Keep/Drop/Increase Sequence
Java (OpenJDK 8), 45 44 bytes



n-&gt;{int i=0;for(;++i&lt;n;n-=i);return~-n+i*i;}


Try it online!

-1 byte thanks to @Nevay

After staring at this for a while, I noticed a pattern. Every time we drop n numbers, the next number in the sequence is a perfect square. Seeing this, I mentally broke the sequence into convenient chunks: [[1],[4,5],[9,10,11],...] Basically, the ith chunk starts with i*i, and iterates upwards for i elements.

To find the nth number in this sequence, we want to find first which chunk the number is in, and then which position in the chunk it occupies. We subtract our increment number i from n until n is less than i (which gives us our chunk), and then simply add n-1 to i*i to get the correct position in the chunk.

Example:

n = 8
n &gt; 1? Yes, n = n - 1 = 7
n &gt; 2? Yes, n = n - 2 = 5
n &gt; 3? Yes, n = n - 3 = 2
n &gt; 4? No, result is 4 * 4 + 2 - 1 = 17