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Search: a211662 -id:a211662
Displaying 1-6 of 6 results found.
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A211668
Number of iterations sqrt(sqrt(sqrt(...(n)...))) such that the result is < 3.
+10
10
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
(list; graph; refs; listen; history; text; internal format)
OFFSET
1,9
COMMENTS
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).
LINKS
Table of n, a(n) for n=1..86.
FORMULA
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).
EXAMPLE
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.
MATHEMATICA
a[n_] := Length[NestWhileList[Sqrt, n, # >= 3 &]] - 1; Array[a, 100] (* Amiram Eldar, Dec 08 2018 *)
PROG
(PARI) 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
(PARI) A211668(n, c=0)={while(n>=3, n=sqrtint(n); c++); c} \\ M. F. Hasler, Dec 07 2018
(Python) from sympy import integer_log
A048766=lambda n: integer_log(n, 3)[0].bit_length() # Nathan L. Skirrow, May 17 2023
CROSSREFS
Cf. A001069, A010096, A211662, A211666, A211670.
KEYWORD
base,nonn
AUTHOR
Hieronymus Fischer, Apr 30 2012
EXTENSIONS
Edited by Michel Marcus, Oct 23 2014 and M. F. Hasler, Dec 07 2018
STATUS
approved
A211666
Number of iterations log_10(log_10(log_10(...(n)...))) such that the result is < 2.
+10
9
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
(list; graph; refs; listen; history; text; internal format)
OFFSET
1,100
COMMENTS
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..infinity} x^(E_{i=1..k} b(i,k)), where b(i,k)=p for i<k and b(i,k)=q for i=k. The explicit first terms of the g.f. are
g(x) = (x^q+x^(p^q)+x^(p^p^q)+x^(p^p^p^q)+…)/(1-x).
LINKS
Table of n, a(n) for n=1..105.
FORMULA
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..infinity} x^(E_{i=1..k} b(i,k)), where b(i,k)=10 for i<k and b(i,k)=2 for i=k.
The explicit first terms of the g.f. are
g(x) = (x^2+x^100+x^(10^100)+...)/(1-x).
EXAMPLE
a(n) = 0, 1, 2, 3 for n = 1, 2, 10^2, 10^10^2 = 1, 2, 100, 10^100.
CROSSREFS
Cf. A001069, A010096, A211662, A211664, A211668, A211670.
KEYWORD
base,nonn
AUTHOR
Hieronymus Fischer, Apr 30 2012
STATUS
approved
A211670
Number of iterations (...f_4(f_3(f_2(n))))...) such that the result is < 2, where f_j(x):=x^(1/j).
+10
8
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
(list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
COMMENTS
Different from A001069, but equal for n < 16.
LINKS
Table of n, a(n) for n=1..90.
FORMULA
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).
EXAMPLE
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.
PROG
(Python)
def A084558(n):
i=1
while n: i+=1; n//=i
return(i-1)
A211670=lambda n: n and A084558(n.bit_length()-1) # Nathan L. Skirrow, May 17 2023
CROSSREFS
Cf. A001069, A010096, A084558, A211662, A211664, A211666, A211668, A211669.
KEYWORD
base,nonn
AUTHOR
Hieronymus Fischer, Apr 30 2012
STATUS
approved
A211669
Number of iterations f(f(f(...(n)...))) such that the result is < 2, where f(x) = cube root of x.
+10
4
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
(list; graph; refs; listen; history; text; internal format)
OFFSET
1,8
COMMENTS
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
LINKS
Table of n, a(n) for n=1..86.
FORMULA
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).
EXAMPLE
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.
MATHEMATICA
Table[Length[NestWhileList[Surd[#, 3]&, n, #>=2&]], {n, 90}]-1 (* Harvey P. Dale, Jan 02 2015 *)
PROG
(PARI) a(n, c=0)={while(n>=2, n=sqrtnint(n, 3); c++); c} \\ M. F. Hasler, Dec 07 2018
CROSSREFS
Cf. A001069, A010096, A023365, A211662, A211664, A211666, A211668, A211670.
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, Apr 30 2012
EXTENSIONS
Edited by M. F. Hasler, Dec 07 2018
STATUS
approved
A211667
Number of iterations sqrt(sqrt(sqrt(...(n)...))) such that the result is < 2.
+10
3
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
(list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
COMMENTS
Different from A001069, but equal for n < 256.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
FORMULA
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
EXAMPLE
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.
MATHEMATICA
a[n_] := Length[NestWhileList[Sqrt, n, # >= 2 &]] - 1; Array[a, 100] (* Amiram Eldar, Dec 08 2018 *)
PROG
(PARI) 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
(PARI) a(n) = if(n<=1, 0, logint(logint(n, 2), 2) + 1); \\ Kevin Ryde, Jan 18 2024
(Python) A211667=lambda n: n and (n.bit_length()-1).bit_length() # Nathan L. Skirrow, May 16 2023
CROSSREFS
Cf. A001069, A001146, A010096, A211662, A211668, A211670.
Cf. A087046 (run lengths).
KEYWORD
nonn,easy
AUTHOR
Hieronymus Fischer, Apr 30 2012
STATUS
approved
A211663
Number of iterations log(log(log(...(n)...))) such that the result is < 1.
+10
2
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
(list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
Same as A211661 for n < 16.
LINKS
Table of n, a(n) for n=1..88.
FORMULA
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).
EXAMPLE
a(n)=1, 2, 3, 4, for n=1, ceiling(e), ceiling(e^e), ceiling(e^e^e), = 1, 3, 16, 3814280, respectively.
CROSSREFS
Cf. A001069, A010096, A211662, A211664, A211666, A211668, A211669.
KEYWORD
base,nonn
AUTHOR
Hieronymus Fischer, Apr 30 2012
STATUS
approved
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