整数作为一个单位中四个四面体数和八个五面体数的和

IF 0.4 4区数学 Q4 MATHEMATICS

American Mathematical Monthly Pub Date : 2023-04-10 DOI:10.1080/00029890.2023.2199652

Benjamin Lee Warren

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引用次数: 0

摘要

1850年，弗雷德里克·波洛克(Frederick Pollock)[1]推测每个正整数可以被写为最多5个四面体数的和，其中第n个四面体数由Tn = 6n(n + 1)(n + 2)给出。今年早些时候，瓦迪姆·波诺马伦科(Vadim Ponomarenko)解决了波洛克猜想[2]的弱版本为Tn−Tn−1−Tn−1 + Tn−2 = n。这个解是弱的，因为它包括减法和加法而不仅仅是加法。下面的恒等式是他在a = 2和k = 1时解的推广。

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Integers as Sums of Four Tetrahedral Numbers and Eight Pentatope Numbers in One Identity

In 1850, Frederick Pollock [1] conjectured that every positive integer can be written as the sum of at most five tetrahedral numbers, where the nth tetrahedral number is given by Tn = 6n(n + 1)(n + 2). Earlier this year, Vadim Ponomarenko solved the weak version of Pollock’s conjecture [2] as Tn − Tn−1 − Tn−1 + Tn−2 = n. This solution is weak in the sense that it includes subtraction as well as addition instead of only addition. The following identity is a generalization of his solution at a = 2 and k = 1.

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来源期刊

American Mathematical Monthly Mathematics-General Mathematics

CiteScore

0.80

自引率

20.00%

发文量

127

审稿时长

6-12 weeks

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