中文

关于Apery多项式之和的可整除性

数论 2011-08-09 v1 组合数学

摘要

对于任意正整数mmα\alpha,我们证明k=0n1ϵk(2k+1)Ak(α)(x)m0(modn),\sum_{k=0}^{n-1}\epsilon^k(2k+1)A_k^{(\alpha)}(x)^m\equiv0\pmod{n}, 其中ϵ{1,1}\epsilon\in\{1,-1\}An(α)(x)=k=0n(nk)α(n+kk)αxk. A_n^{(\alpha)}(x)=\sum_{k=0}^n\binom{n}{k}^{\alpha}\binom{n+k}{k}^{\alpha}x^k.

关键词

引用

@article{arxiv.1108.1546,
  title  = {On divisibility of sums of Apery polynomials},
  author = {Hao Pan},
  journal= {arXiv preprint arXiv:1108.1546},
  year   = {2011}
}

备注

This is a preliminary draft