Spectrum of Permanent’s Values and Its Extremal Magnitudes in Λ³_𝑛 and Λ_𝑛(𝛼,𝛽,𝛾)

Abstract

Let Λ^𝑘_𝑛 denote the class of (0,1) square matrices containing in each row and in each column exactly 𝑘 1’s. The minimal value of 𝑘, for which the behavior of the permanent in Λ^𝑘_𝑛 is not quite studied, is 𝑘=3. We give a simple algorithm for calculation of upper magnitudes of permanent in Λ³_𝑛 and consider some extremal problems in a generalized class Λ_𝑛(𝛼,𝛽,𝛾), the matrices of which contain in each row and in each column nonzero elements 𝛼, 𝛽, and 𝛾 and 𝑛−3 zeros.

1. Introduction

The definition of permanent of a square matrix 𝐴=(𝑎_𝑖,𝑗) of order 𝑛 is per𝐴=∑∏^𝑛_𝑖=1𝑎_{𝑖,𝜎(𝑖)}, where sum is over all permutations 𝜎 of numbers 1,…,𝑛. This definition is very combinatorial. For example, if 𝐴 is (0,1)-matrix, then per𝐴 is the number of arrangements of 𝑛 nonattacking each other rooks on positions of 1’s of 𝐴. Therefore, the most natural applications of the permanent are in combinatorics: enumerating the Latin rectangles and squares, permutations with restricted positions, and different problems in the theory of graphs. Many interesting examples of applications of the permanent one can find in [1], chapter 8 (and not only in Mathematics!). To applications of the permanent there promoted celebrate proofs of the best known van der Waerden’s and Minc’s conjectures by Egorychev [2], Falikman [3], and Brègman [4], respectively. For example, Alon and Friedland [5] excellently used the Minc-Bregman inequality for permanent of (0,1)-matrices for finding the maximum number of perfect matchings in graphs with a given degree sequence.

Let Λ^𝑘_𝑛 denote the class of (0,1) square matrices containing in each row and in each column exactly 𝑘 1’s. If 𝐴∈Λ^𝑘_𝑛, then matrix 𝑘⁻¹𝐴 is doubly stochastic, since all its row and column sums equal 1. Therefore, Λ^𝑘_𝑛-matrices are also called doubly stochastic (0,1)-matrices. These matrices have especially simple and attractive structure and have many applications. It is important that especially class Λ³_𝑛 in permanent’s history was a nontrivial test area for different researches of the permanent. For example, one can mention a remarkable Merriell research [6] of the maximum of permanent on class Λ³_𝑛. An author’s research [7] of the permanent on Λ³₇-matrices (see Section 9) has led to the disproof of an important Balasubramanian conjecture [8] which would yield a full proof of 1967 Ryser hypothesis [9] on transversals of Latin squares.

In the present paper we consider many other problems for Λ³_𝑛. We study also the following natural generalization of Λ³_𝑛. For given real or complex nonzero numbers 𝛼, 𝛽, and 𝛾, denote by Λ_𝑛(𝛼,𝛽,𝛾) the class of square matrices containing every number from {𝛼,𝛽,𝛾} exactly one time in each row and in each column, such that the other elements are 0’s. It is clear that, if 𝐴∈Λ_𝑛(𝛼,𝛽,𝛾) and 𝛼+𝛽+𝛾≠0, then the matrix (𝛼+𝛽+𝛾)⁻¹𝐴 is doubly stochastic.

Definition 1. One calls 𝑝-spectrum in Λ^𝑘_𝑛 (denoting it 𝑝𝑠[Λ^𝑘_𝑛]) the set of all the values which are taken by the permanent in Λ^𝑘_𝑛.

Note that 𝑝-spectrum in Λ¹_𝑛 trivially is {1}. It is known (cf. Tarakanov [10]) that 𝑝𝑠Λ²_𝑛=2,2²,2³,…,2^⌊𝑛/2⌋. (1)

But, for 𝑘≥3,  𝑝-spectrum of Λ^𝑘_𝑛, generally speaking, is unknown. Greenstein (cf. [1], point 8.4, Problem 3) put the problem of describing the 𝑝-spectrum in Λ³_𝑛. In this paper we give an algorithm for calculation of upper values of 𝑝-spectrum in Λ³_𝑛. We also obtain several results for a generalized class Λ_𝑛(𝛼,𝛽,𝛾) with real nonzero numbers 𝛼, 𝛽, and 𝛾. Some results of the present paper were announced by the author in [11].

2. What Is Known about Λ³_𝑛?

(1) Explicit formula for |Λ³_𝑛| (cf. Stanley [12], chapter 1) is Λ³_𝑛=6^−𝑛_{𝑘1+𝑘2+𝑘3=𝑛,𝑘𝑖≥0}(−1)^𝑘2𝑛!²𝑘₂+3𝑘₃!2^𝑘13^𝑘2𝑘₁!𝑘₂!𝑘₃!²6^𝑘3. (2)

(2) Asymptotic formula for |Λ³_𝑛| (cf. O’Neil [13]) is Λ³_𝑛=(3𝑛)!(36)^𝑛𝑒⁻²1+𝑂𝑛^−1+𝜀, (3) where 𝜀>0 is arbitrarily small for sufficiently large 𝑛.

In addition, note that |Λ_𝑛(𝛼,𝛽,𝛾)| with different 𝛼, 𝛽, and 𝛾 is, evidently, the number of 3-rowed Latin rectangles of length 𝑛 such that Λ_𝑛𝛼,𝛽,𝛾=𝑛!𝐾_𝑛, (4) where 𝐾_𝑛 is the number of reduced 3-rowed Latin rectangles with the first row {1,2,…,𝑛}. It is known (Riordan [14], pp. 204–210) that 𝐾_𝑛=^⌊𝑛/2⌋_𝑘=0𝑛𝑘𝐷_𝑛−𝑘𝐷_𝑘𝑈_𝑛−2𝑘, (5) where 𝐷_𝑛 is subfactorial: 𝐷₀=1,𝐷_𝑛=𝑛𝐷_𝑛−1+(−1)^𝑛,𝑛≥1, (6) and {𝑈_𝑛} is sequence of Lucas numbers of the Ménage problem which is defined by Cayley recursion (cf. [14], p. 201) 𝑈₀=1,𝑈₁=−1,𝑈₂𝑈=0,_𝑛=𝑛𝑈_𝑛−1+𝑛𝑈𝑛−2^𝑛−2()+4−1^𝑛𝑛−2,𝑛≥3 (7) (see [15], sequences A102761, A000186).

Denote, furthermore, by Λ³_𝑛 the set of matrices in Λ³_𝑛 with 1’s on the main diagonal. Note that 𝑝𝑠Λ³_𝑛=𝑝𝑠Λ³_𝑛. (8)

Indeed, it is well known that every Λ³_𝑛-matrix 𝐴 has a diagonal of ones (i.e., a set of 1’s not two in the same row or column). Let 𝑙 be such a diagonal. There exists a permutation of rows and columns 𝜋 such that 𝜋(𝑙) will be the main diagonal of 𝜋(𝐴). Nevertheless, per(𝜋(𝐴))=per𝐴 and (8) follows.

(3) A known explicit formula for |Λ³_𝑛| (Shevelev [16]) has a close structure to (5): Λ³_𝑛=^⌊𝑛/2⌋_𝑘=0𝑛𝑘𝑆_𝑛−𝑘𝑆_𝑘𝑈_𝑛−2𝑘, (9) where sequence {𝑆_𝑛} is defined by recursion 𝑆₀=1,𝑆₁=0,𝑆_𝑛=(𝑛−1)𝑆_𝑛−1+12𝑆_𝑛−2,𝑛≥2. (10)

(4) Asymptotic formula for |Λ³_𝑛| (Shevelev [16]) is Λ³_𝑛=𝐶√𝑛𝑛𝑒^2𝑛1+𝑂𝑛^−1+𝜀, (11) where 𝐶=2√𝜋𝑒⁻⁵=0.29098… (12) and 𝜀>0 is arbitrarily small for sufficiently large 𝑛.

(5) Denote by Λ³_𝑛 the set of symmetric matrices in Λ³_𝑛. 𝑃-spectrum on Λ³_𝑛 is given by the following theorem (Shevelev [11]).

Theorem 2. Let 𝑅(𝑛;3) denote the set of all partitions of 𝑛 with parts more than or equal to 3. To every partition 𝑟∈𝑅(𝑛;3)∶𝑛=𝑛₁+𝑛₂+⋯+𝑛_𝑚, 𝑚=𝑚(𝑟), put in a correspondence the number 𝐻(𝑟)=^𝑚_𝑖=1𝑎𝑛_𝑖, (13) where sequence {𝑎(𝑛)} is defined by the recursion 𝑎(3)(4)𝑎(𝑛)()()=6,𝑎=9,=𝑎𝑛−1+𝑎𝑛−2−2,𝑛≥5. (14) Then one has 𝑝𝑠Λ³_𝑛={𝐻(𝑟)∶𝑟∈𝑅(𝑛;3)}. (15)

(6) The maximal value 𝑀(𝑛) of permanent in Λ³_𝑛 was found by Merriell [6].

Theorem 3. If 𝑛≡ℎ(mod3), ℎ=0,1,2, then 𝑀(𝑛)=6^{(𝑛−ℎ)/3}32^ℎ. (16)

Note that the case ℎ=0 of (16) easily follows from the Minc-Bregman inequality for permanent of (0,1)-matrices.

(7) Put 𝑀(𝑛)=𝑀⁽¹⁾(𝑛). In case of 𝑛≡0(mod3), Bolshakov [17] showed that the second maximal 𝑀⁽²⁾<𝑀⁽¹⁾(𝑛) of permanent in Λ³_𝑛 (such that interval (𝑀⁽²⁾,𝑀⁽¹⁾) is free from values of permanent in Λ³_𝑛) equals 𝑀⁽²⁾(𝑛)=20,if𝑛=6,120,if9𝑛=9,616^𝑛/3,if𝑛≥12. (17)

Note that both 𝑀⁽¹⁾(𝑛) and 𝑀⁽²⁾(𝑛) are attained in Λ³_𝑛.

(8) Denote by 𝑚(𝑛) the minimal value of permanent in Λ³_𝑛. In 1979, Voorhoeve [18] obtained a beautiful lower estimate for 𝑚(𝑛): 𝑚(𝑛)4≥63^𝑛−3. (18)

This estimate remains the best even after proof by Egorychev [2] and Falikman [3], the famous van der Waerden conjectural lower estimate per𝐴≥𝑛!/𝑛^𝑛 for every 𝑛×𝑛 doubly stochastic matrix 𝐴. Indeed, this estimate yields only 𝑚(𝑛)≥3^𝑛(𝑛!/𝑛^𝑛), such that (18) is stronger for 𝑛≥4.

(9) Bolshakov [19] found 𝑝-spectrum in Λ³_𝑛 in cases 𝑛≤8. Namely, he added to the evident 𝑝-spectrums 𝑝𝑠[Λ³₃]={6} and 𝑝𝑠[Λ³₄]={9} also the following 𝑝-spectrums: 𝑝𝑠Λ³₅={12,13},𝑝𝑠Λ³₆={},Λ17,18,20,36𝑝𝑠³₇={},Λ24,25,26,27,30,31,32,54𝑝𝑠³₈={}.33,34,35,36,37,38,39,40,41,42,44,45,48,49,52,72,78,81 (19)

3. A Generalization of Theorem 2 on Matrices of Class Λ_𝑛(𝛼,𝛽,𝛾) with Symmetric Positions of Elements

Denote by Λ_𝑛(𝛼,𝛽,𝛾) the set of matrices in Λ_𝑛(𝛼,𝛽,𝛾) with 𝛽’s on the main diagonal. It is clear that, together with (4), Λ_𝑛𝛼,𝛽,𝛾=𝐾_𝑛. (20)

Note that, as for sets Λ³_𝑛, Λ³_𝑛, we have 𝑝𝑠Λ_𝑛𝛼,𝛽,𝛾=𝑝𝑠Λ_𝑛𝛼,𝛽,𝛾. (21)

Denote, furthermore, by Λ_𝑛(𝛼,𝛽,𝛾) the set of matrices 𝑀={𝑚_𝑖,𝑗} in Λ_𝑛(𝛼,𝛽,𝛾) with symmetric positions of elements: 𝑚_𝑖,𝑗=𝛼 if and only if 𝑚_𝑗,𝑖=𝛾.

𝑃-spectrum on Λ_𝑛(𝛼,𝛽,𝛾) is given by the following theorem.

Theorem 4. If to every partition 𝑟∈𝑅(𝑛;3), 𝑛=𝑛₁+𝑛₂+⋯+𝑛_𝑚, 𝑚=𝑚(𝑟), corresponds the number 𝐻_{𝛼,𝛽,𝛾}(𝑟)=^𝑚_𝑖=1𝑎𝑛_𝑖, (22) where sequence {𝑎(𝑛)=𝑎(𝛼,𝛽,𝛾;𝑛} is defined by the recursion 𝑎(3)=𝛼³+𝛽³+𝛾³𝑎(4)+3𝛼𝛽𝛾,=𝛼⁴+𝛽⁴+𝛾⁴+4𝛼𝛽²𝛾+2𝛼𝛾²,𝑎(𝑛)=𝛽𝑎(𝑛−1)+𝛼𝛾𝑎(𝑛−2)+𝛼^𝑛−1𝛼−𝛽−𝛾+𝛾^𝑛−1𝛾−𝛽−𝛼,𝑛≥5, (23) then one has 𝑝𝑠Λ_𝑛𝛼,𝛽,𝛾=𝐻_{𝛼,𝛽,𝛾}(𝑟)∶𝑟∈𝑅(𝑛;3). (24)

Proof. Let 𝑆_𝑛 be the symmetric permutation group of elements 1,…,𝑛. Two positions (𝑖₁,𝑖₂), (𝑗₁,𝑗₂) are called independent if 𝑖_𝑘≠𝑗_𝑘, 𝑘=1,2. We will say that in the 𝑛×𝑛 matrix 𝑀={𝑚_𝑖𝑗}, a weight 𝑚_𝑖𝑗 is appropriated to the position (𝑖,𝑗). Let 𝑠∈𝑆_𝑛 have no cycle of length less than 𝑛. Consider a map 𝜎∶𝑖,𝑗⟼𝑠^𝑖(1),𝑠^𝑗(1), (25) appropriating to the position (𝑠^𝑖(1),𝑠^𝑗(1)) the weight 𝑚_𝑖𝑗.
Now we need several lemmas.
Lemma A. (1) The map 𝜎 is bijective; (2) if 𝐸 is a set of pairwise independent positions, then 𝜎(𝐸) is also a set of pairwise independent positions.
Proof. (a) Consider two distinct positions 𝑖₁,𝑗₁,𝑖₂,𝑗₂, (26) such that, at least, one of the two inequalities holds 𝑖₁≠𝑖₂,𝑗₁≠𝑗₂. (27) Let 𝑖₁≠𝑖₂ such that, say, 𝑖₁>𝑖₂. Show that 𝑠^𝑖1(1)≠𝑠^𝑖2(1). Indeed, if we suppose that 𝑠^𝑖1(1)=𝑠^𝑖2(1); then 𝑠^{𝑖1−𝑖2}=1;  that  is, 𝑠 has a cycle of length 𝑖₁−𝑖₂<𝑛 in spite of the condition. Conversely, if 𝑠^𝑖1(1)≠𝑠^𝑖2(1), then 𝑖₁≠𝑖₂, since 𝑠⁻¹ has not any cycle of length less than 𝑛 as well.
(b) Let positions (26) be independent. Both inequalities (27) hold and, as in (a), we have 𝑠^𝑖1(1)≠𝑠^𝑖2(1), 𝑠^𝑗1(1)≠𝑠^𝑗2(1); that is, the positions 𝜎((𝑖₁,𝑗₁)), 𝜎((𝑖₁,𝑗₁)) are independent as well.
Lemma B. Let 𝑠∈𝑆_𝑛 have no cycle of length less than 𝑛. Then (0,1)-matrix 𝑆 having 1’s on only positions𝑠¹(1),𝑠²(1),𝑠²(1),𝑠³(1)𝑠,…,_𝑛−1(1),𝑠^𝑛(1),𝑠^𝑛(1),𝑠¹(1) (28)is an incidence matrix of 𝑠.
Proof. Since 𝑠 has no cycles of length less than 𝑛, then {𝑠¹(1),…,𝑠^𝑛(1)} is a permutation of numbers {1,…,𝑛}. Thus the set of positions of 1’s of matrix 𝑆 coincides with the set of 1’s of the incidence matrix of 𝑠∶(1,𝑠(1)),…,(𝑛,𝑠(𝑛)).
Let 𝑃=𝑃_𝑛 be (0,1)-matrix with 1’s on positions (1,2),(2,3),…,(𝑛−1,𝑛),(𝑛,1) only.
Lemma C. Let 𝑠∈𝑆_𝑛 have not got any cycle of length less than 𝑛. If 𝑆 and 𝑆⁻¹ are the incidence matrices of 𝑠 and 𝑠⁻¹, then one has𝜎⁻¹(𝑆)=𝑃,𝜎⁻¹𝑆⁻¹=𝑃⁻¹. (29)Proof. Both formulas follow from Lemma B.
Noting that 𝜎(𝐼)=𝐼, where 𝐼 is the identity matrix, we conclude that 𝑆⁻¹+𝐼+𝑆=𝜎𝑃⁻¹+𝐼+𝑃. (30) Moreover, since, by the bijection 𝜎, to every diagonal (i.e., to every set of 𝑛 pairwise independent positions) of the matrix 𝛼𝑆⁻¹+𝛽𝐼+𝛾𝑆 corresponds one and only one diagonal of the matrix 𝛼𝑃⁻¹+𝛽𝐼+𝛾𝑃 with the same products of weights, then we have per𝛼𝑆⁻¹+𝛽𝐼+𝛾𝑆=per𝛼𝑃⁻¹+𝛽𝐼+𝛾𝑃. (31) Note that from the definition it follows that, for every matrix 𝑀∈Λ_𝑛(𝛼,𝛽,𝛾), we have a representation 𝑀=𝛼𝑆⁻¹+𝛽𝐼_𝑛+𝛾𝑆, (32) where 𝑆 is the incidence matrix of a substitution 𝑠. In case when 𝑠 has not got any cycle of length less than 𝑛, the matrix 𝑀 is a completely indecomposable matrix in Λ_𝑛(𝛼,𝛽,𝛾). Thus, by (31), all completely indecomposable matrices of Λ_𝑛(𝛼,𝛽,𝛾) have the same permanent, equal to per(𝛼𝑃⁻¹+𝛽𝐼_𝑛+𝛾𝑃).
In general, a substitution 𝑠 with the incidence matrix 𝑆 in (32) cannot have cycles of length less than 3. Indeed, if, for some 𝑖, we have either 𝑠(𝑖)=𝑖 or 𝑠(𝑠(𝑖))=𝑖, then in both cases 𝑠(𝑖)=𝑠⁻¹(𝑖) which means coincidence of positions 1’s of the matrices 𝑆 and 𝑆⁻¹ in the 𝑖th row.
Let 𝑠∈𝑆_𝑛 be an arbitrary substitution with cycles of length more than 2. Let 𝑠=^𝑟_𝑗=1𝑠_𝑗, (33) where 𝑠_𝑗∈𝑆_𝑙𝑗, 𝑙_𝑗≥3, and ∑^𝑟_𝑗=1𝑙_𝑗=𝑛, be the decomposition of 𝑠 in a product of cycles. Then the matrix 𝑀=𝛼𝑆⁻¹+𝛽𝐼_𝑛+𝛾𝑆 is a direct sum of the matrices 𝑀_𝑗=𝛼𝑆_𝑙⁻¹_𝑗+𝛽𝐼+𝛾𝑆_𝑙𝑗 such that, by (31), per𝑀_𝑗=per(𝛼𝑃⁻¹+𝛽𝐼_𝑙𝑗+𝛾𝑃) and we have per𝑀=^𝑟_𝑗=1per𝑀_𝑗=^𝑟_𝑗=1per𝛼𝑃⁻¹+𝛽𝐼_𝑙𝑗+𝛾𝑃. (34) It is left to notice that Minc [20] found a recursion (23) for per(𝛼𝐼_𝑛+𝛽𝑃+𝛾𝑃²) and, as well known, the multiplication of an 𝑛×𝑛 matrix by 𝑃⁻¹ does not change its permanent. Therefore, per(𝛼𝑃⁻¹+𝛽𝐼_𝑙𝑗+𝛾𝑃)=per(𝛼𝐼_𝑛+𝛽𝑃+𝛾𝑃²).

Example 5. Let us find 𝑝𝑠[Λ₁₁(−1,3,2)].

We have the following partitions of 11 with the parts not less than 3: 11=8+3=7+4=6+5=3+4+4=3+3+5. (35) According to (23), for 𝑎(𝑛)=𝑎(−1,3,2;𝑛), we have 𝑎(3)=16, 𝑎(4)=34 and for 𝑛≥5𝑎(𝑛)=3𝑎(𝑛−1)−2𝑎(𝑛−2)+6(−1)^𝑛. (36)

Using induction, we find 𝑎(𝑛)=2^𝑛+1,if𝑛isodd,2^𝑛+1+2,if𝑛iseven. (37)

Therefore, 𝑝𝑠Λ₁₁(=𝑎()(3)𝑎(8)(5)𝑎(6),𝑎−1,3,2)11,𝑎,𝑎₂(3)𝑎(5),𝑎(3)𝑎²(4={}.)4096,8224,8320,8704,16384,18496 (38)

In the following examples we calculate 𝑝-spectrum for arbitrary 𝑛.

Example 6. Let us find 𝑝𝑠[Λ_𝑛(−1,2,1)].

By induction, for 𝑎(𝑛)=𝑎(−1,2,1;𝑛), we have 𝑎(𝑛)=2,if𝑛isodd,4,if𝑛iseven. (39)

Further, again using induction, one can find that, if 𝑛 is even, then 𝑝𝑠Λ_𝑛(−1,2,1)=4,4²,…,4^{⌊(𝑛/4)⌋} (40) and, if 𝑛 is odd, then 𝑝𝑠Λ_𝑛(−1,2,1)=2,2⋅4,2⋅4²,…,2⋅4^{⌊((𝑛−3)/4)⌋}. (41)

Example 7. Analogously, in case of Λ_𝑛(−1,1,1), for 𝑎(𝑛)=𝑎(−1,1,1;𝑛), we have 𝑎(𝑛)=4,if𝑛≡0(mod6),−2,if𝑛≡3(mod6),1,otherwise,𝑝𝑠Λ_𝑛(=()−1,1,1)1,−2,4,…,−2^{⌊(𝑛−3)/3⌋},if(),()𝑛≡1,2mod31,−2,4,…,−2^{⌊(𝑛−6)/3⌋},(−2)^⌊𝑛/3⌋,if𝑛≡0(mod3). (42)(43)

It is interesting that, in case of 𝑛 multiple of 3, the permanent omits the value (−2)^{⌊(𝑛−3)/3⌋}.

4. Merriell-Type Theorems in a Subclass of Λ_𝑛(𝛼,𝛽,𝛾)

Note that in class Λ_𝑛(𝛼,𝛽,𝛾) the Minc-Bregman inequality and the Merriell theorem, generally speaking, do not hold even for positive 𝛼, 𝛽,  and  𝛾. Nevertheless, some restrictions on 𝛼, 𝛽,  and  𝛾 allow proving some analogs of the Merriell theorem. Recall that 𝑀(𝑛) (16) is attained in Λ³_𝑛. Denote by 𝑀_𝑛(𝛼,𝛽,𝛾) the maximal value of permanent in Λ_𝑛(𝛼,𝛽,𝛾).

Theorem 8. Consider a class Λ_𝑛(𝛼,𝛽,𝛾) with the numbers 𝛼, 𝛽, and 𝛾 satisfying “triangle inequalities” 0≤𝛼≤𝛽+𝛾,0≤𝛾≤𝛼+𝛽, (44) and the following additional conditions: 𝑎³(4)≤𝑎⁴(3),𝛼𝛾+𝛽(𝑎(3))^1/3≤(𝑎(3))^2/3, (45) where sequence {𝑎(𝑛)} is defined by recursion (23). Then, for 𝑛 multiple of 3, one has 𝑀_𝑛𝛼,𝛽,𝛾=(𝑎(3))^𝑛/3. (46)

Proof. Note that conditions (44)-(45) are satisfied, for example, in case 𝛼=𝛽=𝛾=1. Using induction, let us prove that 𝑎(𝑛)≤(𝑎(3))^𝑛/3. (47)
Indeed, for 𝑛=3, this inequality is trivial, while, for 𝑛=4, it follows from the first condition (45). Let it hold for 𝑛≤𝑚−1. Then, according to (23), we have 𝑎(𝑚)=𝛽𝑎(𝑚−1)+𝛼𝛾𝑎(𝑚−2)+𝛼^𝑚−1𝛼−𝛽−𝛾+𝛾^𝑚−1(𝑎(3𝛾−𝛼−𝛽≤𝛽))^(𝑚−1)/3+𝛼𝛾(𝑎(3))^(𝑚−2)/3=(𝑎(3))^(𝑚−2)/3𝛽(𝑎(3))^1/3≤(𝑎(3+𝛼𝛾))^(𝑚−2)/3(𝑎(3))^2/3=(𝑎(3))^𝑚/3. (48)
Note that, according to Theorem 4, the equality in (47) holds in a direct sum of (3×3)-matrices of Λ₃(𝛼,𝛽,𝛾) which corresponds to the partition 𝑛=3+3+⋯+3. Let now 𝐴∈Λ_𝑛(𝛼,𝛽,𝛾). By Theorem 4, there exists a partition of 𝑛 with the parts not less than 3, 𝑛=𝑛₁+⋯+𝑛_𝑚, such that per𝐴=^𝑚_𝑖=3𝑎𝑛_𝑖 (49) and, in view of (47), we have per𝐴≤^𝑚_𝑖=3𝑎(3)^𝑛𝑖/3=(𝑎(3))^𝑛/3. (50) This proves (46).

Example 9. Consider case 𝛽=𝛾−𝛼.

Let us find the values of 𝛼, depending on the magnitude of 𝛾, for which the conditions of Theorem 8 are satisfied. According to (23), we have 𝑎(3)=𝛼³+𝛾−𝛼³+𝛾³+3𝛼𝛾−𝛼𝛾=2𝛾³,𝑎(4)=𝛼⁴+𝛾−𝛼⁴+𝛾⁴+4𝛼𝛾−𝛼²𝛾+2𝛼𝛾²=2𝛼⁴+𝛾⁴. (51)

Thus the condition 𝑎³(4)≤𝑎⁴(3) means that 𝛼⁴+𝛾⁴≤2^1/3𝛾⁴ or 0≤𝛼≤2^1/3−1^1/4𝛾=0.7140199⋯𝛾. (52) And it is easy to verify that the second condition in (45) is satisfied as well. As a corollary, we obtain the following result.

Theorem 10. If (52) holds, then, for 𝑛 multiple of 3, one has 𝑀_𝑛𝛼,𝛾−𝛼,𝛾=2^𝑛/3𝛾^𝑛. (53)

Simple forms of sequence {𝑎(𝑛)} in Examples 5 and 6 allow supposing that in case 𝛽=𝛾−𝛼 (or symmetrical case 𝛽=𝛼−𝛾) sequence {𝑎(𝑛)} keeps a sufficiently simple form. We find this form in the following lemma.

Lemma 11. If 𝛽=𝛾−𝛼, then sequence {𝑎(𝑛)} which is defined by recursion (23) has the form 𝑎(𝑛)=2𝛾^𝑛2𝛼,𝑖𝑓𝑛𝑖𝑠𝑜𝑑𝑑,_𝑛+𝛾^𝑛,𝑖𝑓𝑛𝑖𝑠𝑒𝑣𝑒𝑛. (54)

Proof. Using induction with the base (51), suppose that (54) holds for 𝑚≤𝑛. Then, by (23), for even 𝑛, we have 𝑎(𝑛+1)=𝛾−𝛼𝑎(𝑛)+𝛼𝛾𝑎(𝑛−1)+2𝛼^𝑛𝛼𝛼−𝛾=2𝛾−𝛼^𝑛+𝛾^𝑛+2𝛼𝛾^𝑛+2𝛼^𝑛𝛼−𝛾=2𝛾^𝑛+1, (55) while, if 𝑛 is odd, then we have 𝑎(𝑛+1)=2𝛾−𝛼𝛾^𝑛+2𝛼𝛾𝛼^𝑛−1+𝛾^𝑛−1+2𝛼^𝑛𝛼−𝛾=2𝛼^𝑛+1+𝛾^𝑛+1. (56)

Let now 𝛼≥2^1/3−1^1/4𝛾=0.7140199⋯𝛾. (57)

Theorem 12. If (57) holds, then, for 𝑛 multiple of 4, one has 𝑀_𝑛𝛼,𝛾−𝛼,𝛾=2𝛼⁴+𝛾^4𝑛/4. (58)

Proof. From (51) and (57), we conclude that 𝑎(3)≤(𝑎(4))^3/4. (59)
Let us show that, for 𝑛≥3, 𝑎(𝑛)≤(𝑎(4))^𝑛/4. (60)
For 𝑛=4, inequality (60) is trivial. For 𝑛≥5, we have 𝛼^𝑛+𝛾^𝑛=𝛼^𝑛𝛾1+𝛼^𝑛≤𝛼^𝑛𝛾1+𝛼^4𝑛/4≤𝛼⁴+𝛾^4𝑛/4 (61) and thus, using Lemma 11, we have 𝑎(𝑛)≤2𝛼^𝑛+𝛾^𝑛<2^𝑛/4𝛼⁴+𝛾^4𝑛/4=(𝑎(4))^𝑛/4,𝑛≥3. (62)
Let now 𝐴∈Λ_𝑛(𝛼,𝛽,𝛾). By Theorem 4, there exists a partition of 𝑛 with the parts not less than 3, 𝑛=𝑛₁+⋯+𝑛_𝑚, such that per𝐴=^𝑚_𝑖=3𝑎𝑛_𝑖 (63) and, in view of (60), we have per𝐴≤^𝑚_𝑖=3𝑎(4)^𝑛𝑖/4=(𝑎(3))^𝑛/4 (64) with the equality in a direct sum of (4×4)-matrices of Λ₃(𝛼,𝛽,𝛾) which corresponds to the partition 𝑛=4+4+⋯+4.

Note that, if 𝛼≠(2^1/3−1)^1/4𝛾, then in Theorem 10 we have only maximizing matrix (up to a permutation of the rows and columns) which corresponds to the partition 𝑛=3+3+⋯+3; in Theorem 12 we also have only maximizing matrix (up to a permutation of the rows and columns) which corresponds to the partition 𝑛=4+4+⋯+4. It is interesting that, only in case of the equality 𝛼=𝜃𝛾, where 𝜃=(2^1/3−1)^1/4, Theorems 10–12 are true for every 𝑛 multiple of 12 with the equality of the maximums (2(𝛼⁴+𝛾⁴))^𝑛/4=2^𝑛/3𝛾^𝑛. Thus, up to a positive factor 𝛾,  the class Λ_𝑛(𝜃,1−𝜃,1),𝜃=2^1/3−1^1/4, (65) possesses an interesting extremal property: it contains (𝑛/12)+1 maximizing matrices (up to a permutation of the rows and columns), instead of only maximizing matrix, if 𝜃≠(2^1/3−1)^1/4.

Indeed, the number of the maximizing matrices (up to a permutation of the rows and columns) is defined by the number of the following partitions of 𝑛≡0 (mod 12): 𝑛=3+3+⋯+3,𝑛=(4+4+4)+(3+⋯+3),…,𝑛=4+⋯+4_3𝑖+3+⋯+3_{𝑛−12𝑖}𝑛,𝑖=0,1,…,.12 (66)

5. Estimate of Cardinality of 𝑃-Spectrum on Circulants in Λ³_𝑛 and Λ_𝑛(𝛼,𝛽,𝛾)

Denote by Δ³_𝑛⊂Λ³_𝑛 the set of the circulants in Λ³_𝑛. Note that a circulant 𝐴∈Δ³_𝑛 has a form 𝐴=𝑃^𝑖+𝑃^𝑗+𝑃^𝑘, 0≤𝑖<𝑗<𝑘≤𝑛, where 𝑃=𝑃_𝑛 is (0,1)-matrix with 1’s on positions (1,2),(2,3),…,(𝑛−1,𝑛),(𝑛,1) only. Multiplying 𝐴 by 𝑃^−𝑖 we obtain circulant 𝐵 of the form 𝐵=𝐼_𝑛+𝑃^𝑟+𝑃^𝑠 (67) with per𝐵=per𝐴. Since 𝐵 is defined by a choice of two different values 0<𝑟<𝑘≤𝑛, then trivially 𝑝𝑠Δ³_𝑛≤𝑛2<𝑛²2. (68)

Now we prove essentially more exact and practically unimprovable estimate.

Theorem 13. One has 𝑝𝑠Δ³_𝑛≤𝑛²+3.12 (69)

Proof. Let us return back to the general form 𝐴=𝑃^𝑖+𝑃^𝑗+𝑃^𝑘,0≤𝑖<𝑗<𝑘≤𝑛. (70)
Note that 𝐴 is defined by a choice of a vector (𝑖,𝑗,𝑘), but its rotation, that is, a passage to a vector of the form (𝑖+𝑙,𝑗+𝑙,𝑘+𝑙)(mod𝑛), does not change the magnitude of per𝐴. Indeed it corresponds to the multiplication 𝐴 by 𝑃^𝑙, and our statement follows from the equality per(𝑃^𝑙𝐴)=per𝐴. Besides, its reflection relative to some diameter of the imaginary circumference of the rotation, by the symmetry, keeps magnitude of the permanent. Since geometrically three points on the imaginary circumference define a triangle, then our problem reduces to a triangle case of the following general problem, posed by Professor Richard H. Reis (South-East University of Massachusetts, USA) in a private communication to Hansraj Gupta in 1978:

Let a circumference is split by the same 𝑛 parts. It is required to find the number 𝑅(𝑛,𝑘) of the incongruent convex 𝑘-gons, which could be obtained by connection of some 𝑘 from 𝑛 dividing points. Two 𝑘-gons are considered congruent if they are coincided at the rotation of one relatively other along the circumference and (or) by reflection of one of the 𝑘-gons relatively some diameter.

In 1979, Gupta [21] gave a solution of the Reis problem in the form 𝑅()=1𝑛,𝑘2𝑛−ℎ_𝑘2𝑘2+1𝑘×_{𝑑∣gcd()𝑘,𝑛}𝜑(𝑑)𝑛𝑑𝑘−1𝑑.−1 (71)
If we denote by Δ^𝑘_𝑛⊂Λ^𝑘_𝑛 the set of the circulants in Λ^𝑘_𝑛, then from our arguments it follows that 𝑝𝑠Δ^𝑘_𝑛≤𝑅(𝑛,𝑘). (72) In case 𝑘=3, from (71)-(72) we find 𝑝𝑠Δ³_𝑛≤𝑛²,12if(),𝑛𝑛≡0mod6₂−1,12if(),𝑛𝑛≡1,5mod6₂−4,12if(),𝑛𝑛≡2,4mod6₂+3,12if𝑛≡3(mod6), (73) and (69) follows.

Example 14. In case 𝑛=5 we have only two incongruent triangles corresponding to circulants 𝐼₅+𝑃+𝑃² and 𝐼₅+𝑃+𝑃³.

Nevertheless, the calculations give per(𝐼₅+𝑃+𝑃²)=per(𝐼₅+𝑃+𝑃³)=13. Thus 𝑝𝑠[Δ³₅]={13}, and |𝑝𝑠[Δ³₅]|=1.

Example 15. In case 𝑛=6 we have three incongruent triangles corresponding to circulants 𝐼₆+𝑃+𝑃², 𝐼₆+𝑃+𝑃³, and 𝐼₆+𝑃²+𝑃⁴.

The calculations give per(𝐼₆+𝑃+𝑃²)=20 and per(𝐼₆+𝑃+𝑃³)=17,  while per(𝐼₆+𝑃²+𝑃⁴)=36. Thus 𝑝𝑠[Δ³₆]={17,20,36}, and |𝑝𝑠[Δ³₆]|=3.

Note that a respectively large magnitude of per(𝐼₆+𝑃²+𝑃⁴) is explained by its decomposability in a direct product of circulants (𝐼₃+𝑃+𝑃²)⊗(𝐼₃+𝑃+𝑃²), such that per(𝐼₆+𝑃²+𝑃⁴)=(per(𝐼₃+𝑃+𝑃²))²=6²=36.

It is clear that, in case of circulants in Λ_𝑛(𝛼,𝛽,𝛾), the upper estimate (69) yields either the same estimate, if 𝛼=𝛽=𝛾, or ⌊(𝑛²+3)/4⌋, if 𝛼=𝛽≠𝛾 (and in the symmetric cases), or ⌊(𝑛²+3)/2⌋, if 𝛼, 𝛽, and 𝛾 are distinct numbers.

Add that, using a bijection, one can apply formula (71) to enumerate the two-color bracelets of 𝑛 beads, 𝑘 of which are black and 𝑛−𝑘 are white (see, e.g., the author’s explicit formulas for sequences A032279-A032282, A005513-A005516 in [15]).

6. Algorithm of Calculations of Upper Magnitudes of the Permanent in Λ³_𝑛

Theorem 2 allows, using some additional arguments, giving an algorithm of explicit expressions of upper magnitudes of the permanent in Λ³_𝑛 via numbers (14). For that we need the following lemma.

Lemma 16. For 𝑛₁,𝑛₂≥3, one has 𝑎𝑛₁+𝑛₂≤𝑎𝑛₁𝑎𝑛₂,𝑛₁,𝑛₂≥3. (74)

Proof. By the usual way, from (14) we find 𝑎(𝑛)=𝜑^𝑛+2+(−1)^𝑛𝜑^−𝑛, (75) where 𝜑=(√5+1)/2 is the golden ratio.
Denote 𝜀(𝑛)=(−1)^𝑛𝜑^−𝑛. Since 𝑛≥3, then |𝜀(𝑛)|<0.24, and, consequently, if 𝑛=𝑛₁+𝑛₂, then (2+𝜀(𝑛₁))(2+𝜀(𝑛₂))>1.76²>3. Therefore, we have 𝑎𝑛₁𝑎𝑛₂=𝜑^𝑛1+2+𝜀𝑛₁𝜑^𝑛2+2+𝜀𝑛₂>𝜑^𝑛1+𝑛2+3>𝜑^𝑛1+𝑛2+2+𝜀𝑛₁+𝑛₂=𝑎𝑛₁+𝑛₂. (76)

Note that, actually, the difference between the hand sides of (76) is more than 1.76(𝜑^𝑛1+𝜑^𝑛2).

Let now 𝑛≡𝑗(mod3), 𝑗=0, 1, and 2, and 𝑡∈𝐍. Let 𝑅(𝑚;𝜈) denote the set of all partitions of 𝑛 with parts more than or equal to 𝜈. For us an important role play cases 𝜈=3, 4. To 𝑟∈𝑅(𝑚;3), 𝜌∈𝑅(𝑚;4) put in a correspondence the sets 𝐻_𝑚;3(𝑟)=Π_{𝑟𝑖∈𝑟}𝑎_𝑟𝑖,𝐻_𝑚;4𝜌=Π_{𝜌𝑖∈𝜌}𝑎_𝜌𝑖. (77)

In case 𝑚=3, when 𝜌=∅, let us agree that 𝐻_3;4 is a singleton {6}.

Consider now the set 𝐿_𝑡^(𝑗)=𝐿_𝑡^(𝑗)(𝑛): 𝐿_𝑡^(𝑗)=^4𝑡+𝑗_𝑖=16^{(𝑛−𝑗−3𝑖)/3}𝑦∶𝑦∈𝐻_3𝑖+𝑗;4𝜌,𝑦≥9^3𝑡+𝑗6^{𝑖−4𝑡−𝑗}. (78)

Theorem 17 (algorithm of calculation of upper magnitudes of the permanent in Λ³_𝑛). If 𝑛≥4(3𝑡+𝑗), then the set 𝐿_𝑡^(𝑗) taken in decreasing order of its elements gives |𝐿_𝑡^(𝑗)| upper magnitudes of the permanent in Λ³_𝑛.

Proof. Note that the proof is the same for every value of 𝑗. Therefore, let us consider, say, 𝑗=0. From (74) it follows that, if 𝑟∈𝑅(𝑛;3) contains 𝜆₃ parts 3 and 𝜆₃≤(𝑛/3)−4𝑡, then, for 𝑦∈𝐻_{𝑛−3𝜆3;4}(𝜌), we have 6^𝜆3𝑦≤6^{(𝑛/3)−4𝑡}9^3𝑡. (79) This means that, for the formation of the list of all upper magnitudes of the permanent in Λ³_𝑛 in the condition 𝑛≥12𝑡, which are bounded from below by 6^{(𝑛/3)−4𝑡}9^3𝑡, it is sufficient to consider only a part of the spectrum containing numbers {6^𝜆3𝑦}, where 𝑦∈𝐻_{𝑛−3𝜆3;4}(𝜌) with the opposite condition 𝜆₃≥(𝑛/3)−4𝑡. From the equality 3𝜆₃+⋯+𝑛𝜆_𝑛=𝑛 with the condition 𝜆₃≥(𝑛/3)−4𝑡, we have 4𝜆₄+⋯+𝑛𝜆_𝑛𝑛≤𝑛−33−4𝑡=12𝑡. (80) Since 12𝑡 does not depend on 𝑛, there is only a finite assembly of such partition for arbitrary 𝑛. This ensures a possibility of the realization of the algorithm.
For the considered 𝑟∈𝑅(𝑛;3), for 𝜆₃≥(𝑛/3)−4𝑡, we have 𝐻_𝑛;3(𝑟)=6^{(𝑛−𝑚)/3}, where 𝑦∈𝐻_𝑚;4(𝜌), and 𝑚 has the form 𝑚=3𝑖, 1≤𝑖≤4𝑡. Thus we should choose only 𝐻_𝑛;3(𝑟)≥6^{(𝑛/3)−4𝑡}9^3𝑡, and this yields 𝑦≥9^3𝑡6^{(𝑛/3)−4𝑡}=9^3𝑡6^𝑖−4𝑡,𝑖=1,2,…,4𝑡. (81)

In order to use Theorem 17 for calculating the upper magnitudes 𝑀⁽¹⁾(𝑛)>𝑀⁽²⁾(𝑛)>⋯ of the permanent in Λ³_𝑛, in case, say, 𝑛≡0(mod3),(1)we write a list of partition of numbers 3𝑖, 𝑖=2,3,…,4𝑡, with the parts not less than 4;(2)we compare the corresponding values of 𝑦 with 9^3𝑡6^𝑖−4𝑡 and keep only 𝑦≥9^3𝑡6^𝑖−4𝑡;(3)after that, we take in decreasing order the numbers {𝑦6^{(𝑛/3)−𝑖}}.

Below we give the first 10 upper magnitudes 𝑀⁽¹⁾>𝑀⁽²⁾>⋯>𝑀⁽¹⁰⁾ of the permanent in Λ³_𝑛, for 𝑛≥24, via numbers {𝑎(𝑛)} (14). Consider 𝑀⁽¹⁾=𝑎(3)^𝑛/3=6^𝑛/3,if(),𝑎(4)𝑎(3)𝑛≡0mod3_(𝑛−4)/3=326^(𝑛−1)/3,if(),𝑎(4)𝑛≡1mod3₂𝑎(3)^(𝑛−8)/3=946^(𝑛−2)/3,if𝑛≡2(mod3). (82)

Formula (82) shows that 𝑀⁽¹⁾(𝑛) is attained in Λ³_𝑛. Moreover 𝑀⁽²⁾=𝑎(4)³𝑎(3)^{(𝑛−12)/3}=9616^𝑛/3,if(),𝑎(7)𝑎(3)𝑛≡0mod3_(𝑛−7)/3=31636^(𝑛−1)/3,if(),𝑎(5)𝑎(3)𝑛≡1mod3_(𝑛−5)/3=1366^(𝑛−2)/3,if𝑛≡2(mod3);𝑀⁽³⁾=𝑎(6)𝑎(3)^(𝑛−6)/3=596^𝑛/3,if(),𝑎(4)𝑛≡0mod3₄𝑎(3)^{(𝑛−16)/3}=27632^(𝑛−1)/3,if(),𝑎(8)𝑎(3)𝑛≡1mod3_(𝑛−8)/3=49636^(𝑛−2)/3,if𝑛≡2(mod3);𝑀⁽⁴⁾=𝑎(4)𝑎(5)𝑎(3)^(𝑛−9)/3=13624^𝑛/3,if(),𝑎(4)𝑎(6)𝑎(3)𝑛≡0mod3_{(𝑛−10)/3}=566^(𝑛−1)/3,if(),𝑎(4)𝑎(7)𝑎(3)𝑛≡1mod3_{(𝑛−11)/3}=31624^(𝑛−2)/3,if𝑛≡2(mod3);𝑀⁽⁵⁾=𝑎(9)𝑎(3)^(𝑛−9)/3=13636^𝑛/3,if(),𝑎(4)𝑛≡0mod3₂𝑎(5)𝑎(3)^{(𝑛−13)/3}=13616^(𝑛−1)/3,if(),𝑎(4)𝑛≡1mod3₅𝑎(3)^{(𝑛−20)/3}=81664^(𝑛−2)/3,if𝑛≡2(mod3);𝑀⁽⁶⁾=𝑎(4)𝑎(8)𝑎(3)^{(𝑛−12)/3}=496144^𝑛/3,if(),𝑎(5)𝑛≡0mod3₂𝑎(3)^{(𝑛−10)/3}=1696216^(𝑛−1)/3,if(),𝑎(4)𝑛≡1mod3₂𝑎(6)𝑎(3)^{(𝑛−14)/3}=546^(𝑛−2)/3,if𝑛≡2(mod3);𝑀⁽⁷⁾=𝑎(4)²𝑎(7)𝑎(3)^{(𝑛−15)/3}=31696^𝑛/3,if(),𝑎()𝑎(3)𝑛≡0mod310^{(𝑛−10)/3}=1256216^(𝑛−1)/3,if(),𝑎(4)𝑛≡1mod3₃𝑎(5)𝑎(3)^{(𝑛−17)/3}=39632^(𝑛−2)/3,if𝑛≡2(mod3);𝑀⁽⁸⁾=𝑎(4)⁶𝑎(3)^{(𝑛−24)/3}=816256^𝑛/3,if(),𝑎(4)𝑎(9)𝑎(3)𝑛≡0mod3_{(𝑛−13)/3}=13624^(𝑛−1)/3,if(),𝑎(5)𝑎(6)𝑛≡1mod3𝑎(3)^{(𝑛−11)/3}=65654^(𝑛−2)/3,if𝑛≡2(mod3);𝑀⁽⁹⁾=𝑎(4)³𝑎(6)𝑎(3)^{(𝑛−18)/3}=5616^𝑛/3,if(),𝑎(5)𝑎(8)𝑎(3)𝑛≡0mod3_{(𝑛−13)/3}=63761296^(𝑛−1)/3,if(),𝑎(5)𝑛≡1mod3₂𝑎(4)𝑎(3)^{(𝑛−14)/3}=1696144^(𝑛−2)/3,if𝑛≡2(mod3);𝑀⁽¹⁰⁾=𝑎(5)𝑎(7)𝑎(3)^{(𝑛−12)/3}=40361296^𝑛/3,if(),𝑎(4)𝑛≡0mod3₃𝑎(7)𝑎(3)^{(𝑛−19)/3}=31664^(𝑛−1)/3,if(),𝑎()𝑎(3)𝑛≡1mod311^{(𝑛−11)/3}=67672^(𝑛−2)/3,if𝑛≡2(mod3). (83)(84)(85)(86)(87)(88)(89)(90)(91)

7. Main Conjectural Inequality for Maximum of Permanent in Completely Indecomposable Λ³_𝑛-Matrices

Denote by Λ³_𝑛,1 the set of completely indecomposable matrices in Λ³_𝑛, that is, the set of Λ³_𝑛-matrices containing no Λ³_𝑚-submatrices. Let 𝜇₁(𝑛) denote the maximum of permanent in Λ³_𝑛,1. Our very plausible conjecture which we call “main conjectural inequality (MCI)” is the following.

Conjecture 18 (cf. Shevelev [11], pp. 165-166). For 𝑛₁,𝑛₂≥3, one has 𝜇₁𝑛₁+𝑛₂≤𝜇₁𝑛₁𝜇₁𝑛₂. (92)

In Lemma 16 we essentially proved that in subclass Λ³_𝑛 the MCI is valid.

Besides, in all known cases MCI holds. Moreover, as we shall see, our algorithm of calculation of the consecutive upper magnitudes (𝑀=𝑀₁>𝑀₂>⋯) of permanent in Λ³_𝑛, which is based on MCI, reproduces all Merriell’s and Bolshakov’s results for 𝑀₁ and 𝑀₂. Note also that, for sufficiently large 𝑛, the number of consecutive upper magnitudes of permanent in Λ³_𝑛 grows very quickly with every step of extension of the list of known 𝑝-spectrums for small 𝑛. For example, using the found by Bolshakov 𝑝𝑠[Λ³_𝑖], 𝑖≤8, we obtain, for sufficiently large 𝑛, 4, 7, and 11 upper values of 𝑝𝑠[Λ³_𝑛] in cases 𝑛=3𝑘, 3𝑘+1, and 3𝑘+2 respectively. Further, after calculating 𝑝𝑠[Λ³₉], the number of upper values of, for example, 𝑝𝑠[Λ³_3𝑘] increases by more than thrice.

8. Algorithm of Calculations of Upper Magnitudes of the Permanent in Λ³_𝑛 Based on MCI

Let 𝑛≡𝑗(mod3), 𝑗=0, 1, 2, and 𝑡∈𝐍. Let 𝑅(𝑚;𝜈) denote the set of all partitions of 𝑛 with parts more than or equal to 𝜈. For us an important role play cases 𝜈=3,4. To 𝑟∈𝑅(𝑚;3), 𝜌∈𝑅(𝑚;4) put in a correspondence sets 𝜋_𝑚;3(𝑟)=Π_{𝑟𝑖∈𝑟}𝑥_𝑟𝑖,𝜋_𝑚;4𝜌=Π_{𝜌𝑖∈𝜌}𝑥_𝜌𝑖, (93) where 𝑥_𝑠 runs through all values of permanent in set Λ³_𝑠,1 of completely indecomposable matrices in Λ³_𝑠 (in case 𝑚=3, when 𝜌=∅, let us agree that 𝜋_3;4 is a singleton {6}).

Consider now the set 𝐸_𝑡^(𝑗)=𝐸_𝑡^(𝑗)(𝑛): 𝐸_𝑡^(𝑗)=^4𝑡+𝑗_𝑖=16^{(𝑛−𝑗−3𝑖)/3}𝑦∶𝑦∈𝜋_3𝑖+𝑗;4𝜌,𝑦≥9^3𝑡+𝑗6^{𝑖−4𝑡−𝑗}. (94)

Theorem 19 (algorithm of calculation of upper magnitudes of the permanent in Λ³_𝑛). If 𝑛≥4(3𝑡+𝑗), then the set 𝐸_𝑡^(𝑗) taken in decreasing order of its elements gives |𝐸_𝑡^(𝑗)| upper magnitudes of the permanent in Λ³_𝑛.

Proof. We need three lemmas.

Lemma 20. For 𝑛≥4, one has 𝜇₁(𝑛)≤3^𝑛/2. (95)

Proof. Let, firstly, 𝑛≡0(mod4). Note that 𝜇₁(4)=𝐷₄=9. Using (92), we find 𝜇₁(𝑛)≤𝜇₁(4)𝜇₁(𝑛−4)≤⋯≤𝜇₁(4)^𝑛/4=3^𝑛/2. (96)
Let, furthermore, 𝑛≡𝑖(mod4), 𝑖=1, 2, 3. Note that, by (19), 𝜇₁(5)≤13<3^2.5. Therefore, using (92), we have 𝜇₁(𝑛)≤𝜇₁(4)^{(𝑛−5𝑖)/4}𝜇₁(5)^𝑖<3^{(𝑛−5𝑖)/2}3^2.5𝑖=3^𝑛/2. (97)

Lemma 21. Let 𝑛=3𝜆₃+4𝜆₄+⋯+𝑛𝜆_𝑛 be a partition of 𝑛 with the parts not less than 3. If 𝜆₃≤𝑙 and 𝑛 has the form 𝑛=3𝑙+4𝑚, then, for completely indecomposable matrices 𝐴_𝑖∈Λ³_𝑖, 𝑖=3,4,…,𝑛, one has ^𝑛_𝑖=3per𝐴_𝑖^𝜆𝑖≤6^𝑙9^𝑚. (98)

Proof. Using Lemma 20, we have ^𝑛_𝑖=3per𝐴_𝑖^𝜆𝑖≤6^𝜆3√3^{4𝜆4+⋯+𝑛𝜆𝑛}≤6^𝑙√3^𝑛−𝜆3=6√3^4𝑚=6^𝑙9^𝑚. (99)

Lemma 22. Let 𝑛=3𝜆₃+4𝜆₄+⋯+𝑛𝜆_𝑛 and 𝜆₃≤𝑛−4𝑗3−4𝑡,𝑛≥43𝑡+𝑗, (100) where 𝑡 is a nonnegative integer and 𝑗 is the residue of 𝑛 modulo 3,  𝑗=0, 1, 2; then, for completely indecomposable matrices 𝐴_𝑖∈Λ³_𝑖, 𝑖=3,4,…,𝑛, one has ^𝑛_𝑖=3per𝐴_𝑖^𝜆𝑖≤6^{((𝑛−4𝑗)/3)−4𝑡}9^3𝑡+𝑗. (101)

Proof. Put 𝑙=((𝑛−4𝑗)/3)−4𝑡, 𝑚=(𝑛−3𝑙)/4=3𝑡+𝑗. Now the lemma follows from Lemma 21.

It is left to note that, after these lemmas, the proof of Theorem 19 is the same as proof of Theorem 17.

Note that the use of this algorithm is based on the upper values of 𝑝-spectrum for smaller values of 𝑛.

Consider, for example, case 𝑡=0, 𝑗=2. According to (94), we have 𝐸₀⁽²⁾=²_𝑖=16^{(𝑛−2−3𝑖)/3}𝑦∶𝑦∈𝜋_3𝑖+𝑗;4𝜌,𝑦≥81⋅6^𝑖−2=6^(𝑛−5)/3per𝐴,𝐴∈Λ³₅∶6per𝐴≥81∪81⋅6^(𝑛−8)/3. (102)

Note that the second set is a simpleton, since, by MCI, 𝜇₁(8,3)≤(𝜇₁(4,3))²=81. Since, by (19), 𝑀(5)=13<81/6, then the first set in (102) is empty. Thus 𝐸₀⁽²⁾=𝐸₀⁽²⁾(𝑛) is simpleton: 𝐸₀⁽²⁾=81⋅6^(𝑛−8)/3 (103) and we have 𝑀⁽¹⁾(𝑛)=81⋅6^(𝑛−8)/3,𝑛≥8, (104) which corresponds to Merriell’s result in case 𝑛≡2(mod3).

Further research of the set (94), using (19), gives the following results.

(1) Consider 𝑗=0, 𝑛≥24: 𝑀⁽¹⁾(𝑛)=6^𝑛/3,𝑀⁽²⁾(𝑛)=9616^𝑛/3,𝑀⁽³⁾(𝑛)=596^𝑛/3,𝑀⁽⁴⁾(𝑛)=13624^𝑛/3. (105)

The continuation of this list requires knowing 𝑝𝑠[Λ³₉]. Note that a more detailed analysis shows that, after calculation of 𝑝𝑠[Λ³₉], in this case one can obtain the first 12+|𝐺| upper magnitudes of the permanent in Λ³_𝑛, where 𝐺=𝑝𝑠[Λ³₉]∩([69,116]⧵{72,78,102,108}).

(2) Consider 𝑗=1, 𝑛≥28: 𝑀⁽¹⁾(𝑛)=326^(𝑛−1)/3,𝑀⁽²⁾(𝑛)=896^(𝑛−1)/3,𝑀⁽³⁾(𝑛)=31636^(𝑛−1)/3,𝑀⁽⁴⁾(𝑛)=27632^(𝑛−1)/3,𝑀⁽⁵⁾(𝑛)=566^(𝑛−1)/3,𝑀⁽⁶⁾(𝑛)=13615^(𝑛−1)/3,𝑀⁽⁷⁾(𝑛)=1696216^(𝑛−1)/3. (106)

It is interesting that in this case 𝑝𝑠[Λ³₉] is not used up to 𝑀⁽⁷⁾, but the continuation of this list requires knowing 𝑝𝑠[Λ³₁₀].

(3) Consider 𝑗=2, 𝑛≥32: 𝑀⁽¹⁾(𝑛)=946^(𝑛−2)/3,𝑀⁽²⁾(𝑛)=1366^(𝑛−2)/3,𝑀⁽³⁾(𝑛)=2⋅6^(𝑛−2)/3,𝑀⁽⁴⁾(𝑛)=1396^(𝑛−2)/3,𝑀⁽⁵⁾(𝑛)=49636^(𝑛−2)/3,𝑀⁽⁶⁾(𝑛)=436^(𝑛−2)/3,𝑀⁽⁷⁾(𝑛)=31624^(𝑛−2)/3,𝑀⁽⁸⁾(𝑛)=81664^(𝑛−2)/3,𝑀⁽⁹⁾(𝑛)=546^(𝑛−2)/3,𝑀⁽¹⁰⁾(𝑛)=1196^(𝑛−2)/3,𝑀⁽¹¹⁾(𝑛)=39632^(𝑛−1)/3. (107)

Note that the method not only gives a possibility to calculate the upper magnitudes of the permanent in Λ³_𝑛, but also indicates those direct products on which they are attained. For example, in (107) 𝑀₉ is attained on direct products of some matrices 𝐴_𝑖∈Λ³_𝑖∶𝐴₈⊗𝐴₃⊗⋯⊗𝐴_{3()𝑛−8/3},𝐴₄⊗𝐴₇⊗𝐴₃⊗⋯⊗𝐴_{3()𝑛−11/3},𝐴₄⊗𝐴₄⊗𝐴₆⊗𝐴₃⊗⋯⊗𝐴_{3(𝑛−14)/3}. (108)

Note also that the comparison of (105)–(107) with (82)–(91) shows that the following calculated 𝑀^(𝑖) are attained in Λ³_𝑛, 𝑛≥32∶

in case 𝑛≡0mod3, 𝑀⁽¹⁾,𝑀⁽²⁾,𝑀⁽³⁾,𝑀⁽⁴⁾; (109) in case 𝑛≡1mod3, 𝑀⁽¹⁾,𝑀⁽³⁾,𝑀⁽⁴⁾,𝑀⁽⁵⁾,𝑀⁽⁶⁾,𝑀⁽⁷⁾ (110) (but 𝑀⁽²⁾ is not attained);

in case 𝑛≡2mod3, 𝑀⁽¹⁾,𝑀⁽²⁾,𝑀⁽⁵⁾,𝑀⁽⁷⁾,𝑀⁽⁸⁾,𝑀⁽⁹⁾,𝑀⁽¹¹⁾ (111) (but 𝑀⁽³⁾, 𝑀⁽⁴⁾, 𝑀⁽⁶⁾, and𝑀⁽¹⁰⁾ are not attained).

9. Algorithm for Testing the Parity of Values of the Permanent in Λ³_𝑛

It seems that, among all known methods of calculation of the permanent, only Ryser’s method (cf. [1], chapter 7) could be used for creating an algorithm for testing the parity of values of the permanent. Let 𝐴 be 𝑛×𝑛-matrix. Let 𝐴_𝑟 be a matrix which is obtained by changing some 𝑟 columns of 𝐴 by zero columns. Denote by 𝑆(𝐴_𝑟) the product of row sums of 𝐴_𝑟. Then, by Ryser’s formula, we have per𝐴=𝑆𝐴₀−𝑆𝐴₁+𝑆𝐴₂+()−⋯−1^𝑛−1𝑆𝐴_𝑛−1. (112)

Let now 𝐴 have integer elements. Introduce the following matrix function: Υ(𝐴)=1,ifallrowsumsof𝐴areodd,0,otherwise. (113)

From (112) we have per𝐴≡Υ𝐴₀−Υ𝐴₁+Υ𝐴₂𝐴+⋯+Υ_𝑛−1(mod2). (114)

Using (114), let us create an algorithm of a search of the odd values of the permanent in Λ³_𝑛. Since, evidently, per𝐴≡det𝐴(mod2), then 𝐴 should have pairwise distinct columns. Note that cases 𝑛≡𝑗(mod 3), 𝑗=0,1,2, are considered by the same way. Suppose, say, 𝑛=3𝑡. According to (114), we are interested in only cases when, after removing 𝑟≥1 columns of 𝐴, all row sums will be odd. Suppose that, after removing 𝑟 columns of 𝐴, we have that 𝑝 sums remain equal to 3 and 𝑛−𝑝 sums equal to 1. This means that the total number of the removed 1’s equals 2(𝑛−𝑝)=6𝑡−2𝑝. Since, removing a column, we remove three 1’s, then the number of the removed columns equals 𝑟=2𝑡−((2𝑝)/3). Thus 𝑝=3𝑚 and 𝑟=2(𝑡−𝑚), 𝑚=0,1,…,𝑡−1. However, if 𝑚=𝑡−1, then 𝑟=2. By the condition, these two columns are distinct; therefore, we conclude that at least one row sum equals 2. The contradiction shows that the testing sequence is 𝑟=4,6,…,2𝑡. In cases 𝑛≡1,2(mod3) we obtain the same testing sequence.

Example 23. Let us check the parities of values of the permanent of circulants in Δ³₇⊂Λ³₇.

In this case 𝑡=⌊7/3⌋=2 and, therefore, the testing sequence contains only term 𝑟=4. Note that matrix 𝐴_𝑟 has all odd rows if and only if one row sum equals 3 and each of the 6 other row sums equals 1. Indeed, let, after removing 4 columns of 𝐴, 𝑝 sums remain equal to 3 and 7−𝑝 sums equal to 1. This means that the total number of the removed 1’s equals 2(7−𝑝) and the number of the removed columns equals 𝑟=4=(14−2𝑝)/3; that is, 𝑝=1. Moreover, since in a circulant all rows are congruent shifts of the first one, it is sufficient to consider the case when precisely the first row sum equals 3 and others equal 1 (the multiplication on 7 does not change the parity of the result). This opens a possibility of a momentary handy test on the parity every circulant of class Δ³₇. This test consists of removing all four columns beginning with 0. If now every row 2,…,7 has one 1, then the permanent is even; otherwise, it is odd. We check now directly that from ⁷₃=35 circulants exactly 21 ones have odd permanent.

Remark 24. In 1967, Ryser [9] did a conjecture that the number of the transversals of a Latin square from elements 1,…,𝑛 (i.e., the number of subsets of its 𝑛 pairwise distinct elements, with none in the same row or column) has the same parity as 𝑛. If 𝑛 is even, then the conjecture has been proved by Balasubramanian [8]. Besides, in [8] Balasubramanian posed a conjecture for the parity of a sum of permanents, such that the truth of this conjecture yields Ryser’s hypothesis for odd 𝑛. In the same year (1990), using the result of Example 23, the author disproved Balasubramanian’s conjecture. It is interesting that soon Parker (see [22], p. 258) indeed found several Latin squares of order 7 with even number of transversals. Add that later ([7]) we found even an infinite set of counterexamples to the Balasubramanian conjecture.

10. Open Problems

(1)To prove the MCI (Section 7).(2)Consider class Λ_𝑛(1,1+𝑎,1+𝑏), where 0≤𝑎≤𝑏<(4𝑒−9)/6 (cf. [23], pp. 171-172). Since Λ_𝑛(1,1,1)=Λ³_𝑛, then Voorhoeve’s lower estimate (81/32)(4/3)^𝑛 for permanent of Λ³_𝑛-matrices trivially holds for matrices in Λ_𝑛(1,1+𝑎,1+𝑏). It is clear that, for 𝑎>0,𝑏>0, an essentially stronger lower estimate should exist. However, using van der Waerden-Egorychev-Falikman theorem to class Λ_𝑛(1/(3+𝑎+𝑏),(1+𝑎)/(3+𝑎+𝑏),(1+𝑏)/(3+𝑎+𝑏)) of doubly stochastic matrices, for the permanent of Λ_𝑛(1,1+𝑎,1+𝑏)-matrices, we obtain even weaker lower estimate of the order 𝐶₁√𝑛((3+𝑎+𝑏)/𝑒)^𝑛≪𝐶(4/3)^𝑛 in the case 𝑎+𝑏∈(0,(4𝑒−9)/3). The problem is to find a stronger lower estimate for the permanent in Λ_𝑛(1,1+𝑎,1+𝑏) in this case.(3)Let 𝑀 be a circulant of order 𝑛 with integer elements (cf. [23], pp. 115-116). We conjecture that, for every integer 𝑚, we have per𝑀≡(−1)^𝑛per(𝑚𝐽_𝑛−𝑀)(mod𝑛), where 𝑛×𝑛-matrix 𝐽_𝑛 consists of 1’s only. In [7] the conjecture was proved for every 𝑚 and prime 𝑛. Thus the question is open in case of arbitrary composite number 𝑛.(4)Let us call two Latin rectangles equivalent if the sets of their elements in the corresponding columns are the same. Note that one can treat numbers |Λ³_𝑛| as the numbers of equivalence classes of 3×𝑛 Latin triangles. Let 𝐴=𝐼_𝑛+𝑃+𝑃². It could be proved that the cardinality of the corresponding equivalent class is 2^𝑛+6+2(−1)^𝑛. To find the cardinality of the equivalent class which is defined by matrix 𝐼_𝑛+𝑃+𝑃³.

Acknowledgment

The author is grateful to both of the anonymous referees for several suggestions which promoted improvement in the text.