n-th term of the rise & reset sequence

Haskell, 27 26 bytes

([z|k<-[1..],z<-[1..k]]!!)

Try it online!

Thanks @DanD. for -1 byte!

This is an anonymous function, creating the infinite sequence an just returning the n-th element thereof: [[1..k]| k<-[1..]] produces an infinite list of list: [[1],[1,2],[1,2,3],[1,2,3,4],...]. To concatenate these we can write [z|k<-[1..],z<-[1..k]] which results in [1,1,2,1,2,3,1,2,3,4,...] and finally (...!!) accepts the input n (pointless notation) and returns the n-th term (0-based).

JavaScript, 29 28 bytes

-1 byte thanks to Arnauld!

f=(n,m)=>n++<m?n:f(n+~m,-~m)

Uses the 0-indexed recursive formula found on OEIS.

When called with 1 argument as expected, the default value of the second, m, will be undefined. However, -~undefined, returns 1, which allows us to get the recursion rolling without an explicit m = 1 in the argument list (thanks @Arnauld!)

Test snippet:

f=(n,m)=>n++<m?n:f(n+~m,-~m)

let examples = [0, 1, 5, 10, 15, 1000];

examples.forEach(function log(x) {
    console.log(x, " => ", f(x))
});

Alternatively, for the same byte count, we can have a curried function like so:

f=n=>m=>n++<m?n:f(n+~m)(-~m)

You can call this with f(5)() - it returns a function, which when called, returns the result, as described in this meta post.

Jelly, 5 bytes, 1-indexed

RRF³ị

Try it online!

Explanation:

                                      (Assume N = 4 for the examples)
R      Generate a list of 1 to N      [1, 2, 3, 4]
 R     Generate new lists for each item on the previous list, with that item as N
                                      [[1], [1,2], ...]
  F    Flatten that list              [1, 1, 2, 1, 2, 3 ...]
   ³ị  Use the input number (³) as index (ị) on the list. 
       This is one-based:             [1, 1, 2, 1, 2, 3 ...] 
                                                ^