User blog:VoidSansXD/The + notation

I'm gonna add more every few hours or days.

Currently:

n++1 = n+1

n++2 = (n++1)++1

n++3 = (n++2)++2

n++4 = (n++3)++3

Overall, n++x = (n++(x-1))++(x-1)

n+++1 = (n++n)++n

n+++2 = (n+++1)+++1

Current maximum growth:  I'm not good at growth rates so idk lol