User blog comment:P進大好きbot/Elementary Large Number/@comment-30754445-20181118000005/@comment-35470197-20181118223254

> \(\textrm{Ord}(2^{(n+1)/2}+1) = \omega^{\textrm{Ord}(n)}\)

Right. Also, addition works in the following way, whhere I denote by \(a\) a canonical opposite arrow of \(\textrm{Ord}\): \begin{eqnarray*} a(0) & = & 1 \\ a(1) & = & a(\omega^0) = 2^{\frac{a(0) + 1}{2}} = 3 \in [2^1,2^2) \\ a(2) & = & a(1 + \omega^0) = a(1) + 2^{1 + \frac{a(0) + 1}{2}} = 7 \in [2^2,2^3) \\ a(3) & = & a(2 + \omega^0) = a(2) + 2^{2 + \frac{a(0) + 1}{2}} = 15 \in [2^3,2^4) \\ a(\omega) & = & a(\omega^1) = 2^{\frac{a(1)+1}{2}} = 5 \in [2^2,2^3)\\ a(\omega+1) & = & a(\omega + \omega^0) = a(\omega) + 2^{2 + \frac{a(0)+1}{2}} = 13 \in [2^3,2^4) \\ a(\omega+2) & = & a(\omega + 1 + \omega^0) = a(\omega+1) + 2^{3 + \frac{a(0)+1}{2}} = 29 \end{eqnarray*} Naemely, "express" the ordinal in an \(\omega\)-adic way, and convert it into a \(2\)-adic integer.

> It probably is, but you won't find many such googologists here. Why add yet another confusing term to topic that's already difficult?

Well, I wrote them expecting one of those explanations would help you. Since you know RH (at least its statement), it is not so strange to expect it, is it? (I note that I do not know which branches of mathematics you studied in a university.)