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Number of iterations log_3(log_3(log_3(...(n)...))) such that the result is < 1.
4

%I #8 Mar 08 2020 19:46:33

%S 1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,

%T 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,

%U 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3

%N Number of iterations log_3(log_3(log_3(...(n)...))) such that the result is < 1.

%C For n<16 same as A211663.

%F With the exponentiation definition E_{i=1..n} c(i) := c(1)^(c(2)^(c(3)^(...(c(n-1)^(c(n))))...))); E_{i=1..0} := 1; example: E_{i=1..4} 3 = 3^(3^(3^3)) = 3^(3^27), we get:

%F a(E_{i=1..n} 3) = a(E_{i=1..n-1} 3)+1, for n>=1.

%F G.f.: g(x)= 1/(1-x)*sum_{k=0..infinity} x^(E_{i=1..k} 3). The explicit first terms of the g.f. are

%F g(x)=(x+x^3+x^27+x^7625597484987+…)/(1-x).

%e a(n)=1, 2, 3, 4, 5 for n=1, 3, 3^3, 3^3^3, 3^3^3^3 =1, 3, 27, 7625597484987, 3^7625597484987

%t Table[Length[NestWhileList[Log[3,#]&,n,#>=1&]],{n,90}]-1 (* _Harvey P. Dale_, Mar 08 2020 *)

%Y Cf. A001069, A010096, A211664, A211666, A211668, A211669.

%K base,nonn

%O 1,3

%A _Hieronymus Fischer_, Apr 30 2012