Inequalities between large numbers?

Question

It's been shown Gaham's Number g₆₄ is way larger than Moser's Number (< g₃), itself larger than Skewes' Number {≈(10↑↑4)34}. How about the position of Grahal g₁ = 3↑↑↑↑3 (or Triteto) with respect to Moser's Number and Skewes' Number ?

Answer 1

Skewes number is much much much smaller than Grahal. Even $$3\uparrow\uparrow\uparrow 3$$ would be much much larger. Mosers number is larger than Grahal. Look at this site https://sites.google.com/site/largenumbers/home for the magnitude of Moser and Skewes number.

Answer 2

More in depth:

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We can define Moser's number with the function

$$M_m(n)=\begin{cases}n^n,&m=3\\M_{m-1}^n(n),&m>3\end{cases}$$

where $f^n$ denotes function iteration. Moser's number is then $M_{M_5(2)}(2)$. We can then see that since $M_3(x)=x^x>n^x$ for $x>n$ we have

\begin{align}M_3(M_3(x))&>n^{n^x}\\M_3(M_3(M_3(x)))&>n^{n^{n^x}}\\&\vdots\\M_3^k(x)&>(n\uparrow)^kx\end{align}

where we use Knuth's up-arrow notation. Applying this, we can bound $M_5(2):$

\begin{align}M_5(2)&=M_4(M_4(2))\\&=M_4(M_3(M_3(2)))\\&=M_4(256)\\&=M_3^{256}(256)\\&>(256\uparrow)^{256}256\\&=256\uparrow^2257\end{align}

One can already see this is much larger than Skewe's number. One can further see that by using the same argument, we have for any $n\ge2:$

\begin{align}M_4(n)&>n\uparrow^2n=n\uparrow^32\\M_5(n)&>n\uparrow^3n=n\uparrow^42\\&\vdots\\M_m(n)&>n\uparrow^{m-2}n=n\uparrow^{m-1}2\end{align}

From this we may finally get the bound:

\begin{align}M_{M_5(2)}(2)&\ge M_{256\uparrow^32}(M_{256\uparrow^32}(2))\\&>M_{256\uparrow^32}(2\uparrow^{256\uparrow^32}2)\\&=M_{256\uparrow^32}(4)\\&>4\uparrow^44\\&>3\uparrow^43=3\uparrow\uparrow\uparrow\uparrow3\end{align}

as desired.