Pascal's Triangle (Sort of)
JavaScript (ES6), 79 bytes

0-indexed.

f=(n,a=[L=1])=&gt;a[n]||f(n-L,[...a.map((v,i)=&gt;k=(x=v)+~~a[i-1-i%2]),L++&amp;1?k:2*x])


Demo



f=(n,a=[L=1])=&gt;a[n]||f(n-L,[...a.map((v,i)=&gt;k=(x=v)+~~a[i-1-i%2]),L++&amp;1?k:2*x])

console.log([...Array(79).keys()].map(n =&gt; f(n)).join(', '))




How?

f = (                       // f = recursive function taking:
  n,                        //   n = target index
  a = [L = 1]               //   a[] = current row, L = length of current row
) =&gt;                        //
  a[n] ||                   // if a[n] exists, stop recursion and return it
  f(                        // otherwise, do a recursive call to f() with:
    n - L,                  //   n minus the length of the current row
    [                       //   an array consisting of:
      ...a.map((v, i) =&gt;    //     replace each entry v at position i in a[] with:
        k =                 //       a new entry k defined as:
        (x = v) +           //       v +
        ~~a[i - 1 - i % 2]  //       either the last or penultimate entry
      ),                    //     end of map()
      L++ &amp; 1 ?             //     increment L; if L was odd:
        k                   //       append the last updated entry
      :                     //     else:
        2 * x               //       append twice the last original entry
    ]                       //   end of array update
  )                         // end of recursive call


This algorithm directly generates the sorted rows of Pascal's Triangle. It updates n according to the length of the previous row until a[n] exists. For instance, 6 iterations are required for n = 19:

 L | n  | a[]
---+----+------------------------
 1 | 19 | [ 1 ]
 2 | 18 | [ 1, 1 ]
 3 | 16 | [ 1, 1, 2 ]
 4 | 13 | [ 1, 1, 3, 3 ]
 5 |  9 | [ 1, 1, 4, 4, 6 ]
 6 |  4 | [ 1, 1, 5, 5, 10, 10 ]
                        ^^