Can a larger base than three create a number greater than Graham's number?

[Mentallic]

I'd like to discuss numbers at "orders of magnitude" around Graham's number.

If any readers haven't heard of this number but are curious to follow, you'll first need to understand Knuth's[/PLAIN] up arrow notation and then read up on Graham's[/PLAIN] number.

My question is, since Graham's number ($G=g_{64}$) is essentially a very large power tower of 3's, can we replace the 3's with a larger (but relatively small) number such that we get a new number $H=h_{63}>G$. So for example, let the larger number be a googol, so
$$\newcommand\up{\mathbin{\uparrow}}$$
$$g_0=h_0=4$$
and
$$g_{n}=3\hspace{1 mm}\underbrace{\up\dots\up}_{g_{n-1}} \hspace{2 mm}3,\hspace{5 mm}n>0$$
$$h_{n}=10^{100}\hspace{1 mm}\underbrace{\up\dots\up}_{h_{n-1}} \hspace{2 mm}10^{100},\hspace{5 mm}n>0$$

So clearly $h_{63}>g_{63}$ but can we relate $h_{63}$ to $g_{64}$? The number of up arrows are by far the most powerful operator in $a\up\dots\up b$, so I'd guess $g_{64}$ is larger, but I can't prove it.

A naturally extension to the question could also be what size the number needs to be such that $h_{63}>g_{64}$.

 

[Deedlit]

Okay, for convenience I will use ##a c$for$a\underbrace{\uparrow\dots\uparrow}_{b} c$when there are a lot of up arrows. I will start by proving that$3 \uparrow\uparrow b+3 > 10^{100} \uparrow\uparrow b$for all$b \geq 1$. I will in fact prove the stronger statement$3 \uparrow\uparrow b+3 > (10^{100} \uparrow\uparrow b)^2$, using induction. Case$b=1$:$3 \uparrow\uparrow 4 = 3^{3^{27}} > 3^{600} = 27^{200} > 10^{200} = (10^{100} \uparrow\uparrow 1)^2$. Now assume the statement is true for b, and prove it for b+1. 3 \uparrow\uparrow b+4 = 3^{3 \uparrow\uparrow b+3} > 3^{(10^{100} \uparrow\uparrow b)^2} = (3^{10^{100} \uparrow\uparrow b})^{10^{100} \uparrow\uparrow b} \geq (3^{10 ^ {100}})^{10^{100} \uparrow\uparrow b} > (10^{10^{99}})^{10^{100} \uparrow\uparrow b} > (10^{200})^{10^{100} \uparrow\uparrow b} = ((10^{100})^{10^{100} \uparrow\uparrow b})^2 = (10^{100} \uparrow\uparrow b+1)^2. Next, we will prove that$3 \uparrow\uparrow\uparrow b+2 > (10^{100} \uparrow\uparrow\uparrow b)+4$for$b \geq 1$. We will use induction again. Case$b = 1$:$3 \uparrow\uparrow\uparrow 3 \geq 3 \uparrow\uparrow 4 > 10 ^ {100}$Next, we assume the statement for b, and prove it for b+1. 3\uparrow\uparrow\uparrow b+3 = 3 \uparrow\uparrow (3\uparrow\uparrow\uparrow b+2) > 3 \uparrow\uparrow ((10^{100} \uparrow\uparrow\uparrow b) + 4) \text{(by induction)} > 10^{100} \uparrow\uparrow (10^{100} \uparrow\uparrow\uparrow b+1) \text{(by the previous proposition)} \geq (10^{100}\uparrow\uparrow\uparrow b+1) + 4. Next, we have$3 [a] b+1 > (10^{100} [a] b) + 3$for$a \geq 4, b \geq 1$. We will prove it using double induction and a and b. Case$b=1$: 3 [a] 2 \geq 3 [4] 2 = 3 \uparrow\uparrow\uparrow 3 > 3 \uparrow\uparrow 4 > 10^{100}. Next assume that statement is true for (4, b), and we prove it for (4, b+1). 3 [4] b+2 = 3 [3] (3 [4] b+1) > 3 [3] ((10^{100} [4] b) + 3) > 10 ^ {100} [3] ((10^{100} [4] b) + 1) > 10^{100} [3] (10^{100} [4] b) + 3 = (10^{100} [4] b) + 3. Finally we prove the statement for (a+1, b+1), given that it is true for (a+1, b) and (a, x) for any x. 3 [a+1] b+2 = 3 [a] (3 [a+1] b+1) > 3 [a] ((10^{100} [a+1] b) + 3) > 10 ^ {100} [a] ((10^{100} [a+1] b) + 2) > 10 ^{100} [a] (10^{100} [a+1] b) + 3 = 10 ^{100} [a+1] (b+1) + 3. Next, we have$3 [a+1]3 > 10^{100} [a] 10^{100}$for$a \geq 3$. Indeed,$3 [a+1] 3 = 3 [a] (3 [a] 3) > 3 [a] (10^{100} + 2) > 10^{100} [a] 10^{100}$. So$h_n < g_{n+1}$, since$h_0 = 4 < 3 [4] 3 = g_1$, and if$h_n < g_{n+1}##, then
$$h_{n+1} = 10^{100} [h(n)] 10^{100} < 3 [h(n)+1] 3 \leq 3[g_{n+1}] 3 = g_{n+2}$$.