Decompose a number into triangles
JavaScript (ES6), 52 bytes

f=(n,t=0)=&gt;t&lt;n?f(n-++t,t):t.toString(36)+(n?f(n):'')


How?

Rather than explicitly computing Ti&nbsp;=&nbsp;1&nbsp;+&nbsp;2&nbsp;+&nbsp;3&nbsp;+&nbsp;…&nbsp;+&nbsp;i, we start with t&nbsp;=&nbsp;0 and iteratively subtract t&nbsp;+&nbsp;1 from n while t&nbsp;&lt;&nbsp;n, incrementing t at each iteration. When the condition is not fulfilled anymore, a total of Tt has been subtracted from n and the output is updated accordingly. We repeat the process until n&nbsp;=&nbsp;0.

Below is a summary of all operations for n = 100.

 n  |  t | t &lt; n | output
----+----+-------+--------
100 |  0 | yes   | ""
 99 |  1 | yes   | ""
 97 |  2 | yes   | ""
 94 |  3 | yes   | ""
 90 |  4 | yes   | ""
 85 |  5 | yes   | ""
 79 |  6 | yes   | ""
 72 |  7 | yes   | ""
 64 |  8 | yes   | ""
 55 |  9 | yes   | ""
 45 | 10 | yes   | ""
 34 | 11 | yes   | ""
 22 | 12 | yes   | ""
  9 | 13 | no    | "d"
----+----+-------+--------
  9 |  0 | yes   | "d"
  8 |  1 | yes   | "d"
  6 |  2 | yes   | "d"
  3 |  3 | no    | "d3"
----+----+-------+--------
  3 |  0 | yes   | "d3"
  2 |  1 | yes   | "d3"
  0 |  2 | no    | "d32"


Test cases



f=(n,t=0)=&gt;t&lt;n?f(n-++t,t):t.toString(36)+(n?f(n):'')

console.log(f(1))   // 1
console.log(f(2))   // 11
console.log(f(3))   // 2
console.log(f(4))   // 21
console.log(f(5))   // 211
console.log(f(6))   // 3
console.log(f(100)) // d32
console.log(f(230)) // k5211
console.log(f(435)) // t
console.log(f(665)) // z731