Find the largest number of distinct integers that sum to n

Question

The Task

Given an input positive integer n (from 1 to your language's limit, inclusively), return or output the maximum number of distinct positive integers that sum to n.

Test Cases

Let f define a valid function according to the task:

The sequence for f, starting at 1:

1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, ...

As a larger test case:

>>> f(1000000000) // Might not be feasible with brute-forcers
44720

Test Code

For any test cases not explicitly given, the output of your code should match the result of the following:

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int x = sc.nextInt();
        System.out.println((int) Math.floor(Math.sqrt(2*x + 1./4) - 1./2));
    }
}

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Answer 1

Pyke, 6 bytes

8*h,te

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8      - Push literal 8.
 *     - Multiply by the input.
  h    - Increment.
   ,   - Square root.
    t  - Decrement.
     e - Floor halve.

Answer 2

Swift, 41 bytes

import Foundation
{Int(sqrt($0*8+1)-1)/2}

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Answer 3

Pushy, 8 bytes

8*hrt2/#

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Uses the closed formula (sqrt(8n + 1) - 1) / 2:

8*          \ Multiply by 8
  h         \ Increment
   r        \ Integer root
    t       \ Decrement
     2/     \ Floordiv by 2
       #    \ Output

I thought I recognised this formula - it's the reverse of the function for a triangle number:

f(x) = (x + 1)(x / 2)
f-1(x) = (sqrt(8x+ 1) - 1) / 2

...which makes sense as we're counting integer sums.

Answer 4

Add++, 12 bytes

L,ßR¬+A€<€!s

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Because closed form solutions are boring

Answer 5

Wolfram Language (Mathematica), 22 bytes

⌊√(1+8*#)/2-.5⌋&

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Answer 6

c++,61

int f(int n){for(int i=1;;i){if(n<i) return i-1; else n-=i}}

Answer 7

Cubix, 16 bytes

.(@sI1-?s.O@s)W\

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    . (
    @ s
I 1 - ? s . O @
s ) W \ . . . .
    . .
    . .

Watch it run

• I1 set up the stack with n and 1 (incrementer)
• -? subtract the incrementer from n and test result
  • if result 0 sO@, swap result with incrementer, output and halt
  • if result negative s(O, swap result with incrementer, decrement, output and via a few commands, halt
  • if result positive \s)W, redirect, swap result with incrementer, increment and redirect back into the subtract/test