Find max matching 3 Triangular numbers

Question

It's well known that any positive number can be expressed through at most 3 Triangular numbers. ( https://oeis.org/A000217 )

Example:

11 := 10 + 1

12 := 10 + 1 + 1

13 := 10 + 3

14 := 10 + 3 + 1

15 := 15

I am searching the representation of the positive number n through at most 3 possible Triangular summands. There can exist more than one representation of n. I am interested in the one with the greatest summands.

Is there a more effective way than 2 decreasing for and 1 increasing for loops to find the summands?

public void printMaxTriangularNumbers(int n){
  int[] tri = createTriangularNumbers(1000);

  lbl: for(int i = tri.length-1; ; i--){
    int tmp = n - tri[i];
    if(tmp == 0){
      System.out.println(tri[i]);
      break;
    }
    for(int j=i; j>0; j--){
      int tmp2 = tmp - tri[j];
      if(tmp2 ==0){
        System.out.println(tri[i]);
        System.out.println(tri[j]);
        break lbl;
      }
      for(int k=1; k <= j;k++){
        if(tmp2 - tri[k] == 0){
          System.out.println(tri[i]);
          System.out.println(tri[j]);
          System.out.println(tri[k]);
          break lbl;
        }
      }
    }
  }
}

public int[] createTriangularNumbers(int n){
  int[] out = new int[n+1];
  for(int i=1,sum=0; i<=n;i++){
    out[i] = sum += i;
  }
  return out;
}

Answer 1

As far as I can see, there is no direct formula. An algorithm is needed. For instance, a greedy method does not work. Take for example the value 90:

• The greatest triangular number not greater than 90 is 78. Remains 12
• The greatest triangular number not greater than 12 is 10. Remains 2
• And now it becomes clear we will need 4 terms which is not acceptable.

So I would propose a recursive/backtracking algorithm, where each recursive call deals with one summand only. Each level in the recursion takes first the highest possible triangular number, but if the recursive call fails, it will go for the second largest and dive into recursion again, until there is an acceptable sum.

We can use the formula mentioned at maths.stackexchange.com:

Let T_m be the largest triangular number less than or equal to c. You can actually get an explicit formula for m, namely:

Here is a snippet that implements the recursion. When running it, you can introduce a value, and the triangular summands will be produced for it.

function getTriangularTerms(n, maxTerms) {
    if (maxTerms === 0 && n > 0) return null; // failure: too many terms
    if (n == 0) return []; // Ok! Return empty array to which terms can be prepended
    // Allow several attempts, each time with a
    //   lesser triangular summand:
    for (let k = Math.floor((Math.sqrt(1+8*n) - 1) / 2); k >= 1; k--) {
        let term = k * (k+1)/2;
        // Use recursion
        let result = getTriangularTerms(n - term, maxTerms - 1);
        // If result is not null, we have a match
        if (result) return [term, ...result]; // prepend term
    }
}

// I/O handling
let input = document.querySelector("input");
let output = document.querySelector("span");

(input.oninput = function () { // event handler for any change in the input
    let n = input.value;
    let terms = getTriangularTerms(n, 3); // allow 3 terms max.
    output.textContent = terms.join("+");
})(); // execute also at page load.

Enter number: <input type="number" value="14"><br>
Terms: <span></span>

Answer 2

Since a triangular number is any number t that satisfies t=x(x+1)/2 for any natural number x, what you're asking is to solve

n = a(a+1)/2 + b(b+1)/2 + c(c+1)/2

and to find the solution (a,b,c) with greatest possible max(a,b,c). You didn't specify that you only allow solutions with 3 triangular numbers, so I will assume you also allow solutions of the form (a,b,c,d) and look for the one with the greatest max(a,b,c,d).

There might be multiple solutions, but one with at most 3 triangular numbers always exists. Since it's possible to form any number with 3 triangular numbers, you can find the largest triangular number t with t<=n, and then it will follow n=t+d, where d=n-t. d is a natural number >=0 and therefore can be composed by 3 triangular numbers itself. Since you're interested in the largest summand, the largest will be t, which can be computed with t=x(x+1)/2 where x=floor((sqrt(1+8n)-1)/2) (obtained by solving the formula n=x(x+1)/2+d).

As a practical example, take n=218. With the formula, we get x=20 and t=210, which indeed is the largest triangular number before 218. The other triangular numbers, in this case, will be 6, 1, 1 since the only way to compute 8 is with those.