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%I #31 Mar 27 2020 10:14:46
%S 4,5,6,8,16,65536
%N a(n) = H_n(2,3) = H_(n-1)(2,4) where H_n is the n-th hyperoperator.
%C Originally named "2 plus 3, twice 3, 2 to the power of 3, etc."
%C For hyperoperator definitions and links, see A054871.
%C For nonnegative n, H_(n)(2,3) = H_(n-1)(2,H_(n-1)(2,2)) = H_(n-1)(2,4) or in the clearer square bracket notation: 2[n]3 = 2[n-1]2[n-1]2 = 2[n-1]4. - _Natan Arie Consigli_, Dec 07 2015
%e H_0(2,3) = 3+1 = 4;
%e H_1(2,3) = 2+3 = 5;
%e H_2(2,3) = 2*3 = 6;
%e H_3(2,3) = 2^3 = 2*2*2 = 2*4 = H_2(2,4) = 8;
%e H_4(2,3) = 2^^3 = 2^2^2 = 2^4 = H_3(2,4) = 16;
%e H_5(2,3) = 2^^^3 = 2^^2^^2 = 2^^4 = H_4(2,4) = 2^2^2^2 = 2^16 = 65536;
%e H_6(2,3) = 2^^^^3 = 2^^^2^^^2 = 2^^^4 = H_5(2,4) = 2^^2^^2^^2 = 2^^65536 = 2^2^...^2^2, with 65536 2's.
%o (Haskell) f a 0 = 2 + a / f 0 1 = 0 / f 0 n = 1 / f a n = f (f (a-1) n) (n-1)
%Y Cf. A054871.
%K nonn,bref
%O 0,1
%A Ashley Yakeley (ashley(AT)yakeley.org), Feb 03 2002
%E Hyperoperator notation, new initial term, and examples by _Danny Rorabaugh_, Oct 14 2015
%E Sequence merged with H_(n)(2,4) by _Natan Arie Consigli_, Dec 07 2015
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