A counterexample to Stanley's conjecture on Smith forms of Jucys–Murphy elements

For a partition λ of n, let S^λZ be the integral Specht module and Xₖ=(1,k)+⋯+(k-1,k) the k-th Jucys–Murphy element. Stanley conjectured in 2016 a specific formula for the invariant factors of Xₖ on S^λZ in terms of the contents of the cells that k occupies in standard Young tableaux of shape λ, and remarked, more weakly, that each is then a product of eigenvalues. Both statements are false. For λ=(3,1,1) and $k=3$ the Smith form is $(1,1,1,1,2,2)$, not the predicted $(1,1,1,1,1,4)$. For λ=(5,2) and $k=7$ the eigenvalues are $0$ and $4$, yet the Smith form contains a $2$. Lean checks both matrix computations, and two independent programs check the matrices from the definitions. From the identity X₃(X₃-(1,2))=2 in ZSₙ, checked in Lean, we prove in writing that for n≥3 the conjecture holds at $k=3$ exactly when λ₁≤2 or λ has at most two rows, on S^λZ and on every full-rank lattice in S^λQ stable under the permutations of \1,2,3\. Written proof, with Lean and computer checks; not yet reviewed by experts. Produced by AI models under the author's direction.