A211662 -id:A211662 - cp's OEIS frontend

A211668 Number of iterations sqrt(sqrt(sqrt(...(n)...))) such that the result is < 3.

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

Hieronymus Fischer, Apr 30 2012

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).

			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.

Cf. A001069, A010096, A211662, A211666, A211670.

a[n_] := Length[NestWhileList[Sqrt, n, # >= 3 &]] - 1; Array[a, 100] (* Amiram Eldar, Dec 08 2018 *)

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

A211668(n, c=0)={while(n>=3, n=sqrtint(n); c++); c} \\ M. F. Hasler, Dec 07 2018

from sympy import integer_log
A048766=lambda n: integer_log(n,3)[0].bit_length() # Natalia L. Skirrow, May 17 2023

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).

Edited by Michel Marcus, Oct 23 2014 and M. F. Hasler, Dec 07 2018

A211666 Number of iterations log_10(log_10(log_10(...(n)...))) such that the result is < 2.

Original entry on oeis.org

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

Offset: 1

Hieronymus Fischer, Apr 30 2012

Different from A004216, A057427 and A185114.
For a general definition like "Number of iterations log_p(log_p(log_p(...(n)...))) such that the result is < q", where p > 1, q > 0, the resulting g.f. is
g(x) = (1/(1-x))*Sum_{k>=1} x^(E_{i=1..k} b(i,k)), where b(i,k)=p for i

			a(n) = 0, 1, 2, 3 for n = 1, 2, 10^2, 10^10^2 (= 1, 2, 100, 10^100).

Cf. A001069, A010096, A211662, A211664, A211668, A211670.

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} c := 1; example: E_{i=1..3} 10 = 10^(10^10) = 10^10000000000, we get:
a(E_{i=1..n} 10) = a(E_{i=1..n-1} 10)+1, for n>=1.
G.f.: g(x) = (1/(1-x))*Sum_{k>=1} x^(E_{i=1..k} b(i,k)), where b(i,k)=10 for iThe explicit first terms of the g.f. are g(x) = (x^2+x^100+x^(10^100)+...)/(1-x).

A211670 Number of iterations (...(f_4(f_3(f_2(n))))...) such that the result is < 2, where f_j(x) := x^(1/j).

Original entry on oeis.org

0, 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, 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

Hieronymus Fischer, Apr 30 2012

Different from A001069, but equal for n < 16.

			a(n)=1, 2, 3, 4, 5 for n=2^(1!), 2^(2!), 2^(3!), 2^(4!), 2^(5!) (=2, 4, 64, 16777216, 16777216^5).

Cf. A001069, A010096, A084558, A211662, A211664, A211666, A211668, A211669.

def A084558(n):
 i=1
 while n: i+=1;n//=i
 return(i-1)
A211670=lambda n: n and A084558(n.bit_length()-1) # Natalia L. Skirrow, May 17 2023

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

G.f.: g(x) = (1/(1-x))*Sum_{k>=1} x^(2^(k!)). The explicit first terms of the g.f. are g(x) = (x^2+x^4+x^64+x^16777216+...)/(1-x).

A211669 Number of iterations f(f(f(...(n)...))) such that the result is < 2, where f(x) = cube root of x.

Original entry on oeis.org

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, 2, 2, 2, 2, 2, 2, 2

Offset: 1

Hieronymus Fischer, Apr 30 2012

nonn

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))
  = (x^q + x^(q^p) + x^(q^(p^2)) + x^(q^(p^3)) + ...)/(1 - x).
The first term that equals 3 is a(512). - Harvey P. Dale, Jan 02 2015

			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.

Cf. A001069, A010096, A023365, A211662, A211664, A211666, A211668, A211670.

Table[Length[NestWhileList[Surd[#,3]&,n,#>=2&]],{n,90}]-1 (* Harvey P. Dale, Jan 02 2015 *)

a(n,c=0)={while(n>=2, n=sqrtnint(n,3); c++);c} \\ M. F. Hasler, Dec 07 2018

a(2^(3^n)) = a(2^(3^(n-1))) + 1, for n >= 1.
G.f.: 1/(1-x)*Sum_{k>=0} x^(2^(3^k))
  = (x^2 + x^8 + x^512 + x^134217728 + ...)/(1 - x).

Edited by M. F. Hasler, Dec 07 2018

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

Original entry on oeis.org

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

Hieronymus Fischer, Apr 30 2012

Different from A001069, but equal for n < 256.

			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.

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

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

Cf. A087046 (run lengths).

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

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

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

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

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

A211663 Number of iterations log(log(log(...(n)...))) such that the result is < 1.

Original entry on oeis.org

1, 1, 2, 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, 3, 3

Offset: 1

Hieronymus Fischer, Apr 30 2012

Same as A211661 for n < 16.

			a(n)=1, 2, 3, 4, for n=1, ceiling(e), ceiling(e^e), ceiling(e^e^e), = 1, 3, 16, 3814280, respectively.

Cf. A001069, A010096, A211662, A211664, A211666, A211668, A211669.

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:
a(ceiling(E_{i=1..n} e)) = a(ceiling(E_{i=1..n-1} e))+1, for n>=1.
G.f.: g(x) = (1/(1-x))*Sum_{k>=0} x^(ceiling(E_{i=1..k} e)). The explicit first terms of the g.f. are g(x) = (x + x^3 + x^16 + x^3814280 + ...)/(1-x).

Showing 1-6 of 6 results.

cp's OEIS Frontend

A211668 Number of iterations sqrt(sqrt(sqrt(...(n)...))) such that the result is < 3.

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A211666 Number of iterations log_10(log_10(log_10(...(n)...))) such that the result is < 2.

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A211670 Number of iterations (...(f_4(f_3(f_2(n))))...) such that the result is < 2, where f_j(x) := x^(1/j).

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A211669 Number of iterations f(f(f(...(n)...))) such that the result is < 2, where f(x) = cube root of x.

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

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A211663 Number of iterations log(log(log(...(n)...))) such that the result is < 1.

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