Sums of Powers of Integers

F. T. Howard
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引用次数: 122

Abstract

and so on. Many different writers have worked on the problem of finding simple formulas for Skn, and many different methods have been used; see [3], [5], [6], [9], [10] for just a small sampling of recent articles. The formulas for Tk n are certainly less well known. In the present paper we use generating functions to find new recurrences for Skf7(a, d) and Tk n(a,d). We also show how Skn(a,d) and Tk n(a,d) can be determined from Sk_ln(a,d) and Tk_lr2(a, d), respectively, and we show how Skn(a, d) and Tkn(ay d) can be expressed in terms of Bernoulli numbers. One of the main results is a new "lacunary" recurrence formula for Skn with gaps of 6 (Theorem 3.1); that is, we can use the formula to find Smn for m = 0,1,..., 5; then, using only Smn, we can find S6+mt„; then, using only Smn and S6+nhn, we can find Si2+m3„, and so on. There is a similar recurrence for Tk^n. There are several motivations for this paper: (1) A recent article by Wiener [10] dealt with equation (1.1) and generalized some well-known properties of Skt„. We show how the formulas of [10] can be derived very quickly and how they can be extended. (2) In a recent article by Howard [4], formulas were found which connected Bernoulli numbers to Tkt„. Evidently, the properties of Tk^n are not well known, so the results of [4] are a stimulus to study Tk and Tkt„(a, d) in some detail. (3) The new lacunary recurrences mentioned above are useful and easy to use, and (in the writer's opinion) they are of considerable interest. In Section 3 we illustrate the formulas by computing S6n, Sl2}„, and^g^; in Section 6 we compute TAn,Tl0n, mdT16n. (4) Perhaps the main purpose of the paper is to show how generating functions provide a simple, unified approach to the study of sums of powers of integers. The many, and often repetitious
整数幂的和
等等......许多不同的作者都致力于寻找Skn的简单公式,并使用了许多不同的方法;参见[3]、[5]、[6]、[9]、[10]获取最近文章的一小部分样本。Tk n的公式当然不太为人所知。在本文中,我们使用生成函数来寻找Skf7(a, d)和Tk n(a,d)的新递归式。我们还展示了Skn(a,d)和Tkn(a,d)如何分别由Sk_ln(a,d)和Tk_lr2(a, d)确定,我们还展示了Skn(a,d)和Tkn(ay d)如何用伯努利数表示。其中一个主要的结果是一个新的间隔为6的Skn的“空缺”递推公式(定理3.1);也就是说,我们可以用公式求出m = 0,1,…5;然后,只使用Smn,我们可以找到S6+mt”;然后,仅使用Smn和S6+nhn,我们可以找到Si2+m3”,以此类推。Tk^n也有类似的递归式。本文的动机如下:(1)Wiener[10]最近的一篇文章处理了方程(1.1)并推广了Skt的一些众所周知的性质。我们展示了[10]的公式是如何很快地推导出来的,以及它们是如何推广的。(2)在Howard最近的一篇文章中[4],发现了将伯努利数与Tkt联系起来的公式。显然,Tk^n的性质还不是很清楚,所以[4]的结果是对Tk和Tkt”(a, d)进行详细研究的刺激。(3)以上提到的新的词典递归法很有用,也很容易使用,而且(在作者看来)它们非常有趣。在第3节中,我们通过计算S6n, Sl2}”和^g^来说明公式;在第6节中,我们计算TAn,Tl0n, mdT16n。(4)本文的主要目的可能是说明生成函数如何为研究整数幂和提供一种简单、统一的方法。许多,而且常常是重复的
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