A211667 Number of iterations sqrt(sqrt(sqrt(...(n)...))) such that the result is < 2.

0, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 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, 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, 3, 3, 3

Offset: 1

Author

Hieronymus Fischer, Apr 30 2012

Keywords

• nonn
• easy

Comments

Different from A001069, but equal for n < 256.

Examples

			a(n) = 1, 2, 3, 4, 5, ... for n = 2^1, 2^2, 2^4, 2^8, 2^16, ..., i.e., n = 2, 4, 16, 256, 65536, ... = A001146.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001069, A001146, A010096, A211662, A211668, A211670.

Cf. A087046 (run lengths).

Programs

Mathematica

a[n_] := Length[NestWhileList[Sqrt, n, # >= 2 &]] - 1; Array[a, 100] (* Amiram Eldar, Dec 08 2018 *)

PARI

apply( A211667(n, c=0)={while(n>=2, n=sqrtint(n); c++); c}, [1..50]) \\ This defines the function A211667. The apply(...) provides a check and illustration. - M. F. Hasler, Dec 07 2018

PARI

a(n) = if(n<=1,0, logint(logint(n,2),2) + 1); \\ Kevin Ryde, Jan 18 2024

Python

A211667=lambda n: n and (n.bit_length()-1).bit_length() # Natalia L. Skirrow, May 16 2023

Formula

a(2^(2^n)) = a(2^(2^(n-1))) + 1, for n >= 1.

G.f.: g(x) = 1/(1-x)*Sum_{k>=0} x^(2^(2^k))

= (x^2 + x^4 + x^16 + x^256 + x^65536 + ...)/(1 - x).

a(n) = 1 + floor(log_2(log_2(n))) for n>=2. - Kevin Ryde, Jan 18 2024