 4. It is clear that two multisets A and A of positive integers are combinatorially equivalent if, and only if, their fundamental words are equal. 3. Alvarez Algorithm and Block Structure Let A = {a1 , a2 , . . . , an } be a multiset with n > 2 positive integers. Consider its fundamental word. The steps below are the Alvarez Algorithm, first described in [3]: Step 1: If there is no (n − 1) in the fundamental word, mark the number n and let S1 be the sum of all the numbers in the fundamental word. If there is (n − 1), find the first occurence of (n − 1) in the fundamental word and mark it. Then, calculate the sum of the numbers of the fundamental word, from the first number to the marked number, and denote this sum by S1 . Starting from the number next to the marked (n−1), calculate the sum of the terms of the resonance sequence until it becomes equal to S1 or (S1 + 1), what happens first, and stop at that point, marking it. Starting from the number next to the last marked number, calculate the sum of the terms of the sequence until the sum becomes equal to S1 or (S1 +1), what happens first, stop at that point and mark it. Repeat the previous procedure until the end of the fundamental word, marking the number n at the end. Those marked numbers divides the fundamental word into blocks. This partition is called the Block Structure of Step 1. For 2 ≤ k ≤ n, let Step k be as follows: Step k: If there is no previously unmarked (n − 1) in the fundamental word, mark the number n and let Sk be the sum of all the numbers in the fundamental word. If there is a previously unmarked (n − 1), find the first occurence of a previously unmarked (n − 1) in the fundamental word and mark it. After this, calculate the sum of the numbers of the fundamental word, from the first number to this marked number, and denote this sum by Sk . Starting from the number next to the last marked (n − 1), calculate the sum of the terms of the resonance sequence until it becomes equal to Sk or (Sk + 1), what
COMPLETE CLASSIFICATION OF 3-MULTISETS UP TO COMBINATORIAL EQUIVALENCE
5
happens first, and stop at that point, marking it. Starting from the number next to the last marked number, calculate the sum of the terms of the sequence until the sum becomes equal to Sk or (Sk +1), what happens first, stop at that point and mark it. Repeat the previous procedure until the end of the fundamental word, marking the number n at the end. Those marked numbers divides the fundamental word into blocks. It is called the Block Structure of Step k. For 1 ≤ i ≤ n, define mi = lcm(a1 , . . . , ai−1 , ai+1 , . . . , an ), and let: (1)
Ki =
lcm(a1 , . . . , an ) lcm(a1 , . . . , an ) = lcm(a1 , . . . , ai−1 , ai+1 , . . . , an ) mi
Finally, consider the numbers Ki sorted in decreasing order. They give the sequence L1 ≥ · · · ≥ Ln . The proof of the next theorem can be found in [4]. Theorem 3.1. The following assertions about the block structures on a fundamental word are true: 1. All blocks end at (n − 1) or n, and when the block ends at n, it is the last one of the block structure. Also, every (n − 1) in the fundamental word is the end of exactly one block in all the block structures; 2. The number of blocks in the block structure of step k is equal to Lk ; Example 3.2. Let A = {6, 5, 4}. The fundamental word of A is: 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3 Applying the Alvarez Algorithm, we have the 3 block structures: (1, 1, 1, 1, 1, 2), (1, 1, 1, 2, 2), (1, 1, 2, 1, 1, 2), (2, 1, 1, 1, 2), (1, 1, 1, 1, 1, 3) (1, 1, 1, 1, 1, 2, 1, 1, 1, 2), (2, 1, 1, 2, 1, 1, 2, 2), (1, 1, 1, 2, 1, 1, 1, 1, 1, 3) (1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2), (1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3) Observe that L1 = 5, L2 = 3, L3 = 2 and K1 = 3 = L2 , K2 = 5 = L1 , K3 = 2 = L3 . Example 3.3. Let A = {1, 2, 6}. The fundamental word of A is: 1, 2, 1, 2, 1, 3 Applying the Alvarez Algorithm, we have the 3 block structures: (1, 2), (1, 2), (1, 3)
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(1, 2, 1, 2, 1, 3) (1, 2, 1, 2, 1, 3) Observe that L1 = 3, L2 = 1, L3 = 1 and K1 = 1 = L2 , K2 = 1 = L3 , K3 = 2 = L1 . 4. Two Preliminary Lemmas Lemma 4.1. The sum of the elements of the fundamental word of the resonance sequence generated by the positive integers {a1 , . . . , an } is equal to:
(2)
n 1 S = lcm(a1 , . . . , an ) a i=1 i
Proof. The sum of the elements in the resonance sequence is equal to the number of the multiples of a1 on the segment (0, lcm(a1 , . . . , an )], plus the number of the multiples of a2 on the same segment and so on. The number of multiples of ai on this interval is equal to lcm(a1 , . . . , an )/ai . This concludes the proof of (2).
We use the notation x for the maximum integer, not greater than x. Lemma 4.2. Let t be a positive integer such that t ≤ Ki . The sum of the elements of the first t blocks of the block structure corresponding to Ki is equal to: (3)
tS Ki
Proof. The sum of the elements in the resonance sequence belonging to the first t blocks is equal to the sum of the numbers of occurences of the multiples of each element from the set {a1 , . . . , an } on the interval ]0, tmi ]. The number of occurences of the multiples of an element x on the segment ]0, y] equals to y/x . Thus, the sum of the elements of the first t blocks of the block structure corresponding to Ki is equal to:
(4)
n tmi j=1
aj
It remains to prove that the expressions (3) and (4) are equal. The equation (1) gives: mi lcm(a1 , . . . , ai−1 , ai+1 , . . . , an ) = ∈ Zf orj = i aj aj The sum in equation (4) is equal to: n n tmi tmi 1 tmi tmi tmi = + = = tmi = + aj ai aj ai aj a j=1 j=1 j j=i j=i
COMPLETE CLASSIFICATION OF 3-MULTISETS UP TO COMBINATORIAL EQUIVALENCE
⎢ ⎛ ⎞⎥ n ⎢ ⎥ 1 ⎢ ⎥ ⎢ ⎜ lcm(a1 , . . . , an ) ⎟⎥ ⎢ ⎜ ⎟ ⎥ a tS j=1 j ⎢ ⎜ ⎟⎥ = ⎢t ⎜ ⎟⎥ = lcm(a1 , . . . , an ) ⎢ ⎜ ⎟⎥ Ki ⎣ ⎝ ⎠⎦ lcm(a1 , . . . , ai−1 , ai+1 , . . . , an )
where in the last equality equations (1) and (2) were used.
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5. Main Theorem and Aplications The key result of the paper is the following: Theorem 5.1. Let {a, b, c} and {a , b , c } be two multisets of positive integers. Let: S = lcm(a, b, c) K1 =
1 1 1 + + a b c
lcm(a, b, c) lcm(a, b, c) lcm(a, b, c) , K2 = , K3 = lcm(b, c) lcm(c, a) lcm(a, b)
and S = lcm(a , b , c ) K1 =
1 1 1 + + a b c
lcm(a , b , c ) lcm(a , b , c ) lcm(a , b , c ) , K , K = = 2 3 lcm(b , c ) lcm(c , a ) lcm(a , b )
Then, the multisets {a, b, c} and {a , b , c } are combinatorially equivalent if, and only if, S = S and {K1 , K2 , K3 } = {K1 , K2 , K3 } as multisets. Proof. If the multisets {a, b, c} and {a , b , c } are combinatorially equivalent, the fundamental word of then are the same. Thus, by Lemma 4.1 we have S = S , and by applying the Alvarez Algorithm, and using Theorem 3.1, we have {K1 , K2 , K3 } = {K1 , K2 , K3 }. Suppose that S = S and {K1 , K2 , K3 } = {K1 , K2 , K3 } as multisets. In order to prove that the multisets {a, b, c} and {a , b , c } are combinatorially equivalent, we will prove that, given S, K1 , K2 , K3 obtained from a multiset with 3 elements, it is possible to determine the fundamental word in a unique way. Let {θj } be the family of the numbers defined on the form tS/Ki , for some i, belonging to the segment ]0, S]. Let us sort the collection of these numbers in the increasing order: θ1 < · · · < θN .
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D.L.A. MEDEIROS AND L. BIRBRAIR
We may reconstruct the fundamental word p1 , . . . , pm in the following way: Let p0 = 0 and, for all k ≥ 0 integer, define: w(k) =
k
pi
i=0
For all 0 < i ≤ N , we will construct pi recursively from p0 , . . . , pi−1 , by setting:
(5)
⎧ if w(i − 1) = θN − 3, ⎪ ⎨ 3, pi = 2, if w(i − 1) = θj − 2, for some 1 ≤ j < N, ⎪ ⎩ 1, otherwise.
We claim that the procedure reconstructs completely the fundamental word. By Lemma 4.2, all the numbers θj indicate the end of each block from any block structure, determining all positions of the numbers two and three in the fundamental word. The elements equals to one are also totally determined, because the number of those elements and their positions are determined by all the other partial sums, different from θj , see equation (5).
This finishes the proof.
Corollary 5.2 (Complete Classification of 3-Multisets up to Combinatorial Equivalence) Let A be a multiset with 3 elements. Then, a multiset A with 3 elements is combinatorially equivalent to A if, and only if, one of the three conditions holds: 1. A and A are totally irrational; 2. A and A are semi-rational, with canonical forms {n1 , n2 , λ} and {n1 , n2 , λ }, respectively, and: n n n1 n 2 + n1 + n2 = 1 2 + n1 + n2 λ λ 3. A and A are totally rational, with canonical forms {a, b, c} and {a , b , c }, respectively, and:
1 1 1 1 1 1 + + = lcm(a , b , c ) lcm(a, b, c) + + a b c a b c lcm(a , b , c ) lcm(a , b , c ) lcm(a , b , c ) lcm(a, b, c) lcm(a, b, c) lcm(a, b, c) , , = , , lcm(a, b) lcm(b, c) lcm(c, a) lcm(a , b ) lcm(b , c ) lcm(c , a )
The following result may be useful for calculation of the non-trivial examples of combinatorially equivalent 3-multisets. Theorem 5.3. Let A = {a, b, c} be a multiset of three positive integers, and suppose gcd(a, b, c) = 1. Define S, K1 , K2 , K3 as in Theorem 5.1. Then, the following identities are true:
COMPLETE CLASSIFICATION OF 3-MULTISETS UP TO COMBINATORIAL EQUIVALENCE
(6)
9
K1 K2 · gcd(a, b) + K2 K3 · gcd(b, c) + K3 K1 · gcd(c, a) = S
(7) a = K1 · gcd(a, b) · gcd(a, c); b = K2 · gcd(b, a) · gcd(b, c); c = K2 · gcd(c, a) · gcd(c, b) Proof. The main idea of the proof is the identities: lcm(x, y, z) =
xyz.gcd(x, y, z) xy ; lcm(x, y) = gcd(x, y).gcd(y, z).gcd(z, x) gcd(x, y)
The proof of those identiities can be found in the paper of Valcan and Bagdasar, [5]. We have: K1 K2 .gcd(a, b) + K2 K3 .gcd(b, c) + K3 K1 .gcd(c, a) = lcm(a, b, c) lcm(a, b, c) gcd(a, b) = = lcm(b, c) lcm(c, a) cyc ⎞ ⎛ ⎞⎛ abc.gcd(a, b, c) ⎜ lcm(a, b, c) ⎟ ⎜ gcd(a, b).gcd(b, c).gcd(c, a) ⎟ lcm(a, b, c) ⎟ gcd(a, b) = ⎜ ⎟⎜ = =S ac ⎠ ⎝ ⎠⎝ bc c cyc cyc gcd(c, a) gcd(b, c) This proves equation (6). To prove the first equation in (7), we have: lcm(a, b, c) K1 .gcd(a, b).gcd(a, c) = .gcd(a, b).gcd(a, c) = lcm(b, c) ⎛ ⎞ abc.gcd(a, b, c) ⎜ gcd(a, b).gcd(b, c).gcd(c, a) ⎟ ⎟ .gcd(a, b).gcd(a, c) = a =⎜ ⎝ ⎠ bc gcd(b, c) The other two equations in (7) can be proved in a similar way.
Example 5.4. Let A = {1, 6, 14}. A direct computation results in: S = 52, K1 = 1, K2 = 3, K3 = 7 Thus, by Theorems 5.1 and 5.3, if A0 = {a, b, c}, with gcd(a, b, c) = 1, is combinatorially equivalent to A, then: 3.gcd(a, b) + 21.gcd(b, c) + 7.gcd(c, a) = 52 We have (gcd(a, b), gcd(b, c), gcd(c, a)) = (1, 2, 1), (8, 1, 1) or (1, 1, 4). The first case implies a = 1·1·1 = 1, b = 3·2·1 = 6, c = 7·1·2 = 14, and thus A0 = {1, 6, 14}. The second case implies a = 1·8·1 = 8, b = 3·1·8 = 24, c = 7·1·1 = 7, and thus A0 = {8, 24, 7}. The third case implies a = 1·1·4 = 4, b = 3·1·1 = 3, c = 7·4·1 = 28, and thus A0 = {3, 4, 28}. It is easy to see that the values of S, K1 , K2 , K3 are the same for any of the three possibilities of A0 . With this, we conclude that the only 3-multisets that are non-trivial
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D.L.A. MEDEIROS AND L. BIRBRAIR
combinatorially equivalent to {1, 6, 14} are {3, 4, 28} and {8, 24, 7}. The next example shows that Theorem 5.1. cannot be directly generalized for multisets generated by more the three elements. Example 5.5. Let A = {1, 5, 5, 5} and B = {1, 2, 4, 4}. A direct computation results in S = 8, K1 = K2 = K3 = K4 = 1 for both A and B. However, the fundamental word of A is 1, 1, 1, 1, 4 and the fundamental word of B is 1, 2, 1, 4. Thus, A and B are not combinatorially equivalent. References [1] M.M. Peixoto, Focal decomposition in geometry, arithmetic and physics, in Geometry, Topology and Physics, eds. B. N. Apanasov, S. B. Bradlow, W. A. Rodrigues, Jr. and K. K. Uhlenbeck (de Gruyter, Berlin 1997), pp.213–231. [2] L. Birbrair, M. Sobolevsky and P. Sobolevskii, Focal decomposition for linear differencial equations of the second order, Abstr. Appl. Anal. 14 (2003), pp. 813–821. [3] S. Alvarez, D. Berend, L. Birbrair and D. Gir˜ao, Resonance sequences and focal decomposition, Israel J. Math. 170 (2009), pp. 269–284. [4] L. Birbrair, M. Gomes, W. Pereira, Resonance sequences and recoverability, International Journal od Number Theory, Vol. 11, No. 2 (2015), pp. 495–506 [5] D. Valcan, O. Bagdasar, Generalizations of some divisibility relations in N, Creative Math.& inf. , 18(1) (2009). 92–99. ´ tica, Universidade Federal do Ceara ´ (UFC), Campus do Pici, Departamento de Matema Bloco 914, Cep. 60455-760. Fortaleza-Ce, Brasil E-mail address: davi [email protected] ´ tica, Universidade Federal do Ceara ´ (UFC), Campus do Pici, Departamento de Matema Bloco 914, Cep. 60455-760. Fortaleza-Ce, Brasil E-mail address: [email protected]
Davi Lopes Alves de Medeiros, Lev Birbrair
PII: DOI: Reference:
S0022-314X(16)30282-7 http://dx.doi.org/10.1016/j.jnt.2016.09.035 YJNTH 5603
To appear in:
Journal of Number Theory
Received date: Revised date: Accepted date:
10 May 2016 27 September 2016 28 September 2016
Please cite this article in press as: D.L.A. de Medeiros, L. Birbrair, Complete classification of 3-multisets up to combinatorial equivalence, J. Number Theory (2017), http://dx.doi.org/10.1016/j.jnt.2016.09.035
This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
COMPLETE CLASSIFICATION OF 3-MULTISETS UP TO COMBINATORIAL EQUIVALENCE DAVI LOPES ALVES DE MEDEIROS* AND LEV BIRBRAIR** Abstract. Let A = {a1 , . . . , ak } be a finite multiset of positive real numbers. Consider the sequence of all positive integers multiples of all ai ’s, and note the multiplicity of each term in this sequence. This sequence of multiplicities is called the resonance sequence generated by {a1 , . . . , ak }. Two multisets are called combinatorially equivalent if they generate the same resonance sequence. The paper gives a complete criterion of classification of multisets with 3 elements up to combinatorial equivalence, by showing this is equivalent to a system of equations depending directly of the numbers in the multisets.
1. Introduction Resonance sequences appear in Geometric Theory of Ordinary Differential Equations as an invariant of the linear equations of the second order under the focal equivalence. In general, the resonance sequence is defined for a finite multiset of arbitrary real numbers, but a consideration of rational multisets leads to a discovery of some interesting arithmetical and combinatorial properties of resonance sequences. The main result of the paper is the classification theorem of resonance sequences generated by three elements (Theorem 5.1). We define 4 functions of the elements of the multisets. The multisets are combinatorially equivalent if and only if the values of these functions are equal. The main ingredient of the proof is the Alvarez Algorithm, defined in [3] and investigated in [4]. In order to obtain the result, we define a sort of the inverse Alvarez algorithm. The results of the paper cannot be directly generalized for multisets generated by more the three elements. The general case needs a deeper investigation. We would like to thank Daniel Berend for useful remarks. We also would like to thank the anonymous referee for extremely useful comments. 2010 Mathematics Subject Classification. 11B30, 05A17 . Key words and phrases. Resonance sequences, combinatorial equivalence, focal classification. *Research supported by CNPq grant 131709/2016-0. **Research supported under CNPq grant and by Capes-Cofecub 302655/2014-0. 1
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D.L.A. MEDEIROS AND L. BIRBRAIR
2. Definitions, Basic Properties and Examples Definition 2.1. Consider a finite multiset of positive real numbers A = {a1 , . . . , an }. Let Ai be the set of all positive integer multiples of ai : Ai = {ai , 2ai , 3ai , . . . }, 1 ≤ i ≤ n Let Q = ∪ni=1 Ai . Q is infinite countable and may be sorted by increasing order: Q = {mj : j ∈ N}, (m1 < m2 < · · · < mr < . . . ) Let ind(mj ) = {1 ≤ i ≤ n : mj ∈ Ai } and set rj = |ind(mj )|. The numbers rj are integers with 1 ≤ rj ≤ n. The sequence Ress(A) = (rj )j∈N is the resonance sequence generated by {a1 , . . . , an }. The number n is called the dimension of the multiset A. Example 2.2. Consider the set A = {3, 5, 6}. The set of positive integer multiples of the elements in A is: Q = {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, . . . } and its resonance sequence is: Ress(A) = 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 3, . . . Example 2.3. Consider the set A = {2,
√
5, 3}. The set of positive integer multiples of
the elements in A is: Q = {2,
√ √ √ √ √ √ 5, 3, 4, 2 5, 6, 3 5, 8, 4 5, 9, 10, 5 5, 12, 6 5 . . . }
and its resonance sequence is: Ress(A) = 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1 . . . √ √ √ Example 2.4. Consider the set A = { 2, 3, 5}. It is easy to see that Ai ∩ Aj = ∅ for all i = j, and then its resonance sequence is: Ress(A) = 1, 1, 1, 1, 1, 1, . . . Those examples motivates us the folowing definition: Definition 2.5. A multiset A = {a1 , . . . , an } is totally rational if αi /αj ∈ Q, for all i, j; / Q, for all i = j; and semi-rational if it is neither totally totally irrational if αi /αj ∈ rational nor totally irrational. The main interest in the paper is the combinatoral equivalence, defined below: Definition 2.6. Given two multisets A and A , they are combinatorially equivalent if Ress(A) = Ress(A ), i.e. they generate the same resonance sequence.
COMPLETE CLASSIFICATION OF 3-MULTISETS UP TO COMBINATORIAL EQUIVALENCE
3
It is easy to see that, for c > 0, A = {a1 , . . . , an } and cA = {ca1 , . . . , can } are combinatorially equivalent. Also, it’s easy to see that A = {a1 , . . . , an } and A = {a1 , . . . , an }, where {a1 , . . . , an } is a permutation of {a1 , . . . , an }, are combinatorially equivalent. The combinatorial equivalence of one of these two forms is considered as a trivial combinatorial equivalence. Our main interest is to study non-trivial combinatorial equivalence of multisets. Consider now multisets with 3 elements. Clearly, for every totally rational multiset A = {a1 , a2 , a3 }, there is a unique number c > 0 such that ni = cai (i = 1, 2, 3) are natural numbers such that gcd(n1 , n2 , n3 ) = 1. Note that {a1 , a2 , a3 } and {n1 , n2 , n3 } are trivially combinatorially equivalent. It is clear that, for every semi-rational mutiset A = {a1 , a2 , a3 }, there is a permutation A = {a1 , a2 , a3 } such that a1 /a2 ∈ Q, and a unique number c > 0 such that n1 = ca1 , n2 = ca2 and λ = ca3 , where n1 , n2 are coprime positive integers, and λ is an irrational number. Note that {a1 , a2 , a3 } and {n1 , n2 , λ} are trivially combinatorially equivalent. These observations motivates the next definiton. Definition 2.7. Let A be a totally rational multiset with 3 elements, trivially combinatorially equivalent to A0 = {n1 , n2 , n3 }, where n1 , n2 , n3 are positive integers with gcd(n1 , n2 , n3 ) = 1. We define the multiset A0 as the canonical form of A. In a similar way, let A be a semi-rational multiset with 3 elements, trivially combinatorially equivalent to A0 = {n1 , n2 , λ}, where n1 , n2 are coprime positive integers and λ ∈ / Q. We define the multiset A0 as the canonical form of A. The next two theorems were proved in [3], and they are a classification of totally irrational multisets and semi-rational multisets up to combinatorial equivalence. Theorem 2.8. [3], pg.272. All totally irrational multisets are combinatorially equivalent, and a totally irrational multiset is combinatorially equivalent only to another totally irrational one. Theorem 2.9. [3], pg.273, 276. Let A be a semi-rational multiset with 3 elements, with canonical form {n1 , n2 , λ}. A multiset with 3 element A is combinatorially equivalent to A if, and only if, A is semi-rational, with canonical form {n1 , n2 , λ }, and: n1 n 2 n n + n1 + n2 = 1 2 + n1 + n2 λ λ In order to complete the cassification of multisets with 3 elements up to combinatorial equivalence, it is necessary to know what happens with totally rational multisets. Since the canonical forms of them are multisets with positive integers, the main object of study in this paper will be the multisets with positive integers. Consider now A = {a1 , . . . , ak }, where all the ai are positive integers, and let Q as in Definition 2.1. It is clear that, for x ∈ N, M = lcm(a1 , . . . , ak ) and t ∈ Z such that
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D.L.A. MEDEIROS AND L. BIRBRAIR
x + tM ∈ N, we have x ∈ Q implies x + tM ∈ Q, and ind(x) = ind(x + tM ). This proves that Ress(A) is a periodic sequence. Also, M is the first number y ∈ Q such that |ind(y)| = n. Thus, Ress(A) is completely determined by the numbers until the first n (after that, the sequence Ress(A) will start to repeat). This motivates a new definition. Definition 2.10. Let A be a multiset with n positive integers. The Fundamental Word of A is the subsequence of Ress(A), starting from the first term and ending at the first occurence of n. Example 2.11. If A = {1, 2, 3, 4}, the fundamental word of A is 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 4. It is clear that two multisets A and A of positive integers are combinatorially equivalent if, and only if, their fundamental words are equal. 3. Alvarez Algorithm and Block Structure Let A = {a1 , a2 , . . . , an } be a multiset with n > 2 positive integers. Consider its fundamental word. The steps below are the Alvarez Algorithm, first described in [3]: Step 1: If there is no (n − 1) in the fundamental word, mark the number n and let S1 be the sum of all the numbers in the fundamental word. If there is (n − 1), find the first occurence of (n − 1) in the fundamental word and mark it. Then, calculate the sum of the numbers of the fundamental word, from the first number to the marked number, and denote this sum by S1 . Starting from the number next to the marked (n−1), calculate the sum of the terms of the resonance sequence until it becomes equal to S1 or (S1 + 1), what happens first, and stop at that point, marking it. Starting from the number next to the last marked number, calculate the sum of the terms of the sequence until the sum becomes equal to S1 or (S1 +1), what happens first, stop at that point and mark it. Repeat the previous procedure until the end of the fundamental word, marking the number n at the end. Those marked numbers divides the fundamental word into blocks. This partition is called the Block Structure of Step 1. For 2 ≤ k ≤ n, let Step k be as follows: Step k: If there is no previously unmarked (n − 1) in the fundamental word, mark the number n and let Sk be the sum of all the numbers in the fundamental word. If there is a previously unmarked (n − 1), find the first occurence of a previously unmarked (n − 1) in the fundamental word and mark it. After this, calculate the sum of the numbers of the fundamental word, from the first number to this marked number, and denote this sum by Sk . Starting from the number next to the last marked (n − 1), calculate the sum of the terms of the resonance sequence until it becomes equal to Sk or (Sk + 1), what
COMPLETE CLASSIFICATION OF 3-MULTISETS UP TO COMBINATORIAL EQUIVALENCE
5
happens first, and stop at that point, marking it. Starting from the number next to the last marked number, calculate the sum of the terms of the sequence until the sum becomes equal to Sk or (Sk +1), what happens first, stop at that point and mark it. Repeat the previous procedure until the end of the fundamental word, marking the number n at the end. Those marked numbers divides the fundamental word into blocks. It is called the Block Structure of Step k. For 1 ≤ i ≤ n, define mi = lcm(a1 , . . . , ai−1 , ai+1 , . . . , an ), and let: (1)
Ki =
lcm(a1 , . . . , an ) lcm(a1 , . . . , an ) = lcm(a1 , . . . , ai−1 , ai+1 , . . . , an ) mi
Finally, consider the numbers Ki sorted in decreasing order. They give the sequence L1 ≥ · · · ≥ Ln . The proof of the next theorem can be found in [4]. Theorem 3.1. The following assertions about the block structures on a fundamental word are true: 1. All blocks end at (n − 1) or n, and when the block ends at n, it is the last one of the block structure. Also, every (n − 1) in the fundamental word is the end of exactly one block in all the block structures; 2. The number of blocks in the block structure of step k is equal to Lk ; Example 3.2. Let A = {6, 5, 4}. The fundamental word of A is: 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3 Applying the Alvarez Algorithm, we have the 3 block structures: (1, 1, 1, 1, 1, 2), (1, 1, 1, 2, 2), (1, 1, 2, 1, 1, 2), (2, 1, 1, 1, 2), (1, 1, 1, 1, 1, 3) (1, 1, 1, 1, 1, 2, 1, 1, 1, 2), (2, 1, 1, 2, 1, 1, 2, 2), (1, 1, 1, 2, 1, 1, 1, 1, 1, 3) (1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2), (1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3) Observe that L1 = 5, L2 = 3, L3 = 2 and K1 = 3 = L2 , K2 = 5 = L1 , K3 = 2 = L3 . Example 3.3. Let A = {1, 2, 6}. The fundamental word of A is: 1, 2, 1, 2, 1, 3 Applying the Alvarez Algorithm, we have the 3 block structures: (1, 2), (1, 2), (1, 3)
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D.L.A. MEDEIROS AND L. BIRBRAIR
(1, 2, 1, 2, 1, 3) (1, 2, 1, 2, 1, 3) Observe that L1 = 3, L2 = 1, L3 = 1 and K1 = 1 = L2 , K2 = 1 = L3 , K3 = 2 = L1 . 4. Two Preliminary Lemmas Lemma 4.1. The sum of the elements of the fundamental word of the resonance sequence generated by the positive integers {a1 , . . . , an } is equal to:
(2)
n 1 S = lcm(a1 , . . . , an ) a i=1 i
Proof. The sum of the elements in the resonance sequence is equal to the number of the multiples of a1 on the segment (0, lcm(a1 , . . . , an )], plus the number of the multiples of a2 on the same segment and so on. The number of multiples of ai on this interval is equal to lcm(a1 , . . . , an )/ai . This concludes the proof of (2).
We use the notation x for the maximum integer, not greater than x. Lemma 4.2. Let t be a positive integer such that t ≤ Ki . The sum of the elements of the first t blocks of the block structure corresponding to Ki is equal to: (3)
tS Ki
Proof. The sum of the elements in the resonance sequence belonging to the first t blocks is equal to the sum of the numbers of occurences of the multiples of each element from the set {a1 , . . . , an } on the interval ]0, tmi ]. The number of occurences of the multiples of an element x on the segment ]0, y] equals to y/x . Thus, the sum of the elements of the first t blocks of the block structure corresponding to Ki is equal to:
(4)
n tmi j=1
aj
It remains to prove that the expressions (3) and (4) are equal. The equation (1) gives: mi lcm(a1 , . . . , ai−1 , ai+1 , . . . , an ) = ∈ Zf orj = i aj aj The sum in equation (4) is equal to: n n tmi tmi 1 tmi tmi tmi = + = = tmi = + aj ai aj ai aj a j=1 j=1 j j=i j=i
COMPLETE CLASSIFICATION OF 3-MULTISETS UP TO COMBINATORIAL EQUIVALENCE
⎢ ⎛ ⎞⎥ n ⎢ ⎥ 1 ⎢ ⎥ ⎢ ⎜ lcm(a1 , . . . , an ) ⎟⎥ ⎢ ⎜ ⎟ ⎥ a tS j=1 j ⎢ ⎜ ⎟⎥ = ⎢t ⎜ ⎟⎥ = lcm(a1 , . . . , an ) ⎢ ⎜ ⎟⎥ Ki ⎣ ⎝ ⎠⎦ lcm(a1 , . . . , ai−1 , ai+1 , . . . , an )
where in the last equality equations (1) and (2) were used.
7
5. Main Theorem and Aplications The key result of the paper is the following: Theorem 5.1. Let {a, b, c} and {a , b , c } be two multisets of positive integers. Let: S = lcm(a, b, c) K1 =
1 1 1 + + a b c
lcm(a, b, c) lcm(a, b, c) lcm(a, b, c) , K2 = , K3 = lcm(b, c) lcm(c, a) lcm(a, b)
and S = lcm(a , b , c ) K1 =
1 1 1 + + a b c
lcm(a , b , c ) lcm(a , b , c ) lcm(a , b , c ) , K , K = = 2 3 lcm(b , c ) lcm(c , a ) lcm(a , b )
Then, the multisets {a, b, c} and {a , b , c } are combinatorially equivalent if, and only if, S = S and {K1 , K2 , K3 } = {K1 , K2 , K3 } as multisets. Proof. If the multisets {a, b, c} and {a , b , c } are combinatorially equivalent, the fundamental word of then are the same. Thus, by Lemma 4.1 we have S = S , and by applying the Alvarez Algorithm, and using Theorem 3.1, we have {K1 , K2 , K3 } = {K1 , K2 , K3 }. Suppose that S = S and {K1 , K2 , K3 } = {K1 , K2 , K3 } as multisets. In order to prove that the multisets {a, b, c} and {a , b , c } are combinatorially equivalent, we will prove that, given S, K1 , K2 , K3 obtained from a multiset with 3 elements, it is possible to determine the fundamental word in a unique way. Let {θj } be the family of the numbers defined on the form tS/Ki , for some i, belonging to the segment ]0, S]. Let us sort the collection of these numbers in the increasing order: θ1 < · · · < θN .
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D.L.A. MEDEIROS AND L. BIRBRAIR
We may reconstruct the fundamental word p1 , . . . , pm in the following way: Let p0 = 0 and, for all k ≥ 0 integer, define: w(k) =
k
pi
i=0
For all 0 < i ≤ N , we will construct pi recursively from p0 , . . . , pi−1 , by setting:
(5)
⎧ if w(i − 1) = θN − 3, ⎪ ⎨ 3, pi = 2, if w(i − 1) = θj − 2, for some 1 ≤ j < N, ⎪ ⎩ 1, otherwise.
We claim that the procedure reconstructs completely the fundamental word. By Lemma 4.2, all the numbers θj indicate the end of each block from any block structure, determining all positions of the numbers two and three in the fundamental word. The elements equals to one are also totally determined, because the number of those elements and their positions are determined by all the other partial sums, different from θj , see equation (5).
This finishes the proof.
Corollary 5.2 (Complete Classification of 3-Multisets up to Combinatorial Equivalence) Let A be a multiset with 3 elements. Then, a multiset A with 3 elements is combinatorially equivalent to A if, and only if, one of the three conditions holds: 1. A and A are totally irrational; 2. A and A are semi-rational, with canonical forms {n1 , n2 , λ} and {n1 , n2 , λ }, respectively, and: n n n1 n 2 + n1 + n2 = 1 2 + n1 + n2 λ λ 3. A and A are totally rational, with canonical forms {a, b, c} and {a , b , c }, respectively, and:
1 1 1 1 1 1 + + = lcm(a , b , c ) lcm(a, b, c) + + a b c a b c lcm(a , b , c ) lcm(a , b , c ) lcm(a , b , c ) lcm(a, b, c) lcm(a, b, c) lcm(a, b, c) , , = , , lcm(a, b) lcm(b, c) lcm(c, a) lcm(a , b ) lcm(b , c ) lcm(c , a )
The following result may be useful for calculation of the non-trivial examples of combinatorially equivalent 3-multisets. Theorem 5.3. Let A = {a, b, c} be a multiset of three positive integers, and suppose gcd(a, b, c) = 1. Define S, K1 , K2 , K3 as in Theorem 5.1. Then, the following identities are true:
COMPLETE CLASSIFICATION OF 3-MULTISETS UP TO COMBINATORIAL EQUIVALENCE
(6)
9
K1 K2 · gcd(a, b) + K2 K3 · gcd(b, c) + K3 K1 · gcd(c, a) = S
(7) a = K1 · gcd(a, b) · gcd(a, c); b = K2 · gcd(b, a) · gcd(b, c); c = K2 · gcd(c, a) · gcd(c, b) Proof. The main idea of the proof is the identities: lcm(x, y, z) =
xyz.gcd(x, y, z) xy ; lcm(x, y) = gcd(x, y).gcd(y, z).gcd(z, x) gcd(x, y)
The proof of those identiities can be found in the paper of Valcan and Bagdasar, [5]. We have: K1 K2 .gcd(a, b) + K2 K3 .gcd(b, c) + K3 K1 .gcd(c, a) = lcm(a, b, c) lcm(a, b, c) gcd(a, b) = = lcm(b, c) lcm(c, a) cyc ⎞ ⎛ ⎞⎛ abc.gcd(a, b, c) ⎜ lcm(a, b, c) ⎟ ⎜ gcd(a, b).gcd(b, c).gcd(c, a) ⎟ lcm(a, b, c) ⎟ gcd(a, b) = ⎜ ⎟⎜ = =S ac ⎠ ⎝ ⎠⎝ bc c cyc cyc gcd(c, a) gcd(b, c) This proves equation (6). To prove the first equation in (7), we have: lcm(a, b, c) K1 .gcd(a, b).gcd(a, c) = .gcd(a, b).gcd(a, c) = lcm(b, c) ⎛ ⎞ abc.gcd(a, b, c) ⎜ gcd(a, b).gcd(b, c).gcd(c, a) ⎟ ⎟ .gcd(a, b).gcd(a, c) = a =⎜ ⎝ ⎠ bc gcd(b, c) The other two equations in (7) can be proved in a similar way.
Example 5.4. Let A = {1, 6, 14}. A direct computation results in: S = 52, K1 = 1, K2 = 3, K3 = 7 Thus, by Theorems 5.1 and 5.3, if A0 = {a, b, c}, with gcd(a, b, c) = 1, is combinatorially equivalent to A, then: 3.gcd(a, b) + 21.gcd(b, c) + 7.gcd(c, a) = 52 We have (gcd(a, b), gcd(b, c), gcd(c, a)) = (1, 2, 1), (8, 1, 1) or (1, 1, 4). The first case implies a = 1·1·1 = 1, b = 3·2·1 = 6, c = 7·1·2 = 14, and thus A0 = {1, 6, 14}. The second case implies a = 1·8·1 = 8, b = 3·1·8 = 24, c = 7·1·1 = 7, and thus A0 = {8, 24, 7}. The third case implies a = 1·1·4 = 4, b = 3·1·1 = 3, c = 7·4·1 = 28, and thus A0 = {3, 4, 28}. It is easy to see that the values of S, K1 , K2 , K3 are the same for any of the three possibilities of A0 . With this, we conclude that the only 3-multisets that are non-trivial
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D.L.A. MEDEIROS AND L. BIRBRAIR
combinatorially equivalent to {1, 6, 14} are {3, 4, 28} and {8, 24, 7}. The next example shows that Theorem 5.1. cannot be directly generalized for multisets generated by more the three elements. Example 5.5. Let A = {1, 5, 5, 5} and B = {1, 2, 4, 4}. A direct computation results in S = 8, K1 = K2 = K3 = K4 = 1 for both A and B. However, the fundamental word of A is 1, 1, 1, 1, 4 and the fundamental word of B is 1, 2, 1, 4. Thus, A and B are not combinatorially equivalent. References [1] M.M. Peixoto, Focal decomposition in geometry, arithmetic and physics, in Geometry, Topology and Physics, eds. B. N. Apanasov, S. B. Bradlow, W. A. Rodrigues, Jr. and K. K. Uhlenbeck (de Gruyter, Berlin 1997), pp.213–231. [2] L. Birbrair, M. Sobolevsky and P. Sobolevskii, Focal decomposition for linear differencial equations of the second order, Abstr. Appl. Anal. 14 (2003), pp. 813–821. [3] S. Alvarez, D. Berend, L. Birbrair and D. Gir˜ao, Resonance sequences and focal decomposition, Israel J. Math. 170 (2009), pp. 269–284. [4] L. Birbrair, M. Gomes, W. Pereira, Resonance sequences and recoverability, International Journal od Number Theory, Vol. 11, No. 2 (2015), pp. 495–506 [5] D. Valcan, O. Bagdasar, Generalizations of some divisibility relations in N, Creative Math.& inf. , 18(1) (2009). 92–99. ´ tica, Universidade Federal do Ceara ´ (UFC), Campus do Pici, Departamento de Matema Bloco 914, Cep. 60455-760. Fortaleza-Ce, Brasil E-mail address: davi [email protected] ´ tica, Universidade Federal do Ceara ´ (UFC), Campus do Pici, Departamento de Matema Bloco 914, Cep. 60455-760. Fortaleza-Ce, Brasil E-mail address: [email protected]