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I think it is hopeless to decide whether the number of digits of Graham's number is even or odd because the only way that I can think of is determining the logarithm with accuracy $0.1$ or even better, if the logaithm happens to be near an integer.

Is that right, or is there any trick to decide the question ?

The definition of Graham's number :

If we have the recursion $G_0=4$ , $G_{n+1}=3\uparrow^{G_n} 3$ for all $n\ge 0 $, where $a\uparrow^b c$ denotes Knut's up-arrow-notation, then Graham's number is $G_{64}$.

Knut's uparrow notation works as follows :

$a\uparrow b=a^b$

$a\uparrow \uparrow b=a\uparrow a\uparrow a\uparrow ...\uparrow a\uparrow a\uparrow a\ $ with $b$ $a's$

$a\uparrow \uparrow \uparrow b=a\uparrow \uparrow a \uparrow \uparrow ... \uparrow \uparrow a\uparrow \uparrow a\ $ with $b$ $a's$

and so on.

$a\uparrow^b c$ means $b$ up-arrows between $a$ and $c$

Note, that the calculation is done from right to left, so for example $3\uparrow 3\uparrow 3=3\uparrow 27=3^{27}$ and not $3\uparrow 3\uparrow 3=27\uparrow 3=27^3=3^9$

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  • $\begingroup$ It seems appropriate to elaborate on Graham's number here. $\endgroup$ – barak manos Apr 21 '16 at 19:37
  • $\begingroup$ To avoid trivialities, you should probably restrict the base for the numerals; e.g., decimal or binary. Note that the ternary numeral for any of the $G_i(i \ge 1)$ has evenly many digits, being just $1$ followed by an odd number of $0$s. (For any $i\ge 1$, there is a positive integer $n_i$ such that $G_{i} = 3^{3^{n_i}}$, so the number of digits in the base-$3$ numeral for $G_i$ is $1+3^{n_i}$, which is necessarily even.) $\endgroup$ – r.e.s. Apr 23 '16 at 5:57
  • $\begingroup$ I did not mention a specail base, so I assumed it is clear that I mean the decimal representation. $\endgroup$ – Peter Apr 24 '16 at 20:38
  • $\begingroup$ Graham's number is an enormous stack of $3$'s in a power tower. Taking a log essentially takes one off the bottom of the stack because $\log 3^n=n \log 3$. You are asking to compute $n \log 3$ accurately, which is really no different from computing $3^n$ $\endgroup$ – Ross Millikan Oct 15 '16 at 3:57

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