A160112 Number of cubefree integers not exceeding 10^n.

1, 9, 85, 833, 8319, 83190, 831910, 8319081, 83190727, 831907372, 8319073719, 83190737244, 831907372522, 8319073725828, 83190737258105, 831907372580692, 8319073725807178, 83190737258070643, 831907372580707771

Offset: 0

Author

Gerard P. Michon, May 02 2009, May 06 2009

Keywords

• easy
• nice
• nonn

Comments

An alternate definition specifying "less than 10^n" would yield the same sequence except for the first 3 terms: 0, 8, 84, 833, 8319, etc. (since powers of 10 beyond 1000 are not cubefree anyhow).

The limit of a(n)/10^n is the inverse of Apery's constant, 1/zeta(3), whose digits are given by A088453.

Examples

			a(0)=1 because 1 <= 10^0 is not a multiple of the cube of a prime.
a(1)=9 because the 9 numbers 1,2,3,4,5,6,7,9,10 are cubefree; 8 is not.
a(2)=85 because there are 85 cubefree integers equal to 100 or less.
a(3)=833 because there are 833 cubefree integers below 1000 (which is not cubefree itself).
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..31 (terms 0..29 from Gerard P. Michon)
• Gerard P. Michon, On the number of cubefree integers not exceeding N.

Crossrefs

A004709 (cubefree numbers). A088453 (limit of the string of digits). A160113 (binary counterpart for cubefree integers). A071172 & A053462 (decimal counterpart for squarefree integers). A143658 (binary counterpart for squarefree integers).

Programs

Maple

with(numtheory): A160112:=n->add(mobius(i)*floor(10^n/(i^3)), i=1..10^n): (1,9,85,seq(A160112(n), n=3..5)); # Wesley Ivan Hurt, Aug 01 2015

Mathematica

Table[ Sum[ MoebiusMu[x]*Floor[10^n/(x^3)], {x, 10^(n/3)}], {n, 0, 18}] (* Robert G. Wilson v, May 27 2009 *)

Python

from sympy import mobius, integer_nthroot
def A160112(n): return sum(mobius(k)*(10**n//k**3) for k in range(1, integer_nthroot(10**n,3)[0]+1)) # Chai Wah Wu, Aug 06 2024

Python

from bitarray import bitarray
from sympy import integer_nthroot
def A160112(n): # faster program
    q = 10**n
    m = integer_nthroot(q,3)[0]+1
    a, b = bitarray(m), bitarray(m)
    a[1], p, i, c = 1, 2, 4, q-sum(q//k**3 for k in range(2,m))
    while i < m:
        j = 2
        while i < m:
            if j==p:
                c -= (b[i]^1 if a[i] else -1)*(q//i**3)
                j, a[i], b[i] = 0, 1, 1
            else:
                t1, t2 = a[i], b[i]
                if (t1&t2)^1:
                    a[i], b[i] = (t1^1)&t2, ((t1^1)&t2)^1
                    c += (t2 if t1 else 2)*(q//i**3) if (t1^1)&t2 else (t2-2 if t1 else 0)*(q//i**3)
            i += p
            j += 1
        p += 1
        while a[p]|b[p]:
            p += 1
        i = p<<1
    return c # Chai Wah Wu, Aug 06 2024

Formula

a(n) = Sum_{i=1..floor(10^(n/3))} A008683(i)*floor(10^n/i^3).