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#36 — Optimal constant for partition differencesno
OPENHARDnumber theoryadditive combinatorics
For all sufficiently large $N$, if $A\sqcup B=\{1,\ldots,2N\}$ is a partition into two equal parts, then there is some $x$ such that the number of solutions to $a-b=x$ with $a\in A$ and $b\in B$ is at least $cN$. Find the optimal constant $c>0$.
Notes: Current best bounds: 0.379005 < c < 0.380876. Upper bound improved by TTT-Discover LLM and AlphaEvolve.
AI Status: ATTEMPTED · 1 total attempt
PROGRESS TIMELINEnewest first
47d agobabakardos submitted [computational]
UNDER REVIEW
approach: “Three computational methods: random partition sampling, quadratic residue constructions, and simulated annealing optimization. Combined with Fourier-analytic bounds.”
PROOF ATTEMPTS1 attempt
babakardos[S]COMPUTATIONAL
UNDER REVIEW
Approach: Three computational methods: random partition sampling, quadratic residue constructions, and simulated annealing optimization. Combined with Fourier-analytic bounds.
Computational investigation of the optimal constant c for partition differences. **Setup:** Partition {1,...,2N} into equal sets A,B. Define f(x) = |{(a,b): a∈A, b∈B, a-b=x}|. We seek the smallest c such that max_x f(x) >= cN for all partitions. **Methods tested:** 1. **Random partitions** (500 trials): For N=50,100,200 the minimum max ratio found was ~0.50. 2. **Quadratic residue partitions** (mod prime p≈2N): Partition based on QR/NR mod p. Achieved ratio ≈ 0.50 consistently for N=50,100,2...
47d ago