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%I #13 Jan 11 2020 15:57:47
%S 0,1,2,4,256
%N a(0) = 0, a(n) = a(n-1)[n-1]a(n-1).
%C a[n]b is the square-bracket notation for n-th hyperoperation of a times b see A054871 for more info.
%e a(0) = 0;
%e a(1) = a(0)[0]a(0) = 1;
%e a(2) = a(1)[1]a(1) = 1+1 = 2;
%e a(3) = a(2)[2]a(2) = 2*2 = 4;
%e a(4) = a(3)[3]a(3) = 4^4 = 256;
%e a(5) = a(4)[4]a(4) = 256^^256 = 256^256^....^256 (256 times).
%Y Cf. A271552.
%K nonn,bref
%O 0,3
%A _Natan Arie Consigli_, Apr 09 2016
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