3
$\begingroup$
  • How big must $n$ approximately be, that

    $$n\rightarrow n \rightarrow n\ \approx \ 3 \rightarrow 3 \rightarrow 3 \rightarrow 3$$

    holds ?

    $n\rightarrow n \rightarrow n$ (conway-chain) is equivalent to $n \uparrow^n n$ (Knuth's up-arrow notation)

  • Is there a general method to calculate approximately $n$ such that

    $$n \rightarrow n \rightarrow n \approx k \rightarrow k \rightarrow k \rightarrow k$$

    for a given $k$ ?

$\endgroup$
  • 1
    $\begingroup$ Be advised that Graham's number lies between $3\to 3\to 64 \to 2$ and $3\to 3 \to 65 \to 2$, while what you're asking for is $3 \to 3\to (3\to 3\to 27 \to 2) \to 2$, which is much larger. So the number $n$ will be quite large. $\endgroup$ – Arthur Nov 29 '14 at 19:06
2
$\begingroup$

$$n \approx 3 \rightarrow 3 \rightarrow ([3 \rightarrow 3 \rightarrow 27 \rightarrow 2] -1) \rightarrow 2$$

Since even $3 \rightarrow 3 \rightarrow 27 \rightarrow 2$ is quite large, this is approximately $3 \rightarrow 3 \rightarrow 3 \rightarrow 3$.


To see this, note that $$3 \rightarrow 3 \rightarrow 3 \rightarrow 3 = 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 27 \rightarrow 2) \rightarrow 2$$

And $$(a \rightarrow a \rightarrow b \rightarrow 2) \uparrow^{(a \rightarrow a \rightarrow b \rightarrow 2)} (a \rightarrow a \rightarrow b \rightarrow 2) \approx a \rightarrow a \rightarrow (b+1) \rightarrow 2$$


For larger $k$, one will have $n \approx k\rightarrow k\rightarrow k\rightarrow k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.