AI 中文总结
证明森德拉姆关于\(F_{n,M}\)舒尔展开式系数非负整数的猜想，通过幂和展开、伯恩斯坦公式等方法，利用杨图划分、特征值分离等技术，还给出相关反例。
AI 中文摘要
对于正整数\(n\)和\(M\)，森德拉姆定义了\(F_{n,M}=\sum_{\substack{d\mid n\d\le M}} Lie_{n/d}[p_d]\)，并猜想其舒尔展开式中的每个系数都是非负整数。我们证明了该猜想。幂和展开分离了恒等类的贡献，证明在内在多数阈值处划分杨图。若第一行或第一列包含超过一半的框，伯恩斯坦生成公式将每个矩形循环特征值分离为一个钩子常数和仅支持短循环的余数。平凡和符号特征对循环子群的限制精确地评估了常数，而均匀二项式收缩同时控制到边界的每个距离。若行和列都没有多数，斯旺森的反钩估计给出卡塔兰规模的维数下界。福明 - 卢洛夫矩形特征估计与截断莫比乌斯权重的均匀界一起表明恒等类占主导。森德拉姆的 plethystic 恒等式因此意味着对于每个固定的\(M\)，\(\prod_{r=1}^M(1 - p_r)^{-1}\)是舒尔正的。我们还记录了一个关于理查德·斯坦利的一个单独同余类乘积猜想的\(12\)次反例。
英文摘要
For positive integers $n$ and $M$, Sundaram defined $ F_{n,M}=\sum_{\substack{d\mid n\\d\le M}} Lie_{n/d}[p_d] $ and conjectured that every coefficient in its Schur expansion is a nonnegative integer. We prove the conjecture. The power-sum expansion isolates the contribution of the identity class, and the proof divides Young diagrams at the intrinsic majority threshold. If the first row or first column contains more than half of the boxes, the Bernstein creation formula separates each rectangular-cycle character value into a hook constant and a remainder supported only on short cycles. The restrictions of the trivial and sign characters to the cyclic subgroup evaluate the constants exactly, whereas a uniform binomial contraction controls every distance from the boundary at once. If neither a row nor a column has a majority, Swanson's opposite-hook estimate gives a Catalan-scale lower bound for the dimension. The Fomin--Lulov rectangular-character estimate, together with uniform bounds for the truncated Möbius weights, then shows that the identity class dominates. Sundaram's plethystic identity consequently implies that $\prod_{r=1}^M(1-p_r)^{-1}$ is Schur-positive for every fixed $M$. We also record a degree-$12$ counterexample to a separate congruence-class product conjecture attributed to Richard Stanley.
