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a(n) is the largest number of coins obtainable by making repeated moves in this puzzle: Start with 1 coin in each of n boxes B(i), i=1..n. One can iterate moves of two types: (1) remove a coin from a nonempty B(i) (i <= n-1) and place two coins in B(i+1); (2) remove a coin from a nonempty B(i) (i <= n-2) and switch the contents of B(i+1) and B(i+2).
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%I #32 Apr 19 2019 10:33:42

%S 1,3,7,28

%N a(n) is the largest number of coins obtainable by making repeated moves in this puzzle: Start with 1 coin in each of n boxes B(i), i=1..n. One can iterate moves of two types: (1) remove a coin from a nonempty B(i) (i <= n-1) and place two coins in B(i+1); (2) remove a coin from a nonempty B(i) (i <= n-2) and switch the contents of B(i+1) and B(i+2).

%C An Ackermann-like function. The underlying puzzle was invented by Hans Zantema. The derivation and proof of the general formula involving a palindromic sequence of up-arrows is by Richard Stong.

%C The next term is too large to include (2^16385, it has 4933 digits).

%H Zuming Feng, Po-Shen Loh, and Yi Sun, <a href="http://yisun.io/papers/imo2010.pdf">51st International Mathematical Olympiad</a>, Math. Mag. 83 (2010), pp. 320-323.

%H Terence Tao, <a href="https://polymathprojects.org/2010/07/08/minipolymath2-project-imo-2010-q5/">Minipolymath2 project: IMO 2010 Q5</a> (2010)

%H A. van den Brandhof, J. Guichelaar, and A. Jaspers, <a href="http://www.maa.org/press/ebooks/half-a-century-of-pythagoras-magazine">Half a Century of Pythagoras Magazine</a>, MAA, 2015, 225

%H Stan Wagon, <a href="http://mathforum.org/wagon/2017/p1233.html">The Generous Automated Teller Machine</a>

%H Stan Wagon, <a href="/A281701/a281701.pdf">Richard Stong's proof of the uparrow formula</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Knuth&#39;s_up-arrow_notation">Knuth's up-arrow notation</a>

%F Let f_n(x) = 2↑↑...↑x, with n Knuth up-arrows, so f_0(x) = 2x, f_1(x) = 2^x, f_2(x) = 2↑↑x = 2^2^...^2 with x copies of 2, etc.

%F Let F_n be the composition of f_0, f_1,...,f_(n-4).

%F Let G_n be the same composition but in the opposite order.

%F Then a(n) = G_n(F_n(7)), a formula due to Richard Stong.

%e a(5) = f_0(f_1(f_1(f_0(7)))) = 2*2^(2^(2*7)) = 2*2^(2^14) = 2^16385.

%Y Cf. A307611.

%K nonn,nice

%O 1,2

%A _Stan Wagon_, Jan 27 2017