Ternary-partition numerator at minus one
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For a partition λ of n, its subsum polynomial is sp(λ,x)=∏ᵢ(1+x^λᵢ). Sum 1/sp over the specified partition family, combine over the denominator ∏ⱼ(1+xʲ)^⌊n/j⌋ for allowed parts j, then cancel the greatest common factor of the individual numerator summands.
For ternary partitions, whose parts are powers of three, is num_T(n,−1)=3^v₃(n!) for every n≥1, where v₃ is the exponent of 3?
Why it matters and what counts as progress
Why it matters: The evaluation predicts an exact prime-power value from the arithmetic of the factorial.
What counts as progress: A proof of the special-value formula for every n, including the initial cases, or a rigorously evaluated exception.
How the problem developed
- 11 May 2026 · Conjecture [1]
Cristina Ballantine, George Beck, Brooke Feigon, and Kathrin Maurischat Introduced reciprocal subsum polynomials and posed special-value conjectures for the associated numerator polynomials, including the ternary case.
- 20 May 2026 · Special value [2]
Evan Chen, Ken Ono, and Jujian Zhang, with AxiomProver-generated Lean proofs Proved the ternary special-value conjecture at x=−1 as part of six formalized coprimality, divisibility, special-value, and recurrence results.
AI contribution
Reciprocals of Partition Polynomials
AxiomProver generated Lean proofs of Ballantine–Beck–Feigon–Maurischat Conjectures 2, 5, 7, 8, 9 and the intended corrected reading of 10, covering coprimality, divisibility, special values and a recurrence. It also found an n=4 counterexample to a printed binary log-concavity claim.
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