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a(n) = 2^(3^(4^...^n)...).
2

%I #16 Oct 10 2018 02:33:31

%S 2,8,2417851639229258349412352

%N a(n) = 2^(3^(4^...^n)...).

%C The next term is too large to include.

%D David Applegate, Marc LeBrun and N. J. A. Sloane, Descending Dungeons and Iterated Base-Changing, in "The Mathematics of Preference, Choice and Order: Essays in Honor of Peter Fishburn", edited by Steven Brams, William V. Gehrlein and Fred S. Roberts, Springer, 2009, pp. 393-402.

%H David Applegate, Marc LeBrun and N. J. A. Sloane, <a href="https://arxiv.org/abs/math/0611293">Descending Dungeons and Iterated Base-Changing</a>, arXiv:math/0611293 [math.NT], 2006-2007.

%H David Applegate, Marc LeBrun, N. J. A. Sloane, <a href="https://www.jstor.org/stable/40391135">Descending Dungeons, Problem 11286</a>, Amer. Math. Monthly, 116 (2009) 466-467.

%e a(4) = 2^(3^4) = 2417851639229258349412352.

%t a[n_] := Fold[#2^#1&, n, Range[2, n-1] // Reverse];

%t Table[a[n], {n, 2, 4}] (* _Jean-François Alcover_, Oct 10 2018 *)

%Y Cf. A014221, A049384, A121263, A121265, A121295, A121296.

%K nonn,bref

%O 2,1

%A _David Applegate_ and _N. J. A. Sloane_, Nov 08 2006