{"title":"欧拉质点生成多项式再探讨","authors":"R. Heffernan, Nick Lord, Des MacHale","doi":"10.1017/mag.2024.11","DOIUrl":null,"url":null,"abstract":"Euler’s polynomial f (n) = n2 + n + 41 is famous for producing 40 different prime numbers when the consecutive values 0, 1, …, 39 are substituted: see Table 1. Some authors, including Euler, prefer the polynomial f (n − 1) = n2 − n + 41 with prime values for n = 1, …, 40. Since f (−n) = f (n − 1), f (n) actually takes prime values (with each value repeated once) for n = −40, −39, …, 39; equivalently the polynomial f (n − 40) = n2 − 79n + 1601 takes (repeated) prime values for n = 0, 1, …, 79.","PeriodicalId":22812,"journal":{"name":"The Mathematical Gazette","volume":"691 1","pages":""},"PeriodicalIF":0.0000,"publicationDate":"2024-02-15","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"Euler’s prime-producing polynomial revisited\",\"authors\":\"R. Heffernan, Nick Lord, Des MacHale\",\"doi\":\"10.1017/mag.2024.11\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"Euler’s polynomial f (n) = n2 + n + 41 is famous for producing 40 different prime numbers when the consecutive values 0, 1, …, 39 are substituted: see Table 1. Some authors, including Euler, prefer the polynomial f (n − 1) = n2 − n + 41 with prime values for n = 1, …, 40. Since f (−n) = f (n − 1), f (n) actually takes prime values (with each value repeated once) for n = −40, −39, …, 39; equivalently the polynomial f (n − 40) = n2 − 79n + 1601 takes (repeated) prime values for n = 0, 1, …, 79.\",\"PeriodicalId\":22812,\"journal\":{\"name\":\"The Mathematical Gazette\",\"volume\":\"691 1\",\"pages\":\"\"},\"PeriodicalIF\":0.0000,\"publicationDate\":\"2024-02-15\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"The Mathematical Gazette\",\"FirstCategoryId\":\"1085\",\"ListUrlMain\":\"https://doi.org/10.1017/mag.2024.11\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"\",\"JCRName\":\"\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"The Mathematical Gazette","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.1017/mag.2024.11","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 0
摘要
欧拉的多项式 f (n) = n2 + n + 41 以连续替换 0、1、...、39 的值时产生 40 个不同的质数而闻名:见表 1。包括欧拉在内的一些学者更倾向于使用多项式 f (n - 1) = n2 - n + 41,其中 n = 1, ..., 40 为质数。由于 f (-n) = f (n - 1),f (n) 在 n = -40,-39,...,39 时实际上取质数值(每个值重复一次);等价多项式 f (n - 40) = n2 - 79n + 1601 在 n = 0,1,...,79 时取(重复)质数值。
Euler’s polynomial f (n) = n2 + n + 41 is famous for producing 40 different prime numbers when the consecutive values 0, 1, …, 39 are substituted: see Table 1. Some authors, including Euler, prefer the polynomial f (n − 1) = n2 − n + 41 with prime values for n = 1, …, 40. Since f (−n) = f (n − 1), f (n) actually takes prime values (with each value repeated once) for n = −40, −39, …, 39; equivalently the polynomial f (n − 40) = n2 − 79n + 1601 takes (repeated) prime values for n = 0, 1, …, 79.