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 $k$ ?
2026-03-25 04:39:17.1774413557
Comparison of $4$-entry-conway-chains and $3$-entry-conway-chains
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1
$$n \approx 3 \rightarrow 3 \rightarrow ([3 \rightarrow 3 \rightarrow 27 \rightarrow 2] -1) \rightarrow 2$$
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 $n \approx k\rightarrow k\rightarrow k\rightarrow k$.