A211666 -id:A211666 - cp's OEIS frontend

A023532 a(n) = 0 if n is of the form m*(m+3)/2, otherwise 1.

0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Clark Kimberling

From Stark: "alpha = 0.101101110111101111101111110 ... is irrational. For if alpha were rational, its decimal expansion would be periodic and have a period of length r starting with the k-th digit of the expansion.
"But by the very nature of alpha, there will be blocks of r digits, all 1, in this expansion after the k-th digit and the periodicity would then guarantee that everything after such a block of r digits would also be all ones.
"This contradicts the fact that there will always be zeros occurring after any given point in the expansion of alpha. Hence alpha is irrational."
a(A000096(n)) = 0; a(A007401(n)) = 1. - Reinhard Zumkeller, Dec 04 2012
Sequence B is called a reverse reluctant sequence of sequence A, if B is triangle array read by rows: row number k lists first k elements of the sequence A in reverse order. A023532 is reverse reluctant sequence of sequence A211666. - Boris Putievskiy, Jan 11 2013
An example of a sequence with infinite critical exponent [Vaslet]. - N. J. A. Sloane, May 05 2013

			From _Boris Putievskiy_, Jan 11 2013: (Start)
As a triangular array written by rows, the sequence begins:
  0;
  1, 0;
  1, 1, 0;
  1, 1, 1, 0;
  1, 1, 1, 1, 0;
  1, 1, 1, 1, 1, 0;
  1, 1, 1, 1, 1, 1, 0;
  ...
(End)

Harold M. Stark, An Introduction to Number Theory, The MIT Press, Cambridge, Mass, eighth printing 1994, page 170.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Elise Vaslet, Critical exponents of words over 3 letters, Electronic Journal of Combinatorics, 18 (2011), #P125.
Index entries for characteristic functions

Cf. A023531, A211666.

Essentially the same sequence as A114607 and A123110. - N. J. A. Sloane, Feb 07 2020

a023532 = (1 -) . a010052 . (+ 9) . (* 8)
a023532_list = concat $ iterate (\rs -> 1 : rs) [0]
-- Reinhard Zumkeller, Dec 04 2012

A023532 := proc(n)
    option remember ;
    local m,t ;
    for m from 0 do
        t := m*(m+3)/2 ;
        if t > n then
            return 1 ;
        elif t = n then
            return 0 ;
        end if;
    end do:
end proc:
seq(A023532(n),n=0..40) ; # R. J. Mathar, May 15 2025

a = {}; Do[a = Append[a, Join[ {0}, Table[1, {n} ] ] ], {n, 1, 13} ]; a = Flatten[a]
Table[PadLeft[{0},n,1],{n,0,20}]//Flatten (* Harvey P. Dale, Jul 10 2019 *)

for(n=1,9,print1("0, ");for(i=1,n,print1("1, "))) \\ Charles R Greathouse IV, Jun 16 2011

a(n)=!issquare(8*n+9) \\ Charles R Greathouse IV, Jun 16 2011

from sympy.ntheory.primetest import is_square
def A023532(n): return bool(is_square((n<<3)+9))^1 # Chai Wah Wu, Feb 10 2023

a(n) = 0 if and only if 8n+9 is a square. - Charles R Greathouse IV, Jun 16 2011
Blocks of lengths 1, 2, 3, 4, ... of ones separated by a single zero.
a(n) = 1 - floor((sqrt(9+8n)-1)/2) + floor((sqrt(1+8n)-1)/2). - Paul Barry, May 25 2004
a(n) = A211666(m), where m = (t^2 + 3*t + 4)/2n - n, t = floor((-1 + sqrt(8*n-7))/2). - Boris Putievskiy, Jan 11 2013
a(n) = [A002262(n) < A003056(n)]. - Yuchun Ji, May 18 2020
a(n) = 1-A023531(n). - R. J. Mathar, May 15 2025

Additional comments from Robert G. Wilson v, Nov 06 2000

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

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

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

A023532 a(n) = 0 if n is of the form m*(m+3)/2, otherwise 1.

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

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