求 2 个五边形数，其和和差产生五边形数

Question

我正在尝试计算两个五边形数，它们的和与差将产生另一个五边形数。在我的主要函数中，我使用五角数定理产生五角数和，产生一个五角数，然后我使用 is_pentagonal 函数检查这两个数的差是否也是五角数。

我用 C++ 编写了以下代码，但由于某种原因没有给出正确的答案，我不确定错误在哪里。

问题是，当我得到答案 d 时，j 和 k 就不是五边形了。 j 和 k 只是超出了数值限制，随机数最终产生了五边形 d，我不明白为什么会这样。谢谢。

bool is_perfect_square(int n) 
{
        if (n < 0) return false;
        int root = sqrt(n);
        return n == root * root;
}

bool is_pentagonal(int n)
{
        if(is_perfect_square(24*n + 1) && (int)sqrt(24*n+1)%6 == 5)return true;
        return false;
}

int main() {
    int j = 0, k = 0, d = 0, n = 1;
    while(!is_pentagonal(d))
    {
        j = (3*n+1)*(3*(3*n+1)-1)/2;
        k = (n*(9*n+5)/2)*(3*n*(9*n+5)/2-1)/2;
        d = k - j;
        ++n;
    }
    cout << d << endl;
    return 0;
}

Answer 1

我在 ideone 中运行这段代码:

#include <iostream>
#include <math.h>

using namespace std;

bool is_perfect_square(unsigned long long int n);
bool is_pentagonal(unsigned long long int n);


int main() {

    // I was just verifying that your functions are correct
    /*
    for (int i=0; i<100; i++) {
        cout << "Number " << i << " is pentagonal? " << is_pentagonal(i) << endl;
    }
    */

    unsigned long long int j = 0, k = 0, d = 0;
    int n = 1;
    while(!is_pentagonal(d))
    {
        j = (3*n+1)*(3*(3*n+1)-1)/2;
        if (!is_pentagonal(j)) {
            cout << "Number j = " << j << " is not pentagonal; n = " << n << endl;
        }
        k = (n*(9*n+5)/2)*(3 *n*(9*n+5)/2-1)/2;
        if (!is_pentagonal(k)) {
            cout << "Number k = " << k << " is not pentagonal; n = " << n << endl;
        }
        d = k - j;
        ++n;
     }

    cout << "D = |k-j| = " << d << endl;

    return 0;
}

bool is_perfect_square(unsigned long long int n) {
    if (n < 0)
        return false;
    unsigned long long int root = sqrt(n);
    return n == root * root;
}

bool is_pentagonal(unsigned long long int n)
{
    if(is_perfect_square(24*n + 1) && (1+(unsigned long long int)sqrt(24*n+1))%6 == 0)return true;
    return false;
}

结果是：

Number k = 18446744072645291725 is not pentagonal; n = 77
Number k = 18446744072702459675 is not pentagonal; n = 78
Number k = 18446744072761861113 is not pentagonal; n = 79
...

如果您将这些数字与 cplusplus.com you will notice you are very close to that. So basically what happens is that your algorithm is correct, but numbers quickly get so big they overflow, and then they are just wrong. See here 中报告的 2^64 = 18 446 744 073 709 551 616 进行比较，以检查您现在有哪些选择。

求 2 个五边形数，其和和差产生五边形数

Finding 2 pentagonal numbers whose sum and difference produce pentagonal number

c++

number-theory