On the Skolem Problem and the Skolem Conjecture

R. Lipton, F. Luca, Joris Nieuwveld, J. Ouaknine, David Purser, J. Worrell
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引用次数: 9

Abstract

It is a longstanding open problem whether there is an algorithm to decide the Skolem Problem for linear recurrence sequences (LRS) over the integers, namely whether a given such sequence has a zero term (i.e., whether un = 0 for some n). A major breakthrough in the early 1980s established decidability for LRS of order 4 or less, i.e., for LRS in which every new term depends linearly on the previous four (or fewer) terms. The Skolem Problem for LRS of order 5 or more, in particular, remains a major open challenge to this day. Our main contributions in this paper are as follows: First, we show that the Skolem Problem is decidable for reversible LRS of order 7 or less. (An integer LRS is reversible if its unique extension to a bi-infinite LRS also takes exclusively integer values; a typical example is the classical Fibonacci sequence, whose bi-infinite extension is ⟨…, 5, −3, 2, −1, 1, 0, 1, 1, 2, 3, 5, …⟩.) Second, assuming the Skolem Conjecture (a central hypothesis in Diophantine analysis, also known as the Exponential Local-Global Principle), we show that the Skolem Problem for LRS of order 5 is decidable, and exhibit a concrete procedure for solving it.
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论Skolem问题和Skolem猜想
是否存在一种算法来决定整数上线性递归序列(LRS)的Skolem问题是一个长期存在的开放问题,即给定的这样的序列是否有零项(即,对于某些n, un是否= 0)。20世纪80年代初的一个重大突破建立了4阶或更低阶LRS的可判决性,即对于LRS中的每个新项线性依赖于前四个(或更少)项。特别是5阶以上LRS的Skolem问题,至今仍是一个重大的公开挑战。本文的主要贡献如下:首先,我们证明了对于7阶或更低阶的可逆LRS, Skolem问题是可判定的。(整数LRS是可逆的,如果它对双无穷LRS的唯一扩展也只取整数值;一个典型的例子是经典斐波那契数列,它的双无穷扩展⟨…,5,−3,2,−1,1,0,1,1,1,2,3,5,…⟩。)其次,假设Skolem猜想(丢芬图分析中的一个中心假设,也称为指数局部-全局原理),我们证明了5阶LRS的Skolem问题是可决定的,并展示了解决它的具体步骤。
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