A001069 -id:A001069 - cp's OEIS frontend

A334789 a(n) = 2^log_2(n) where log_2(n) = A001069(n) is the number of log_2(log_2(...log_2(n))) iterations needed to reach < 2.

1, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset: 1

Kevin Ryde, May 10 2020

Differs from A063511 for n>=256. For example a(256)=8 whereas A063511(256)=16. The respective exponent sequences are A001069 (for here) and A211667 (for A063511) which likewise differ for n>=256.

2^log*(n) arises in computational complexity measures for Fürer's multiplication algorithm.

Kevin Ryde, Table of n, a(n) for n = 1..8192
Martin Fürer, Faster integer multiplication, Proceedings of the 39th Annual ACM Symposium on Theory of Computing, 11-13 June 2007. And in SIAM Journal of Computing, volume 30, number 3, 2009, pages 979-1005.
Index to divisibility sequences

Cf. A001069, A014221 (indices of new highs), A063511, A211667.

a(n)=my(t);while(n>1,n=log(n+.5)\log(2);t++);2^t \\ Charles R Greathouse IV, Apr 09 2012

a(n) = my(c=0); while(n>1, n=logint(n,2);c++); 1<Kevin Ryde, May 18 2020

a(n) = 2^A001069(n).

a(n) = 2^lg*(n), where lg*(x) = 0 if x <= 1 and 1 + lg*(log_2(x)) otherwise. - Charles R Greathouse IV, Apr 09 2012

A084558 a(0) = 0; for n >= 1: a(n) = largest m such that n >= m!.

Original entry on oeis.org

0, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5

Offset: 0

Antti Karttunen, Jun 23 2003

For n >= 1, a(n) = the number of significant digits in n's factorial base representation (A007623).
After zero, which occurs once, each n occurs A001563(n) times.
Number of iterations (...(f_4(f_3(f_2(n))))...) such that the result is < 1, where f_j(x):=x/j. - Hieronymus Fischer, Apr 30 2012
For n > 0: a(n) = length of row n in table A108731. - Reinhard Zumkeller, Jan 05 2014

			a(4) = 2 because 2! <= 4 < 3!.

F. Smarandache, "f-Inferior and f-Superior Functions - Generalization of Floor Functions", Arizona State University, Special Collections.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Yi Yuan and Zhang Wenpeng, On the Mean Value of the Analogue of Smarandache Function, Smarandache Notions J., Vol. 15.

A dual to A090529.
Cf. A084555, A084556, A084557.
Cf. A001069, A010096.
Cf. A000142, A001563, A055089, A060130, A111095, A211664, A211670, A108731, A212598, A220656, A220657, A220658, A220659, A231716, A235224, A257510.

a084558 n = a090529 (n + 1) - 1  -- Reinhard Zumkeller, Jan 05 2014

0, seq(m$(m*m!),m=1..5); # Robert Israel, Apr 27 2015

Table[m = 1; While[m! <= n, m++]; m - 1, {n, 0, 104}] (* Jayanta Basu, May 24 2013 *)
Table[Floor[Last[Reduce[x! == n && x > 0, x]]], {n, 120}] (* Eric W. Weisstein, Sep 13 2024 *)

a(n)={my(m=0);while(n\=m++,);m-1} \\ R. J. Cano, Apr 09 2018

def A084558(n):
  i=1
  while n: i+=1; n//=i
  return(i-1)
print(list(map(A084558,range(101)))) # Natalia L. Skirrow, May 28 2023

From Hieronymus Fischer, Apr 30 2012: (Start)
a(n!) = a((n-1)!)+1, for n>1.
G.f.: 1/(1-x)*Sum_{k>=1} x^(k!).
The explicit first terms of the g.f. are: (x+x^2+x^6+x^24+x^120+x^720...)/(1-x).
(End)
Other identities:
For all n >= 0, a(n) = A090529(n+1) - 1. - Reinhard Zumkeller, Jan 05 2014
For all n >= 1, a(n) = A060130(n) + A257510(n). - Antti Karttunen, Apr 27 2015
a(n) ~ log(n^2/(2*Pi)) / (2*LambertW(log(n^2/(2*Pi))/(2*exp(1)))) - 1/2. - Vaclav Kotesovec, Aug 22 2025

Name clarified by Antti Karttunen, Apr 27 2015

A010096 log2*(n) (version 1): number of times floor(log_2(x)) is used in floor(log_2(floor(log_2(...(floor(log_2(n)))...)))) = 0.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 1

Leonid Broukhis

From Hieronymus Fischer, Apr 08 2012: (Start)
A possibly simpler definition could be: "Number of iterations log_2(log_2(log_2(...(n)...))) such that the result is < 1".
Changing "< 1" to "<= 1" produces version 3, A230864.
With the only difference in the termination criterion, the definition is essentially the same as version 2, A001069. If we change the definition to "floor(log_2(... = 1" we get A001069. Therefore we get A001069 when subtracting 1 from each term. (End)

			Becomes 5 at 65536, 6 at 2^65536, etc.

Cf. A063510, A000523, A001069 (version 2), A230864 (version 3).

a010096 = length . takeWhile (/= 0) . iterate a000523
-- Reinhard Zumkeller, Mar 16 2012

f[n_] := Length@ NestWhileList[ Log[2, #] &, n, # >= 1 &] - 1; Array[f, 105] (* Robert G. Wilson v, Apr 19 2012 *)

a(n)=if(n<1,0,1+a(log(n)\log(2))) \\ Charles R Greathouse IV, Apr 17 2012

a(n)=if(n<1,0,1+a(logint(n,2))) \\ Charles R Greathouse IV, Oct 23 2015

From Hieronymus Fischer, Apr 08 2012: (Start)
a(n) = A001069(n) + 1.
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} 2 = 2^(2^(2^2)) = 2^16, we get:
a(E_{i=1..n} 2) = a(E_{i=1..n-1} 2) +1, for n >= 1.
G.f.: g(x) = 1/(1-x)*Sum_{k>=0} x^(E_{i=1..k} 2).
The explicit first terms of this g.f. are
g(x) = (x + x^2 + x^4 + x^16 + x^65536 + ...)/(1-x). (End)

Edited by Hieronymus Fischer, Apr 08 2012

Edited by N. J. A. Sloane, Nov 03 2013

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

Original entry on oeis.org

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).

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

A230864 log2*(n) (version 3): number of iterations log_2(log_2(log_2(...(n)...))) required for the result to be <= 1.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 1

N. J. A. Sloane, Nov 03 2013

nonn

This is the definition of log2* as given by Wikipedia on Nov 03 2013.

If a(n) = k, then n = 2^2^2^...^2^x, where x is in the range 0 < x <= 1 and there are k 2's in the tower. For example a(5)=3 and 5 = 2^2^2^.28134014520...

Wikipedia, Iterated Logarithm

Cf. A010096 (version 1), A001069 (version 2).

a(1)=0; thereafter a(n) = A010096(n-1).

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).

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cp's OEIS Frontend

A334789 a(n) = 2^log_2*(n) where log_2*(n) = A001069(n) is the number of log_2(log_2(...log_2(n))) iterations needed to reach < 2.

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A084558 a(0) = 0; for n >= 1: a(n) = largest m such that n >= m!.

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A010096 log2*(n) (version 1): number of times floor(log_2(x)) is used in floor(log_2(floor(log_2(...(floor(log_2(n)))...)))) = 0.

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

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A334789 a(n) = 2^log_2(n) where log_2(n) = A001069(n) is the number of log_2(log_2(...log_2(n))) iterations needed to reach < 2.