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I am wondering, is the number of digits in Graham's number greater than the number of protons in the known universe (~10^122)?

Or is there some other 'big' lower bound to the number of digits in G64?

C Shreve
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    It is greater. Because Graham‘s number can not nearly be represented by power notation (even repeated powers) the logarithm ($log_{10} a$ is approximately the number of digits of a) won‘t nearly be representable in power notation as well (it would just remove one layer of repeated exponentiation). – Saha Mar 01 '22 at 20:36
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    Far far far far far far far far far far far far far far far far far far far far far far far greater – Henry Mar 01 '22 at 20:36
  • Graham's number is much greater than $10^{10^{10^{10^{\dots}}}}$. – Crostul Mar 01 '22 at 20:38
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    I'm not sure you'd have enough protons even if you included the imaginary universes. – CyclotomicField Mar 01 '22 at 20:47
  • The number of digits of Graham's number cannot be distinguished from Graham's number itself , although technically the number of digits is of course much smaller. Even the magnitude of $$G_1=3\uparrow \uparrow \uparrow \uparrow 3$$ is utterly incomprehensible. – Peter Mar 02 '22 at 08:31
  • The way to $G_1$ is as follows : Step $1$ is $3$ . Step $2$ is $3\uparrow 3\uparrow 3=3^{27}$. Step $3$ is a power tower of $3^{27}$ threes (call this $N$). So, the number of threes in the power tower is iterated. Continue this way until step $N$. This is $G_1$. The leap to $G_2$ is impossible to grasp and this is just the beginning to the journey to $G_{64}$ – Peter Mar 02 '22 at 08:33

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Say that the "$1$-proton number" is defined as the number of protons in the observable universe, the "$2$-proton number" is the number of protons that would result if you replaced each proton in the observable universe with its own tiny copy of the observable universe, the $3$-proton number is the resulting proton count when replacing each proton there with an even tinier copy of our observable universe, etc. (In general the "$n+1$-proton number" is defined as the resulting proton count when doing this replacement for the universe that is used for totalling the $n$-proton number.) Then Graham's number is greater than the $1,000,000$-proton number, in fact it's even greater than the $a$-proton number where $a$ is the $1,000,000$-proton number.

The $1$-proton number is estimated to be around $10^{80}$ (also known as the Eddington number), so after replacing each proton with its own miniature observable universe containing $\approx 10^{80}$ protons, the resulting proton count (the $2$-proton number) will be $\approx 10^{80}\times 10^{80}$, or $10^{160}$. After replacing each of these $10^{160}$ protons with $10^{80}$ each, the proton count (the $3$-proton number) will be around $10^{240}$, etc. In general for $n>1$ the $n$-proton number is around $10^{80\times n}$.

If our goal is to show that the number of digits in $G_{64}=\textrm{Graham's number}$ is greater than the $a$-proton number where $a$ is itself the $1,000,000$-proton number, it would be enough to show that $G_{64}$ has more than $10^{80\times 10^{80,000,000}}$ digits, i.e. $G_{64}>10^{10^{80\times 10^{80,000,000}}-1}$.

Even $G_1$, the first step in the construction of Graham's number, is much larger than this lower bound. $G_1$ is $3\uparrow\uparrow\uparrow\uparrow 3$, which evaluates to $3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow 3)$. The function $3\uparrow\uparrow\uparrow x$ is increasing, so it's sufficient to show that the even smaller number $3\uparrow\uparrow\uparrow 3$ is greater than the lower bound (this is a decrease of the right argument from $(3\uparrow\uparrow\uparrow 3)$ to just $3$.) $3\uparrow\uparrow\uparrow 3$ evaluates to $3\uparrow\uparrow(3\uparrow\uparrow 3)=3\uparrow\uparrow 7,625,597,484,987$. But the following calculation shows even $3\uparrow\uparrow 5$ is greater than $10^{10^{80\times 10^{80,000,000}}-1}$:

$3^{3^{3^{3^{3^3}}}}$

$=10^{\log(3)\times 10^{\log(3)\times 10^{\log(3)\times 10^{\log(3)\times 3^3}}}}$

$=10^{\log(3)\times 10^{\log(3)\times 10^{\log(3)\times 10^{12.88227387743\ldots}}}}$

$=10^{\log(3)\times 10^{\log(3)\times 10^{10^{12.56090264130034\ldots}}}}$

$=10^{\log(3)\times 10^{10^{\log(\log(3))+10^{12.56090264130034\ldots}}}}$

$=10^{\log(3)\times 10^{10^{10^{12.56090264130026\ldots}}}}$

$=10^{10^{\log(\log(3))+10^{10^{12.56090264130026\ldots}}}}$

$=10^{10^{10^{10^{12.56090264130026\ldots}}}}$, while $10^{10^{80\times 10^{80,000,000}}-1}$ is only $10^{10^{10^{10^{7.903089997\ldots}}}}$.

C7X
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