Problem 45 | Project Euler |

[sx349]

https://pe-cn.github.io/45/

Problem 45 Triangular, pentagonal, and hexagonal Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:       Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15,

[shangkelingxiang]

注意到第三个包含于第一个

3b^2-b-4c^2+2c=0

解出来枚举判断判别式是否是整数即可

# 注意到 Tn 包含 Hn
'''
for i in range(1,100000000):
    fuck=(1+(48*i*i-24*i+1)**0.5)/6
    # print(fuck)
    if(fuck==int(fuck)):
        print(fuck)
'''
print(31977*(31977*3-1)//2)

[p-gp]

#include <stdio.h>

_Bool isPen(long long int n);
float qsqrt(float n);

int main(){

    int i=1;
    long long int   num=1;

    while(1){
        num+=(4*i+1);
        if(num>40755&&isPen(num)){
            printf("%lld", num);
            break;
        }
        ++i;
    }

    return 0;
}

_Bool isPen(long long int n){
    long long int x=(long long int)qsqrt((float)(n*24+1));
    if(x*x==n*24+1&&((1+x)%6==0)){
        return 1;
    }
    return 0;
}

float qsqrt(float n){
    float nhalf=0.5f*n;

    int i=*(int *)&n;
    i=0x5f375a86-(i>>1);
    n=*(float *)&i;
    n=n*(1.5f-nhalf*n*n);
    n=n*(1.5f-nhalf*n*n);

    return 1/n;
}

[SunMoonTrain]

Select[Range@100000, IntegerQ[1/6 + 1/6 Sqrt[1 - 24 # + 48 #^2]] &]
n (2 n - 1) /. n -> %[[3]]