login
A211668
Number of iterations sqrt(sqrt(sqrt(...(n)...))) such that the result is < 3.
10
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, 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, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3
OFFSET
1,9
COMMENTS
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))
= (x^q + x^(q^p) + x^(q^(p^2)) + x^(q^(p^3)) + ...)/(1-x).
FORMULA
a(3^(2^n)) = a(3^(2^(n-1))) + 1, for n >= 1.
G.f.: g(x) = 1/(1-x)*Sum_{k >= 0} x^(3^(2^k))
= (x^3 + x^9 + x^81 + x^6561 + x^43946721 + ...)/(1 - x).
EXAMPLE
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.
MATHEMATICA
a[n_] := Length[NestWhileList[Sqrt, n, # >= 3 &]] - 1; Array[a, 100] (* Amiram Eldar, Dec 08 2018 *)
PROG
(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
(PARI) A211668(n, c=0)={while(n>=3, n=sqrtint(n); c++); c} \\ M. F. Hasler, Dec 07 2018
(Python) from sympy import integer_log
A048766=lambda n: integer_log(n, 3)[0].bit_length() # Nathan L. Skirrow, May 17 2023
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Hieronymus Fischer, Apr 30 2012
EXTENSIONS
Edited by Michel Marcus, Oct 23 2014 and M. F. Hasler, Dec 07 2018
STATUS
approved