论文标题

将Zeckendorf的定理扩展到此非恒定复发关系上的非恒定复发和Zeckendorf游戏

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

论文作者

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

论文摘要

Zeckendorf的定理指出,每个正整数都可以唯一地表示为非阳性斐波那契数的总和,从$ 1、2、3、5,\ ldots $索引。许多作者已将其推广,尤其是恒定系数固定深度线性复发,具有阳性(或在某些情况下)系数。在这项工作中,我们将此结果扩展到具有非恒定系数的复发,$ a_ {n + 1} = n a_ {n} + a_ {n-1} $。分解定律每$ m $都具有唯一的分解为$ \ sum s_i a_i $带有$ s_i \ le i $,其中,如果$ s_i = i $,则$ s_ s_ {i-1} = 0 $。类似于Zeckendorf的原始证明,我们使用贪婪算法。我们表明,汇率之间的几乎所有差距(随着$ n $接近无穷大)的长度为零,并给出了一种启发式,即汇总数量的分布趋向于高斯。此外,我们基于这种复发关系建立了一个游戏,将游戏推广到斐波那契数字上。给定固定的整数$ n $和$ n = na_1 $的初始分解,玩家通过使用与复发关系相关的动作来替代,无论谁移动最后一场胜利。我们证明该游戏是有限的,以$ n $的独特分解结束,并且任何一个玩家都可以在两人游戏中获胜。我们发现达到最短游戏的策略以及这款最短游戏的长度。然后,我们表明,在有三个以上玩家的广义游戏中,没有球员有获胜的策略。最后,我们演示了两个玩家游戏中的一个玩家如何迫使游戏取得优势。

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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