A211668 -id:A211668 - cp's OEIS frontend

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).

A211664 Number of iterations (...(log_4(log_3(log_2(n))))...) such that the result is < 1.

Original entry on oeis.org

1, 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, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

Offset: 1

Hieronymus Fischer, Apr 30 2012

			a(n)=1, 2, 3, 4, 5 for n=1, 2, 2^3, 2^3^4, 2^3^4^5 (=1, 2, 8, 2417851639229258349412352, 2^3^1024).

Cf. A001069, A010096, A084558, A211661, A211666, 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} := 1; example: E_{i=1..4} 3 = 3^(3^(3^3)) = 3^(3^27), we get:
a(E_{i=1..n} (i+1)) = a(E_{i=1..n-1} (i+1))+1, for n>=1.
G.f.: g(x) = (1/(1-x))*Sum_{k>=0} x^(E_{i=1..k} (i+1)).
The explicit first terms of the g.f. are g(x) = (x+x^2+x^(2^3)+x^(2^3^4)+x^(2^3^4^5)+...)/(1-x) =(x+x^2+x^8+x^2417851639229258349412352+...)/(1-x).

A211662 Number of iterations log_3(log_3(log_3(...(n)...))) such that the result is < 2.

Original entry on oeis.org

0, 1, 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, 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

			Records a(n)=0, 1, 2, 3, 4, for n=1, 2, 3^2, 3^3^2, 3^3^3^2 (=1, 2, 9, 3^9 = 19683, 3^19683).

Cf. A001069, A010096, A211661, 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(E_{i=1..n} 3) = a(E_{i=1..n-1} 3)+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)=3 for i

A211661 Number of iterations log_3(log_3(log_3(...(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, 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

Offset: 1

Hieronymus Fischer, Apr 30 2012

For n<16 same as A211663.

			a(n)=1, 2, 3, 4, 5 for n=1, 3, 3^3, 3^3^3, 3^3^3^3 (=1, 3, 27, 7625597484987, 3^7625597484987).

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

Table[Length[NestWhileList[Log[3,#]&,n,#>=1&]],{n,90}]-1 (* Harvey P. Dale, Mar 08 2020 *)

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(E_{i=1..n} 3) = a(E_{i=1..n-1} 3)+1, for n>=1.
G.f.: g(x) = (1/(1-x))*Sum_{k>=0} x^(E_{i=1..k} 3). The explicit first terms of the g.f. are g(x) = (x+x^3+x^27+x^7625597484987+...)/(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).

A211665 Minimal number of iterations of log_10 applied to n until the result is < 1.

Original entry on oeis.org

1, 1, 1, 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, 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

Different from A055642 and A138902, cf. Example.

Instead the real-valued log function one can consider only the integer part (i.e., A004216), since log_b(x) < k <=> x < b^k <=> floor(x) < b^k for any integer k >= 0; that's also why the first 2, 3, 4, ... appears exactly for 10, 10^10, 10^(10^10) etc. - M. F. Hasler, Dec 12 2018

			a(n) = 1, 2, 3, 4 for n = 1, 10, 10^10, 10^(10^10), i.e., n = 1, 10, 10000000000, 10^10000000000.
a(n) = 2 for all n >= 10, n < 10^10.

Cf. A001069, A010096, A211661, A211663, A211666, A211668, A211670.

a[n_] := Length[NestWhileList[Log10, n, # >= 1 &]] - 1; Array[a, 100] (* Amiram Eldar, Dec 08 2018 *)

a(n,i=1)={while(n=logint(n,10),i++);i} \\ M. F. Hasler, Dec 07 2018

With 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..3} 10 = 10^(10^10) = 10^10000000000, we have:
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>=0} x^(E_{i=1..k} 10) = (x + x^10 + x^(10^10) + ...)/(1-x).

Name reworded by M. F. Hasler, Dec 12 2018

Showing 1-9 of 9 results.

cp's OEIS Frontend

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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A211664 Number of iterations (...(log_4(log_3(log_2(n))))...) such that the result is < 1.

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A211662 Number of iterations log_3(log_3(log_3(...(n)...))) such that the result is < 2.

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A211661 Number of iterations log_3(log_3(log_3(...(n)...))) such that the result is < 1.

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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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A211665 Minimal number of iterations of log_10 applied to n until the result is < 1.

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