A160113 Number of cubefree integers not exceeding 2^n.

1, 2, 4, 7, 14, 27, 54, 107, 214, 427, 854, 1706, 3410, 6815, 13629, 27259, 54521, 109042, 218080, 436158, 872318, 1744638, 3489278, 6978546, 13957092, 27914186, 55828364, 111656716, 223313428, 446626866, 893253744, 1786507472, 3573014938, 7146029910, 14292059832

Offset: 0

Author

Gerard P. Michon, May 02 2009

Keywords

• easy
• nice
• nonn

Comments

An alternate definition specifying "less than 2^n" would yield the same sequence except for the first 3 terms: 0,1,3,7,14,27,54,107, etc. (since powers of 2 beyond 8 are not cubefree).

The limit of a(n)/2^n is the inverse of Apery's constant, 1/zeta(3) [see A088453].

Examples

			a(0)=1 because there is just one cubefree integer (1) not exceeding 2^0 = 1.
a(3)=7 because 1,2,3,4,5,6,7 are cubefree but 8 is not.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..103 (terms 0..80 from Gerard P. Michon)
• Gerard P. Michon, On the number of cubefree integers not exceeding N.

Crossrefs

Cf. A004709 (cubefree numbers), A160112 (decimal counterpart for cubefree integers), A143658 (binary counterpart for squarefree integers), A071172 & A053462 (decimal counterpart for squarefree integers).

Cf. A060431.

Programs

Haskell

a160113 = a060431 . (2 ^)  -- Reinhard Zumkeller, Jul 27 2015

Mathematica

a[n_] := Sum[ MoebiusMu[i]*Floor[2^n/i^3], {i, 1, 2^(n/3)}]; Table[a[n], {n, 0, 32}] (* Jean-François Alcover, Dec 20 2011, from formula *)
Module[{nn=20,mu},mu=Table[If[Max[FactorInteger[n][[All,2]]]<3,1,0],{n,2^nn}];Table[Total[Take[mu,2^k]],{k,0,nn}]] (* The program generates the first 20 terms of the sequence. To get more, increase the value (constant) for nn, but the program may take a long time to run. *) (* Harvey P. Dale, Aug 13 2021 *)

Python

from sympy import mobius, integer_nthroot
def A160113(n): return sum(mobius(k)*((1<Chai Wah Wu, Aug 06 2024

Python

from bitarray import bitarray
from sympy import integer_nthroot
def A160113(n): # faster program
    q = 1<Chai Wah Wu, Aug 06 2024

Formula

a(n) = Sum_{i=1..2^(n/3)} A008683(i)*floor(2^n/i^3).