如何计算素数斐波那契数？

Question

（这是我之前问题的扩展）。

How to print a text notification beside prime fibonacci numbers?

任务：

我遇到困难的作业部分如下：

• 确定前 20 个斐波那契数中的哪一个是素数。
• 在基本挑战的打印输出中添加“这是一个主要的”文本通知。
• 将 FibPrimes 存储在一个名为 FibPrimes 的数组中。

问题：

我似乎无法弄清楚如何获取每个单独的斐波那契数，确定它是素数，将素数斐波那契数放入一个数组中，然后打印输出。我之前能够制作一个工作程序，但我没有使用代码来计算素数，我手动将它们输入到一个数组中（参见底部的先前尝试）。

尝试：

这是我尝试过的：

与本题相关的代码用cmets强调。

package fibonaccinumbers;

public class FibonacciNumbers {

    public static void main(String[] args) {

        // Creation of Fibonacci Numbers Array.
        // Let i represent the index number.
       int [] FibNums = new int[20];
        FibNums[0] = 0;
        FibNums[1] = 1;

        // Will add on each successive FibNums.
        // Will be used to calculate average.
        int FibSum = 1;

        // RELEVANT TO QUESTION.
        // Creation if Fibonacci Primes Array.
        // Let n represent the index number.
        int [] FibPrimes = new int[7];
        int n=0;

        // Printing first two fibonacci numbers.
        System.out.println(0);
        System.out.println(1 + "*");

        // Printing remaining fibonacci numbers up to 20th term.
        for (int i=2; i<FibNums.length;i++){ // Begin number generation loop.
            FibNums[i] = FibNums[i-1] + FibNums[i-2];

            // Checks if the fibonacci number is odd.
            // A number is not odd if two divides into it evenly.
            boolean oddcheck = true;
            if (FibNums[i]%2==0){
                oddcheck = false;
            }

            // RELEVANT TO QUESTION.
            // A prime number can only be divided by 1 and inself.
            // Divide FibNums[i] by every number inbetween.
            // If a number divides evenly, it is not a prime (exception: 2). 
            // Else, the number is a prime.
            boolean primecheck;
            for (int divisor = 2; divisor < FibNums[i]/2; divisor++){
                if (FibNums[i] % divisor == 0){
                    primecheck = false;
                } else {
                    primecheck = true;
                }

            // REVELANT TO QUESTION.
            // Add FibNums[i] to the FibPrimes[n] array if it is a prime.
            if (primecheck == false){
                FibPrimes[n] = FibNums[i];
                n = n + 1;
            }

            // RELEVANT TO QUESTION.
            // If any element in the FibPrimes array is equal to the FibNums 
            // array, then the number is a prime.
            for (n=0; n<FibPrimes.length; n++){
                if (FibNums[i] == FibPrimes[n]){
                    System.out.print("This is a prime." + " ");
                }
            }

            // Prints odd fibonacci numbers with a star beside it.
            // Prints even fibonacci numbers with no star beside it.
            if (oddcheck == true){
            System.out.println(FibNums[i] + "*");
            } else {
            System.out.println(FibNums[i]);    
            }

            FibSum = FibSum + FibNums[i];

        } // End number generation loop.

        System.out.print ( "The average is" + " " + FibSum/20);

    }

}

输出：

0
1*
The average is 0The average is 0The average is 0The average is 0This is a prime. 8
This is a prime. 8
The average is 013*
13*
13*
13*
The average is 321*

这是不正确的。

上一次尝试：

这个解决方案“有效”，但为了不走“懒惰路线”，我必须使用代码进行计算。仅显示相关的 sn-ps：

// Creation if Fibonacci Primes Array.
// Ideally, this should be caulculated.
int [] FibPrimes = new int[7];
FibPrimes[0] =  2;
FibPrimes[1] =  3;
FibPrimes[2] =  5;
FibPrimes[3] =  13;
FibPrimes[4] =  89;
FibPrimes[5] =  233;
FibPrimes[6] =  1597;

    // If any element in the FibPrimes array is equal to the FibNums 
    // array, then the number is a prime.
    for (int n=0; n<FibPrimes.length; n++){
        if (FibNums[i] == FibPrimes[n]){
            System.out.print("This is a prime." + " ");
        }
    }

输出：

0
1*
1*
This is a prime. 2
This is a prime. 3*
This is a prime. 5*
8
This is a prime. 13*
21*
34
55*
This is a prime. 89*
144
This is a prime. 233*
377*
610
987*
This is a prime. 1597*
2584
4181*
The average is 547

这是所需的输出！ 但是，我不能使用它，因为我必须使用代码计算素数斐波那契数。这就是“懒惰路线”。

谢谢。

Answer 1

在谷歌上只需 10 分钟，您将能够创建简单而快速的东西。

使用下面的数学公式：

您可以在paper 找到更多您可以制作您的素数斐波那契数列。此外，您将需要一种方法来检查序列中的哪些数字是素数，因此快速方法是通过AKS algorithm

这是上述所有内容的完整示例：

public class FibonacciSequence {

    public static void main(String[] args) {
        for (int i = 1; i <= 20; i++) {
            int num = (int) ((getY_1(i) - getY_2(i)) / Math.sqrt(5));
            if (isPrime(num)) {
                System.out.print("This is a prime : ");
            }
            System.out.println(num);
        }
    }

    public static double getY_1(int N) {
        return Math.pow((1 + Math.sqrt(5.0)) / 2.0, N);
    }

    public static double getY_2(int N) {
        return Math.pow((1 - Math.sqrt(5.0)) / 2.0, N);
    }

    public static boolean isPrime(int num) {
        if (num == 2 || num == 3)
            return true;

        if (num % 2 == 0 || num % 3 == 0) {
            return false;
        }

        int i = 5;
        int s = 2;

        while (i * i <= num) {
            if (num % i == 0) {
                return false;
            }

            i += s;
            s = 6 - s;
        }
        return true;
    }
}

当然，如果您不想将值 (1) 识别为质数，您可以排除它：P。

输出：

This is a prime : 1
This is a prime : 1
This is a prime : 2
This is a prime : 3
This is a prime : 5
8
This is a prime : 13
21
34
55
This is a prime : 89
144
This is a prime : 233
377
610
987
This is a prime : 1597
2584
4181
6765

P.S：我有空闲时间：P