yes.
x,a = aa
x,a,a = x,(x,a)
x,a,a,a = x,(x,(x,a))
x,,a = x,a,a,a...a times...a
x,,,a = x,,a,,a,,a...a times...a
x,b@a = x,,,,,,,,.......b times.........,,,,,,a
x,b/a = x,(x,(x,(...b times...(x,a@a)...)))
x,b//a = x,(x,(x,(...(x,b/a)...)/a)/a)/a
x,b/(c)a = x,b/////...c times.../////a
x1,a = x,a/(x,a/(x,a/(...a times...(x,a/(a)a)...))
x2,a = x,a/(x,a/(x,a/(...a times...(x,a/(a)a)...)) where a = x1,b
xb,a = x,a/(x,a/(x,a/(...a times...(x,a/(a)a)...)) where a = x(m-1),b
x[b],a = x(x(x(x(...b times...(xa,a)...)a)a)a)a
x{b}a = x[x[x[x[...b times...[x[b]a]...]a]a]a]a
xLVL3(b)a = x{x{x{x{...b times.{x{b}a}