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arXiv 2009.12475math.NT
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将Zeckendorf定理推广到非常数常返关系及这种非常数常返关系上的Zeckendorf对策

Extending Zeckendorf's Theorem to a Non-constant Recurrence and the Zeckendorf Game on this Non-constant Recurrence Relation

Elżbieta Bołdyriew, Anna Cusenza, Linglong Dai, Pei Ding, Aidan Dunkelberg, John Haviland, Kate Huffman, Dianhui Ke, Daniel Kleber, Jason Kuretski, John Lentfer… 展开作者

Elżbieta Bołdyriew, Anna Cusenza, Linglong Dai, Pei Ding, Aidan Dunkelberg, John Haviland, Kate Huffman, Dianhui Ke, Daniel Kleber, Jason Kuretski, John Lentfer, Tianhao Luo, Steven J. Miller, Clayton Mizgerd, Vashisth Tiwari, Jingkai Ye, Yunhao Zhang, Xiaoyan Zheng, Weiduo Zhu

摘要:

Zeckendorf定理指出,每个正整数都可以唯一地表示为从$1, 2, 3, 5,\ldots$索引的非相邻斐波那契数的和。这已经被许多作者推广,特别是常系数固定深度线性递归与积极的(或在某些情况下非负)系数。在这项工作中,我们将这一结果推广到一个经常性的系数,$a_{n+1} = n a_{n} + a_{n-1}$。分解法则变为每个$m$都有一个唯一的分解,即$\sum s_i a_i$和$s_i \le i$,其中如果$s_i = i$,则$s_{i-1} = 0$。与Zeckendorf的原始证明类似,我们使用贪婪算法。我们发现,几乎所有的间隙之间的被加数,作为$n$接近无穷大,长度为零,并给出一个启发式的被加数的分布趋于高斯。此外,我们建立了一个游戏的基础上,这种递归关系,推广游戏的斐波那契数。给定一个固定的整数$n$和$n= na_1$的初始分解,参与者通过使用与递归关系相关的移动进行交替,最后移动的人获胜。我们表明,游戏是有限的,并在唯一的分解$n$结束,任何一个球员都可以赢得一个两人游戏。我们找到了实现最短博弈的策略,以及这个最短博弈的长度。然后,我们证明了在这个广义博弈中,当有三个以上的球员,没有球员有获胜的策略。最后,我们演示了如何在两个玩家的游戏中的一个玩家可以迫使游戏的进展,他们的优势。

英文摘要:

Zeckendorf's Theorem states that every positive integer can be uniquely represented as a sum of non-adjacent Fibonacci numbers, indexed from $1, 2, 3, 5,\ldots$. This has been generalized by many authors, in particular to constant coefficient fixed depth linear recurrences with positive (or in some cases non-negative) coefficients. In this work we extend this result to a recurrence with non-constant coefficients, $a_{n+1} = n a_{n} + a_{n-1}$. The decomposition law becomes every $m$ has a unique decomposition as $\sum s_i a_i$ with $s_i \le i$, where if $s_i = i$ then $s_{i-1} = 0$. Similar to Zeckendorf's original proof, we use the greedy algorithm. We show that almost all the gaps between summands, as $n$ approaches infinity, are of length zero, and give a heuristic that the distribution of the number of summands tends to a Gaussian. Furthermore, we build a game based upon this recurrence relation, generalizing a game on the Fibonacci numbers. Given a fixed integer $n$ and an initial decomposition of $n= na_1$, the players alternate by using moves related to the recurrence relation, and whoever moves last wins. We show that the game is finite and ends at the unique decomposition of $n$, and that either player can win in a two-player game. We find the strategy to attain the shortest game possible, and the length of this shortest game. Then we show that in this generalized game when there are more than three players, no player has the winning strategy. Lastly, we demonstrate how one player in the two-player game can force the game to progress to their advantage.

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