[Paper Review] A Note on Boolean Lattices and Farey Sequences
This paper establishes monotone bijections between the standard Farey sequence 𝒻ₘ and the left and right halfsequences of the Farey subsequence ℱ(ℬ(2m), m) associated with rank-m elements in the Boolean lattice of subsets of a 2m-set. It derives novel combinatorial identities involving binomial coefficients indexed by fractions in these sequences, revealing deep structural connections between Boolean lattices and Farey sequences through rank-based subset enumeration and modular arithmetic constraints.
We establish monotone bijections between the Farey sequences of order m and the halfsequences of Farey subsequences associated with the rank m elements of the Boolean lattice of subsets of a 2m-set. We also present a few related combinatorial identities.
Motivation & Objective
- To establish a structural correspondence between the standard Farey sequence 𝒻ₘ and the halfsequences of the Farey subsequence ℱ(ℬ(2m), m) in the Boolean lattice of subsets of a 2m-element set.
- To explore the combinatorial properties of ℱ(ℬ(2m), m), particularly its partitioning into rank-based subsets and the enumeration of such subsets via binomial coefficients.
- To derive new combinatorial identities by leveraging the bijections and symmetry properties of the Farey subsequence ℱ(ℬ(2m), m), especially under order-reversing maps.
- To extend these results to the standard Farey sequence 𝒻ₘ via the established bijections, yielding analogous identities for 𝒻ₘ.
Proposed method
- Constructs a monotone bijection between the Farey sequence 𝒻ₘ and the left and right halfsequences of ℱ(ℬ(2m), m), using modular inverses to define predecessor and successor fractions.
- Applies the order-reversing map 𝒻(ℬ(n), m) → 𝒻(ℬ(n), n−m) defined by h/k ↦ (k−h)/k, which preserves bijectivity and symmetry.
- Uses the poset structure of the Boolean lattice ℬ(n), particularly order ideals and filters generated by rank-m elements, to partition the subposet ℱ(𝕀(a′) ∩ ℬ(n)⁽¹⁾).
- Derives a key identity by equating two expressions for the cardinality 2ⁿ − 2ⁿ⁻ᵐ, one via subset enumeration and the other via summing binomial coefficients over fractions in ℱ(ℬ(n), m).
- Applies the symmetry between ℱ(ℬ(n), m) and ℱ(ℬ(n), n−m) to derive Proposition 2, which equates two sums of binomial coefficients to 2ⁿ − 2ᵐ − 2ⁿ⁻ᵐ + 1.
- Specializes the general case to n = 2m, leading to Proposition 7 and Corollary 8, which present symmetric identities for ℱ(ℬ(2m), m) and 𝒻ₘ involving sums over fractions in specific intervals and binomial coefficient products.
Experimental results
Research questions
- RQ1How are the Farey sequence 𝒻ₘ and the halfsequences of the Farey subsequence ℱ(ℬ(2m), m) related via monotone bijections?
- RQ2What combinatorial identities emerge from the partitioning of the subposet ℱ(𝕀(a′) ∩ ℬ(n)⁽¹⁾) in the Boolean lattice ℬ(n)?
- RQ3How does the symmetry between ℱ(ℬ(n), m) and ℱ(ℬ(n), n−m) via the map h/k ↦ (k−h)/k lead to balanced binomial coefficient sums?
- RQ4What identities for the standard Farey sequence 𝒻ₘ can be derived from the bijections established for ℱ(ℬ(2m), m)?
- RQ5What role do modular inverses and floor functions play in determining the predecessor and successor of a fraction in ℱ(ℬ(2m), m)?
Key findings
- A monotone bijection exists between the Farey sequence 𝒻ₘ and the left and right halfsequences of ℱ(ℬ(2m), m), with predecessor and successor fractions defined via modular inverses of h modulo (k−h).
- The sum of binomial coefficients ∑_{h/k ∈ ℱ(ℬ(2m), m), 0 < h/k < 1} ∑_{s ≤ ⌊min{m/h, m/(k−h)}⌋} ₑ(m, sh)⋅₇(m, s(k−h)) equals 2²ᵐ − 2ᵐ⁺¹ + 1.
- The sum over fractions with h/k < 1/2 equals the sum over h/k > 1/2, and both equal 2²ᵐ⁻¹ − 2ᵐ − ½⋅₇(2m, m) + 1.
- The sum over h/k ∈ (0, 1/3) and (1/3, 1/2) of terms involving ₑ(m, s·h) + ₑ(m, s·(k−2h)) equals the sum over h/k ∈ (1/2, 2/3) and (2/3, 1), with the common value 2²ᵐ⁻¹ − 2ᵐ − ½⋅₇(2m, m) − ∑_{t≤⌊m/2⌋} ₑ(m, 2t)⋅₇(m, t) + 1.
- The standard Farey sequence 𝒻ₘ satisfies the identity ∑_{h/k ∈ 𝒻ₘ, 0 < h/k < 1} ∑_{s ≤ ⌊m/k⌋} ₑ(m, sh)⋅₇(m, sk) = 2²ᵐ⁻¹ − 2ᵐ − ½⋅₇(2m, m) + 1.
- For 𝒻ₘ, the sum over h/k < 1/2 of ₑ(m, sk)(ₑ(m, sh) + ₑ(m, s(k−h))) equals the sum over h/k > 1/2, and both equal 2²ᵐ⁻¹ − 2ᵐ − ½⋅₇(2m, m) − ∑_{t≤⌊m/2⌋} ₑ(m, 2t)⋅₇(m, t) + 1.
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This review was created by AI and reviewed by human editors.