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a(n) = H_n(4,2) where H_n is the n-th hyperoperator.
3

%I #21 Jan 11 2020 15:57:46

%S 3,6,8,16,256

%N a(n) = H_n(4,2) where H_n is the n-th hyperoperator.

%C See A054871 for definitions and key links.

%C This sequence is also a(0)=0, a(1)=1, a(n)=H_{n-1)(4,4) for n>1.

%e a(0) = H_0(4,2) = 2+1 = 3

%e a(1) = H_1(4,2) = 4+2 = 6.

%e a(2) = H_2(4,2) = 4*2 = 8.

%e a(3) = H_3(4,2) = 4^2 = 16.

%e a(4) = H_4(4,2) = 4^^2 = 4^4 = 256.

%e a(5) = H_5(4,2) = 4^^^2 = 4^^4 = 4^4^4^4 = 4^4^256 = ...

%Y Cf. A054871, A067652.

%K nonn,less,bref

%O 0,1

%A _Natan Arie Consigli_, May 02 2015

%E First term corrected and hyperoperator notation implemented by _Danny Rorabaugh_, Oct 20 2015