User blog comment:AJZajac/How fast does this sequence grow (problem in ordinal arithmetic)?/@comment-30754445-20171212210639

Whether you use n or a(n), the difference is completely negligible.

Let's write:

X=fω2(2)

And let's call the two versions of your function a(n) and b(n).

Then:

a(2) = b(2) = X

a(3) = fω+X(a(2)) = fω+X(X) = fω2(X)

b(3) = fω+X(3)

Note that we can expand b(3) further and bound it by:

b(3) = fω+X(3) = fω+X-1(fω+X-1(fω+X-1(3))) >> fω+X-1(X-1) = fω2(X-1)

In other words the difference between a(3) and b(3) is smaller than the difference between fω2(X) and fω2(X-1)

And since X is a huge number (it's way bigger than Graham's number btw) then the difference between fω2(X-1) and fω2(X) is completely negligible.

The same argument can be used to show that a(n) and b(n) are very close to one another for n>3 as well.