Triangular, Pentagonal, and Hexagonal

// https://projecteuler.net/problem=45

// Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

// Triangle    Tn=n(n+1)/2    1, 3, 6, 10, 15, ...

// Pentagonal    Pn=n(3n-1)/2    1, 5, 12, 22, 35, ...

// Hexagonal    Hn=n(2n-1)    1, 6, 15, 28, 45, ...

// It can be verified that T285 = P165 = H143 = 40755.

// Find the next triangle number that is also pentagonal and hexagonal.

// Answer: 1533776805

// Authored by: Tim Varley 💘

#include <iostream>
#include <cmath>

long long triangular_pentagonal_hexagonal() {
    for (long long n = 144; ; n++) {
        long long h = n * (2 * n - 1);
        double k = (1 + sqrt(1 + 24.0 * h)) / 6;
        if (k == floor(k)) {
            return h;
        }
    }
    return 0;
}

#if ! defined UNITTEST_MODE
int main(int argc, char const *argv[]) {
    std::cout << triangular_pentagonal_hexagonal() << std::endl;
}
#endif // UNITTEST_MODE

package org.tvarley.euler.solutions;

import org.tvarley.euler.Solution;

public class Solution045 implements Solution {
  public String solve() {
    long n = 144;
    while (true) {
      long hn = n * (2 * n - 1);
      if (isPentagonal(hn) && isTriangular(hn)) {
        return Long.toString(hn);
      }
      n++;
    }
  }

  private boolean isPentagonal(long n) {
    long k = (long) ((1 + Math.sqrt(1 + 24 * n)) / 6);
    return k * (3 * k - 1) / 2 == n;
  }

  private boolean isTriangular(long n) {
    long k = (long) (( -1 + Math.sqrt(1 + 8 * n)) / 2);
    return k * (k + 1) / 2 == n;
  }
}

const isPentagonal = (n) => {
  const k = (1 + Math.sqrt(1 + 24 * n)) / 6;
  return k === Math.floor(k);
};

module.exports = {
  answer: () => {
    let m = 144;
    while (true) {
      const h = m * (2 * m - 1);
      if (isPentagonal(h)) {
        return h;
      }
      m++;
    }
  }
};

def solve():
    """
    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, ...

    Pentagonal     Pn=n(3n−1)/2     1, 5, 12, 22, 35, ...

    Hexagonal     Hn=n(2n−1)     1, 6, 15, 28, 45, ...

    It can be verified that T285 = P165 = H143 = 40755.

    Find the next triangle number that is also pentagonal and hexagonal.
    https://projecteuler.net/problem=45
    """
    def is_pentagonal(n):
        disc = 1 + 24 * n
        sqrt_d = int(disc ** 0.5 + 0.5)
        if sqrt_d * sqrt_d == disc:
            k = (1 + sqrt_d) // 6
            return k * (3 * k - 1) // 2 == n
        return False

    def is_hexagonal(n):
        disc = 1 + 8 * n
        sqrt_d = int(disc ** 0.5 + 0.5)
        if sqrt_d * sqrt_d == disc:
            k = (1 + sqrt_d) // 4
            return k * (2 * k - 1) == n
        return False

    n = 286
    while True:
        t = n * (n + 1) // 2
        if is_pentagonal(t) and is_hexagonal(t):
            return t
        n += 1

package main

import "fmt"

import "math"

func isPentagonal(n int) bool {
  disc := 1 + 24*n
  k := int((1 + math.Sqrt(float64(disc))) / 6)
  return k*(3*k-1)/2 == n
}

func isHexagonal(n int) bool {
  disc := 1 + 8*n
  k := int((1 + math.Sqrt(float64(disc))) / 4)
  return k*(2*k-1) == n
}

func main() {
  n := 286
  for {
    t := n * (n + 1) / 2
    if isPentagonal(t) && isHexagonal(t) {
      fmt.Println(t)
      return
    }
    n++
  }
}

// https://projecteuler.net/problem=45
//
// Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
//
// Triangle     Tn=n(n+1)/2     1, 3, 6, 10, 15, ...
// Pentagonal     Pn=n(3n-1)/2     1, 5, 12, 22, 35, ...
// Hexagonal     Hn=n(2n-1)     1, 6, 15, 28, 45, ...
//
// It can be verified that T285 = P165 = H143 = 40755.
// Find the next triangle number that is also pentagonal and hexagonal.
//
// Answer: 1533776805

pub fn triangular_pentagonal_hexagonal() -> u64 {
    // Hexagonal numbers are also triangular: Hn = n(2n-1) = Tn for some n
    // Every hexagonal is also triangular.
    // So need Hn that is also pentagonal.
    // Hn = n(2n-1)
    // Pn = m(3m-1)/2
    // Set equal: n(2n-1) = m(3m-1)/2
    // Solve for integers n,m >143 (since H143=40755)
    let mut n = 144;
    loop {
        let h = n * (2 * n - 1);
        // Check if h is pentagonal
        // m(3m-1)/2 = h => 3m^2 - m - 2h = 0
        let discriminant = 1 + 24 * h;
        let sqrt_disc = (discriminant as f64).sqrt() as u64;
        if sqrt_disc * sqrt_disc == discriminant && (sqrt_disc + 1) % 6 == 0 {
            return h;
        }
        n += 1;
    }
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn euler_045() {
        assert_eq!(triangular_pentagonal_hexagonal(), 1533776805);
    }
}