A211670 -id:A211670 - cp's OEIS frontend

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

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

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).

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).

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

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

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

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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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A211664 Number of iterations (...(log_4(log_3(log_2(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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A211665 Minimal number of iterations of log_10 applied to n until the result is < 1.

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