Odlyzko’s Sign-Alternation Theorem — Complete Proof presents a self-contained reconstruction of the Stanton–Zeilberger–O’Hara route to the sign alternation theorem attributed to Odlyzko. The manuscript is written as a complete proof package: it derives the KOH identity from the corrected O’Hara–Goodman product-decomposition corridor, turns that corridor into a q-enumeration recurrence, expands the recurrence into the Stanton–Zeilberger multiplicity form, proves the finite sign theorem by a sign-cone argument, and finally passes to the coefficientwise limit giving Odlyzko’s reciprocal q-factorial product. The theorem proved is the following sign-alternation statement. For every integer k ≥ 0, the Maclaurin coefficients of 1 / ( (1 + q) (1 + q + q²) ⋯ (1 + q + ⋯ + qᵏ⁻¹) ) alternate in sign. Equivalently, with (q) ₖ = ∏₁≤j≤k (1 − qʲ), the target is (1 − q) ᵏ / (q) ₖ ∈ 𝒜, where 𝒜 is the cone of formal power series 𝒜 = f (q) =∑ₙ≥0 aₙqⁿ: (−1) ⁿaₙ ≥ 0 for every n. The manuscript treats the result as a known theorem in the Stanton–Zeilberger–O’Hara lineage, and its contribution is a complete reconstruction of the proof corridor in one coherent manuscript. The proof does not merely quote the KOH identity as an external black box: it reconstructs the corrected product-decomposition mechanism that produces KOH, then proves the sign theorem and limiting passage explicitly. The proof spine is: corrected O’Hara–Goodman corridor⇒ q-enumeration recurrence⇒ KOH identity⇒ finite sign theorem⇒ coefficientwise limit⇒ Odlyzko sign alternation. The first structural layer is the alternating sign cone. The cone 𝒜 is shown to be closed under finite sums, Cauchy products, even shifts q²ʳ, multiplication by powers (1 − q) ˢ for s ≥ 0, and coefficientwise limits. These closure properties are the formal algebraic engine used in the final sign argument. The second layer introduces Gaussian polynomials G (n, k) = n+k choose kq= ∏₁≤i≤k (1 − qⁿ⁺ⁱ) / ∏₁≤i≤k (1 − qⁱ), with the convention G (n, k) =0 for na, the corrected spread is spread (p) = max₁≤j≤a+2 (pⱼ − pⱼ₋₂). The active set is M (p) = j: pⱼ − pⱼ₋₂ = spread (p), and the degree is computed by decomposing M (p) into maximal consecutive intervals D and setting deg (p) = ∑D floor ( (|D|+1) /2). The corrected endpoint convention is essential. Endpoints 1 and a+2 may attach to active intervals, but the deletion operation must use the internal active index h = max (M (p) ∩ 2, …, a+1). This internal-active deletion avoids the endpoint ambiguity created by the corrected spread. The manuscript proves that the corrected endpoint convention is regression-safe, that active intervals transform correctly under deletion, and that the auxiliary boundary states U (0, −1) and U (0, −2) are exactly the terminal singleton states required by the corrected decomposition. The central product-decomposition theorem is the rank-shifting bijection U (b, a, m, d) ≅ U (b−md, a−2d, ≤m−1) × U (am+2m−2b, d), with rank shift |p| = |q| + |r| + 2bd − md (d+1). Here U (b, a, m, d) denotes the set of partitions in U (b, a) with corrected spread m and degree d. The proof gives the deletion map, the insertion inverse, the monotonicity of the deleted charge tuple 0 ≤ r₁ ≤ r₂ ≤ ⋯ ≤ rd ≤ am + 2m − 2b, the active-interval transport table for the degree drop, the tail-maximality lemma ensuring that insertion creates the maximal internal active index, and the full rank-shift calculation. Taking ranks in this product decomposition gives the q-enumeration recurrence. If v (b, a;m) = ∑∈ₔ (₁, ₀, ≤₌) q^|p|, then for a, b ≥ 0 and m ≥ 1, v (b, a;m) = v (b, a;m−1) ∑₃≥₁ q²ᵇᵈ⁻ᵐᵈ⁽ᵈ⁺¹⁾ G (am+2m−2b, d) v (b−md, a−2d;m−1). The d=0 branch is treated separately as the lower-layer term v (b, a;m−1), avoiding any ambiguity in the Gaussian convention. Iterating this recurrence from G (n, k) = G (k, n) = v (k, n;k) gives the KOH identity in Stanton–Zeilberger multiplicity form. If λ = (1^d₁ 2^d₂ ⋯ k^dₖ) ⊢ k, and Dᵣ = ∑ₒ≥ₑ dₛ, then 2n (λ) = ∑₁≤r≤k (Dᵣ² − Dᵣ), and the recurrence expands to G (n, k) = ∑⏚⊢₊ q²ⁿ⁽λ⁾ ∏₁≤s≤k G (Aₛ, dₛ), where Aₛ = sn − 2 (k−s) + 2∑ₓ>ₒ (t−s) dₜ. This is the precise KOH identity needed for the sign-alternation proof. The finite sign theorem is then proved from KOH. If Nk is even, then (1 − q) ᵐ G (N, k) ∈ 𝒜, where m = min (⌈k/2⌉, ⌊ (N+1) /2⌋). For N=2r, each KOH summand is analyzed through the alpha deficit α (λ) = m − ∑ₛ⌈dₛ/2⌉. For r ≥ ⌈k/2⌉, a packet argument gives α (λ) ≥ 0 for every λ⊢k. For r d. Taking N through even integers and using cone closure under coefficientwise limits gives (1 − q) ^⌈k/2⌉ / (q) ₖ ∈ 𝒜. Multiplication by the remaining cone element (1 − q) ^k−⌈k/2⌉ yields (1 − q) ᵏ / (q) ₖ ∈ 𝒜, which is exactly Odlyzko’s sign-alternation statement. The appendix records the sharpness of the exponent threshold. For k≥1 and e≥0, (1 − q) ᵉ / (q) ₖ ∈ 𝒜 ⇔ e ≥ ⌈k/2⌉. Sufficiency follows from the finite sign theorem and the coefficientwise limit. Necessity follows from the dominant pole at q=1: if e<⌈k/2⌉, then q=1 is the unique dominant pole of order k−e, while every other root of unity contributes pole order at most ⌊k/2⌋. Therefore the coefficients are eventually positive, including along odd degrees, contradicting membership in 𝒜. This manuscript is intended as a polished proof reconstruction and publication-candidate proof package. Its mathematical core is the corrected O’Hara–Goodman corridor, internal-active product decomposition, q-enumeration recurrence, explicit KOH iteration, finite sign-cone theorem, and coefficientwise limiting exit.
Maximus Shlygin (Tue,) studied this question.