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I have been playing around with the Ackermann function a bit and realized that it gets very big very fast. (Im going to use $A$ for $Ackermann$ throughout this question)

Already $A(5,1)$ is (according to WolframAlpha) an integer too large to represent. It also presents me with a representation that looks like this:

$$ A(5,1) = 2 \uparrow \uparrow \uparrow 4 - 3 = 2 \uparrow^{3} 4 - 3 $$

After playing around a little I found out that it always represents it like this:

$$ A(n,m) = 2 \uparrow^{n-2} (m+3)-3 $$

After seeing this I started to wonder what would happen if you use Graham's number as the arguments. Since the value of it would uncomprehensible I tried to find something to compare it against. Remembering how the number is defined I asked myself if $G_{G_{64}}$ is bigger than $A(G_{64},G_{64})$?

Or in other words:

$$ A(G_{64},G_{64}) = 2 \uparrow^{G_{64} - 2} (G_{64} + 3) - 3 $$ vs $$ G_{G_{64}} = 3 \uparrow^{G_{G_{64}} - 1} 3 $$

I personally supspect it is $G_{G_{64}}$ since it has a lot more arrows but I'm not entirely sure.

BrainStone
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    The first number is approximately $G_{65}$, so the second number is much much bigger. – Peter Oct 15 '16 at 18:40
  • @Peter I thought so too. Is there a decent way to prove that? (Especially $A(G_{64},G_{64}) \approx G_{65}$ (and therefore $A(G_{64},G_{64}) <<< G_{G_{64}}$)) – BrainStone Oct 15 '16 at 18:45
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    Saibian proved a theorem which I used to prove $3\uparrow^{n+1}3>n\uparrow^n n$ for $n\ge 1$. – Peter Oct 15 '16 at 18:52
  • The theorem states $(a\uparrow^m b)\uparrow^m c>a\uparrow^{m} (b+c)$. Look at "Large number Site" for more details. – Peter Oct 15 '16 at 18:55
  • @Peter care to make that an answer? (I understand why what you said is true. But having stuff like that as an answer would be better I guess.) – BrainStone Oct 15 '16 at 19:02
  • I think my prove would not be rigorous. Deedlit seems to be an expert for such questions. – Peter Oct 15 '16 at 19:40
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    @Deedlit Here is a question you probably like. Do you have a rigorous proof for $G_{G_{64}}>A(G_{64},G_{64})$ ? – Peter Oct 15 '16 at 19:41
  • @Peter Of course he does.... – Simply Beautiful Art Jul 25 '17 at 22:32

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In fact, $G_{65}$ is already greater than $Ack(G_{64}, G_{64})$. Using monotonicity of Knuth arrows, we have

$G_{65} = 3 \uparrow^{G_{64}} 3 = 3 \uparrow^{G_{64}-1} (3 \uparrow ^{G_{64}-1} 3) > 2 \uparrow^{G_{64}-2} (G_{64}+3) = Ack(G_{64},G_{64})+3$.

To see that $3 \uparrow^{G_{64}-1} 3 > G_{64}+3$, observe that $3 \uparrow^1 3 = 27$, so again by monotonicity of Knuth arrows we have $3 \uparrow^b 3 \ge b+26$, and in particular $3 \uparrow^{G_{64}-1} 3 \ge G_{64}+25$.

Deedlit
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