A089579 Total number of perfect powers > 1 below 10^n.

3, 11, 39, 123, 365, 1109, 3393, 10489, 32668, 102229, 320988, 1010194, 3184136, 10046919, 31723590, 100216743, 316694003, 1001003330, 3164437423, 10004650116, 31632790242, 100021566155, 316274216760, 1000100055682, 3162493192563, 10000464300849, 31623776828239, 100002154796112

Offset: 1

Author

Martin Renner, Dec 29 2003

Keywords

• nonn

Comments

k is a perfect power <=> there exist integers a and b, b > 1, and k = a^b.

From Robert G. Wilson v, Jul 17 2016: (Start)

Limit_{n->oo} a(n)/sqrt(10^n) = 1.

A089580(n) - a(n) = A275358(n).

The four terms which make up the difference between A089580(2) - a(2) are: 16 = 2^4 = 4^2, 64 = 2^6 = 4^3 = 8^2 and 81 = 3^4 = 9^2; one for 16, two for 64 and one for 81 making a total of 4. See A117453.

(End)

Examples

			For n=2, the 11 perfect powers > 1 below 10^2 = 100 are: 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81. - _Michael B. Porter_, Jul 18 2016
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..100

Crossrefs

Cf. A001597, A070428, A089580, A275358.

Programs

Mathematica

Table[lim=10^n-1; Sum[ -(Floor[lim^(1/k)]-1)*MoebiusMu[k], {k,2,Floor[Log[2,lim]]}], {n,30}] (* T. D. Noe, Nov 16 2006 *)

Python

from sympy import mobius, integer_nthroot
def A089579(n): return int(sum(mobius(x)*(1-integer_nthroot(10**n,x)[0]) for x in range(2,(10**n).bit_length())))-1 if n>1 else 3 # Chai Wah Wu, Aug 13 2024

SageMath

def A089579(n):
    gen = (p for p in srange(2, 10^n) if p.is_perfect_power())
    return sum(1 for _ in gen)
print([A089579(n) for n in range(1, 7)])  # Peter Luschny, Sep 15 2023

Formula

a(n) = A070428(n) - 2 for n >= 2.

Extensions

a(9)-a(10) from Martin Renner, Oct 02 2004

More terms from T. D. Noe, Nov 16 2006

More precise name by Hugo Pfoertner, Sep 15 2023