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%I #17 Dec 10 2018 02:58:40
%S 0,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,
%T 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,
%U 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2
%N Number of iterations f(f(f(...(n)...))) such that the result is < 2, where f(x) = cube root of x.
%C For the general case of "Number of iterations f(f(f(...(n)...))) such that the result is < q, where f(x) = x^(1/p)", with p > 1, q > 1, the resulting g.f. is g(x) = 1/(1 - x)*Sum_{k>=0} x^(q^(p^k))
%C = (x^q + x^(q^p) + x^(q^(p^2)) + x^(q^(p^3)) + ...)/(1 - x).
%C The first term that equals 3 is a(512). - _Harvey P. Dale_, Jan 02 2015
%F a(2^(3^n)) = a(2^(3^(n-1))) + 1, for n >= 1.
%F G.f.: 1/(1-x)*Sum_{k>=0} x^(2^(3^k))
%F = (x^2 + x^8 + x^512 + x^134217728 + ...)/(1 - x).
%e a(n) = 1, 2, 3, 4, 5, ... for n = 2^1, 2^3, 2^9, 2^27, 2^81, ..., i.e., n = 2, 8, 512, 134217728, 2417851639229258349412352, ... = A023365.
%t Table[Length[NestWhileList[Surd[#,3]&,n,#>=2&]],{n,90}]-1 (* _Harvey P. Dale_, Jan 02 2015 *)
%o (PARI) a(n,c=0)={while(n>=2, n=sqrtnint(n,3); c++);c} \\ _M. F. Hasler_, Dec 07 2018
%Y Cf. A001069, A010096, A023365, A211662, A211664, A211666, A211668, A211670.
%K nonn
%O 1,8
%A _Hieronymus Fischer_, Apr 30 2012
%E Edited by _M. F. Hasler_, Dec 07 2018
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