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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. 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.

Sequence in context: A280560 A044932 A211662 * A065687 A300403 A077433

Adjacent sequences:  A211666 A211667 A211668 * A211670 A211671 A211672

KEYWORD

nonn

AUTHOR

Hieronymus Fischer, Apr 30 2012

EXTENSIONS

Edited by M. F. Hasler, Dec 07 2018

STATUS

approved

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Last modified January 21 13:55 EST 2020. Contains 331113 sequences. (Running on oeis4.)