OFFSET
0,2
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.
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.
FORMULA
a(n) = Sum_{i=1..floor(10^(n/3))} A008683(i)*floor(10^n/i^3).
EXAMPLE
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).
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 *)
PROG
(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
CROSSREFS
KEYWORD
easy,nice,nonn
AUTHOR
Gerard P. Michon, May 02 2009, May 06 2009
STATUS
approved