Curtz's six-residue denominator conjecture

OEIS A189733, revision 36, states:

a(n) = denominator(B(0,n)). Conjecture: a(6n)=1, a(1+6n)=1+3n, a(2+6n)=1+2n, a(3+6n)=2+3n, a(4+6n)=1, a(5+6n)=6+6n. a(n) = 2*a(n-6) - a(n-12).

All indices are natural numbers. Define the actual rational triangle by T(m,0)=0, T(m,1)=(-1)^m/(m+1), and T(m,k+2)=T(m,k+1)+T(m+1,k), with boundary cases taking precedence. The public triangle implements these three branches, and the public a is exactly its zeroth-row reduced denominator. In particular, a zero rational has denominator one. The theorem result is the equality of the six-tuples for every natural k, including zero, with no extra premise.

This is Paul Curtz's externally named OEIS conjecture. Its exact statement and quantifiers appear in issue #25342. The proof concerns the six formulas; no separate theorem about the entry's empirical generating function or the lower-triangular extension is claimed.

The directly read source JSON, revision 36, still labels the six formulas as a conjecture and supplies no proof annotation. The bounded literature screen recorded in #25342 checked arXiv and related entries A189731 and A242926. No prior exact settlement was found by identifier and structural searches in the repository, including private Lean declarations, or by authenticated GitHub Lean code search for A189733. Pinned Mathlib provides the finite polynomial sums, derivative rules and reduced-rational denominator lemmas, but no source-triangle identity. Unindexed proofs and worldwide priority remain unverified; this is not an exhaustive literature claim.

Use Q(0)=0, Q(1)=1, Q(n+2)=Q(n+1)-X Q(n). The finite coefficient moment J(m,P) sums coeff(P,i)/(m+i+1) over the support. Multiplication by X increments the moment index, proving T(m,n)=(-1)^m J(m,Q(n)) by the literal recursion.

The companion polynomials start at L(0)=2 and L(1)=1 with the same recurrence. Their identity L(n+1)=Q(n+1)-2X Q(n) proves L'(n+1)=-(n+1)Q(n). Applying J(0,.) to the derivative yields the endpoint difference. Since L(n+1)(0)=1, this gives T(0,n)=(1-L(n+1)(1))/(n+1).

The recurrence at X=1 has period six, with values 2,1,-1,-2,-1,1. Consequently the zeroth-row values are respectively 0, 1/(3k+1), 1/(2k+1), 1/(3k+2), 0, and -1/(6k+6). All displayed nonzero denominators are positive, so reduction gives the six claimed denominators.

A natural index k for which the six actual reduced denominators differ from the displayed tuple would refute the assertion. The zeroth-row values and denominators are taken from the literal recursive array, including the denominator-one convention for zero.

The source-triangle moment identity and the companion derivative identity are intermediate propositions on the proof path to the public theorem D5/S1/Recurrence/Residue/OEISA189733DenominatorConjecture.result. The argument derives an exact formula for every natural index before extracting reduced denominators; no finite table supplies the universal claim.

Theorem: the target consists of all six formulas, universally over Nat. The associated formal assertion has no cutoff and no additional hypothesis.

No exhaustive literature coverage or worldwide priority is asserted. Unindexed proofs remain outside the bounded source and literature screen recorded in issue #25342. The generating-function comment and the lower-triangular extension are outside this resolution.