中文

多项式乘以广义斐波那契数之和

数论 2025-05-12 v1

摘要

给定 a,bNa,b\in\mathbb Nc0,c1Zc_0,c_1\in\mathbb Z,且 (c0,c1)(0,0)(c_0,c_1)\neq (0,0),以及一个广义斐波那契数列 (sn)n0(s_n)_{n\geq 0},其中 s0=c0s_0 = c_0s1=c1s_1 = c_1,且对所有正整数 nn 都有 sn+1=asn+bsn1s_{n+1}=as_{n}+bs_{n-1}。本文我们得到的结果是:对于每一个具有实系数的多项式 P(x)P(x),我们总能找到三个实系数多项式 F1(x),G1(x),H1(x)F_1(x), G_1(x), H_1(x)(不一定是不同的),满足恒等式:  2k=1nP(k)sk1=F1(n)sn+1+G1(n)sn+H1(n),  nN\;2\sum_{k=1}^{n}P(k)s_{k-1} = F_1(n)s_{n+1} + G_1(n)s_n + H_1(n), \;\forall n\in\mathbb N。此外,我们对 (sn)n0(s_n)_{n\geq 0} 提出两个约束:一个约束意味着存在无限多个三元组 (F1(x),G1(x),H1(x))(F_1(x), G_1(x), H_1(x)) 满足恒等式   2k=1nP(k)sk1=F1(n)sn+1+G1(n)sn+H1(n),  nN\;2\sum_{k=1}^{n}P(k)s_{k-1} = F_1(n)s_{n+1} + G_1(n)s_n + H_1(n), \;\forall n\in\mathbb N,而另一个约束意味着只有一个三元组 (F1(x),G1(x),H1(x))(F_1(x), G_1(x), H_1(x)) 满足该恒等式。

关键词

引用

@article{arxiv.2505.05734,
  title  = {On Sum of a Polynomial Multiplied by Generalized Fibonacci Numbers},
  author = {Ivan Hadinata},
  journal= {arXiv preprint arXiv:2505.05734},
  year   = {2025}
}

备注

This is a preprint version, updated on 17 October 2024. The final version has been published in Jurnal Matematika Integratif (link: https://jurnal.unpad.ac.id/jmi/article/view/58753) with the new title "On Sums Involving Polynomials and Generalized Fibonacci Sequences"