To search for an arithmetic prime number sequence with a length of 10, you can enumerate the tolerance d from small to large, and then enumerate the first item p from small to large, and check whether p, p+d, p+2d, ..., p+9d are all prime numbers .
import java.util.*;
public class PrimeArithmetic {
// 判断是否为素数
public static boolean isPrime(int n) {
if (n < 2) {
return false;
}
int sqrt_n = (int) Math.sqrt(n);
for (int i = 2; i <= sqrt_n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
public static void main(String[] args) {
int length = 10;
int max_d = 1000; // 公差最大值
for (int d = 1; d <= max_d; d++) {
for (int p = 2; p <= 10000; p++) {
boolean is_arithmetic = true;
for (int i = 0; i < length; i++) {
int num = p + i * d;
if (!isPrime(num)) {
is_arithmetic = false;
break;
}
}
if (is_arithmetic) {
System.out.println("公差为 " + d + "，首项为 " + p);
return;
}
}
}
System.out.println("未找到长度为 " + length + " 的等差素数列。");
}
}
The tolerance of an arithmetic prime sequence must be relatively prime to the leading term.
Explanation: Assume that the first item of the arithmetic prime sequence is a, the tolerance is d, and a and d have a common factor p greater than 1. Then, the second term in this arithmetic prime sequence is a+d, which is also divisible by p. Therefore, the second term in this arithmetic prime sequence is not a prime number. Therefore, if the tolerance of an arithmetic prime sequence is not coprime with the first term, then the arithmetic prime sequence cannot be all prime numbers.