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a(n) = 2*n^(2*n^2 + 2).
6

%I #22 Jan 11 2020 15:57:47

%S 0,2,2048,6973568802,590295810358705651712,

%T 4440892098500626161694526672363281250,

%U 7659889842507589786155370254176860349292085605349868961792,6468953019249515982689295538200433621714406397809250801867790662783382919273856120002

%N a(n) = 2*n^(2*n^2 + 2).

%C Next term is 128 * 8^128 = 2^391, which has 118 digits.

%C This function is a particular instance of the fast-iteration hierarchy function F[k]_n(x). This sequence is also a(n) = F[n]_3(2). See A275000 for details and definitions and formulae.

%C a(n) is an even function.

%F a(n) = f_3(2), with f_0(x) = x+n, f_k+1(x) = (f_k)^x(x).

%F a(n) = a(-n) because 2*h^(2*h^2 + 2) = 2*(-h)^(2*(-h)^2 + 2).

%t Table[2 n^(2 n^2 + 2), {n, 0, 7}] (* _Michael De Vlieger_, Nov 30 2016 *)

%K nonn

%O 0,2

%A _Natan Arie Consigli_, Nov 09 2016