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%I #28 Jun 30 2023 03:48:13
%S 0,0,1,4,7625597484987
%N Herbert-Ackermann numbers: a(n) = Hack(n,n,n) = H_{n+1}(n,n), where Hack is the Herbert-Ackermann function.
%C The Herbert-Ackermann function is defined as follows:
%C Hack(0,y,z) := y+z;
%C Hack(x,y,0) := 0, x > 0;
%C Hack(x,y,1) := y, x > 0;
%C Hack(x,y,z) := Hack(x-1,y,Hack(x,y,z-1)), x > 0.
%C This is an Ackermann function variant: the recurrence relation is also satisfied by the original (3-argument) Ackermann function.
%C Write a[n]b as H_n(a,b), the n-th hyperoperator of a times b (See A054871 for more information). We have Hack(0,0,0) = 0 and, for x > 0, Hack(x,y,1) = y = y[x+1]1. Suppose Hack(x-1,y,z-1) = y[x](z-1). By induction on z, Hack(x,y,z) = Hack(x-1,y,Hack(x,y,z-1)) = y[x]y[x+1](z-1) = y[x+1]z; so Hack(n,n,n) = n[n+1]n for nonnegative n.
%H Robert S. Boyer, <a href="http://www.cs.utexas.edu/users/boyer/ftp/nqthm/nqthm-1992/examples/basic/peter.events">Versions of Ackermann functions</a>
%F a(-1) = 0;
%F a(n) = Hack(n,n,n), n >= 0.
%F a(n) = H_{n+1}(n,n) = n[n+1]n, n >= -1.
%e a(-1) = (-1)[0](-1) = 0; (successor of -1)
%e a(0) = 0[1]0 = 0+0 = 0;
%e a(1) = 1[2]1 = 1*1 = 1;
%e a(2) = 2[3]2 = 2^2 = 4;
%e a(3) = 3[4]3 = 3^^3 = 3^3^3 = 3^27 = 7625597484987;
%e a(4) = 4[5]4 = 4^^^4 = 4^^4^^4^^4 = 4^^4^^(4^4^4^4) = ... (where 4^4^4^4 = 10^(8.0723... × 10^153), thus 4^^4^^(4^4^4^4) is humongous!)
%e Recursively:
%e a(0) = Hack(0,0,0) = 0+0 = 0;
%e a(1) = Hack(1,1,1) = Hack(0,1,Hack(1,1,0)) = Hack(0,1,0) = 1+0 = 1;
%e a(2) = Hack(2,2,2) = Hack(1,2,Hack(2,2,1)) = Hack(1,2,2) =
%e Hack(0,2,Hack(1,2,1)) = Hack(0,2,2) = 2+2 = 4;
%e a(3) = Hack(3,3,3) = Hack(2,3,Hack(3,3,2)) = Hack(2,3,Hack(2,3,Hack(3,3,1))) = Hack(2,3,Hack(2,3,3)) = ... (the number of recursions is already exploding...)
%Y Cf. Sequences involving 2-argument Ackermann function variants: A001695, A046859, A074877, A126333, A143796, A143797.
%Y For sequences involving 3-argument Ackermann function variants see A054871.
%Y Cf. A004231.
%K nonn
%O -1,4
%A _Natan Arie Consigli_, Dec 23 2015
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