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Let m be the n-th hyperoperation applied to n (see A189896). a(n) is the m-th hyperoperation applied to n and m.
1

%I #19 Nov 28 2021 03:12:01

%S 1,2,65536

%N Let m be the n-th hyperoperation applied to n (see A189896). a(n) is the m-th hyperoperation applied to n and m.

%C Let bn=A189896(n), that is, bn=H_n(n,n), where H_n is the n-th hyperoperation.

%C Then

%C a(n) = H_bn(n,bn), hence

%C a(0) = H_b0(0,b0) = H_1(0,1) = 0+1 = 1,

%C a(1) = H_b1(1,b1) = H_2(1,2) = 1*2 = 2,

%C a(2) = H_b2(2,b2) = H_4(2,4) = 2^2^2^2 = 65536, and

%C a(3) is too large to include here.

%F a(n) = H_H_n(n,n)(n,H_n(n,n)).

%e a(0) = H_H_0(0,0)(0,H_0(0,0)) = H_1(0,1) = 0+1 = 1.

%e a(1) = H_H_1(1,1)(1,H_1(1,1)) = H_2(1,1+1) = 1*(1+1) = 2.

%e a(2) = H_H_2(2,2)(2,H_2(2,2)) = H_4(2,2*2) = H_3(2,H_3(2,H_3(2,2))) = 2^2^2^2 = 2^2^4 = 2^16 = 65536.

%Y Cf. A189896.

%K nonn,bref

%O 0,2

%A _Jens Ahlström_, Jun 16 2012