Yes, and you can generalize this to BBN(v, N), which means N applications of BB to v. If we replace N with v, it becomes a single parameter BBN(v). Then we ca

Yes, and you can generalize this to BBN(v, N), which means N applications of BB to v. If we replace N with v, it becomes a single parameter BBN(v). Then we can apply this thing to itself, and so on. So, with any finite storage space, we can represent an arbitrarily large number in this fashion. Thus, finite storage does not set an upper bound on the number that can be represented.