|
%I #20 May 17 2023 08:35:28
%S 0,0,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,
%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 2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3
%N Number of iterations sqrt(sqrt(sqrt(...(n)...))) such that the result is < 3.
%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), 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).
%F a(3^(2^n)) = a(3^(2^(n-1))) + 1, for n >= 1.
%F G.f.: g(x) = 1/(1-x)*Sum_{k >= 0} x^(3^(2^k))
%F = (x^3 + x^9 + x^81 + x^6561 + x^43946721 + ...)/(1 - x).
%e a(n) = 1, 2, 3, 4, 5 for n = 3^1, 3^2, 3^4, 3^8, 3^16, i.e., n = 3, 9, 81, 6561, 43946721.
%t a[n_] := Length[NestWhileList[Sqrt, n, # >= 3 &]] - 1; Array[a, 100] (* _Amiram Eldar_, Dec 08 2018 *)
%o (PARI) a(n) = {my(nbi = 0); if (n < 3, return (nbi)); r = n; nbi= 1; while ((nr = sqrt(r)) >= 3, nbi++; r = nr); return (nbi);} \\ _Michel Marcus_, Oct 23 2014
%o (PARI) A211668(n, c=0)={while(n>=3, n=sqrtint(n); c++); c} \\ _M. F. Hasler_, Dec 07 2018
%o (Python) from sympy import integer_log
%o A048766=lambda n: integer_log(n,3)[0].bit_length() # _Nathan L. Skirrow_, May 17 2023
%Y Cf. A001069, A010096, A211662, A211666, A211670.
%K base,nonn
%O 1,9
%A _Hieronymus Fischer_, Apr 30 2012
%E Edited by _Michel Marcus_, Oct 23 2014 and _M. F. Hasler_, Dec 07 2018
|
