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%I #11 May 19 2021 14:57:04
%S 3,3,4,9,256
%N H_n(n, 2) where H_c(a, b) is the hyperoperation function with operator c.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Hyperoperation">Hyperoperation</a>
%H <a href="/index/Ho">Index Section Ho-Hy</a>
%F a(n) = H_n(n, 2)
%F H_c(a, b) = {b + 1 if c = 0; a if c = 1, b = 0; 0 if c = 2, b = 0; 1 if c >= 3, b = 0; H_{c-1}(a, H_c(a, b - 1)) otherwise}
%e a(0) = H_0(0, 2) = 2 + 1 = 3
%e a(1) = H_1(1, 2) = 1 + 2 = 3
%e a(2) = H_2(2, 2) = 2 * 2 = 4
%e a(3) = H_3(3, 2) = 3 ^ 2 = 9
%e a(4) = H_4(4, 2) = 4 ^^ 2 = 4 ^ 4 = 256
%e a(5) = H_5(5, 2) = 5 ^^^ 2 = 5 ^^ 5 = 5 ^ 5 ^ 5 ^ 5 ^ 5 =~ 10 ^ (10 ^ (10 ^ (2184.1257...)))
%o (Python)
%o def H(a, b, c):
%o if c == 0: return b + 1
%o if c == 1 and b == 0: return a
%o if c == 2 and b == 0: return 0
%o if c >= 3 and b == 0: return 1
%o return H(a, H(a, b - 1, c), c - 1)
%o for n in range(5): print(H(n, 2, n))
%Y For H_n(x,2) with fixed x, cf. A054871 (x=3), A253855 (x=4), A255176 (x=2), A256131 (x=10), A261143 (x=1). - _Danny Rorabaugh_, Oct 20 2015
%K nonn
%O 0,1
%A _Grant Garcia_, Sep 06 2010
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