A275358 -id:A275358 - cp's OEIS frontend

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

Martin Renner, Dec 29 2003

nonn

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)

			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

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

Cf. A001597, A070428, A089580, A275358.

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 *)

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

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

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

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

A089580 Total number of perfect powers > 1 below 10^n, counting multiple representations separately.

Original entry on oeis.org

3, 15, 49, 143, 406, 1174, 3507, 10674, 32965, 102716, 321797, 1011533, 3186389, 10050743, 31730134, 100228040, 316713623, 1001037546, 3164497349, 10004755374, 31632975598, 100021893194, 316274794666, 1000101078148, 3162495003352, 10000467510247, 31623782520064, 100002164895587

Offset: 1

Martin Renner, Dec 29 2003

nonn

From Robert G. Wilson v, Jul 17 2016: (Start)
a(n) ~ sqrt(10^n).
a(n) - A089579(n) = A275358(n).
The four terms which make up the difference a(2) - A089579(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)
This sequence correlates (see Link) to A006880 via a power fit A*x^B. For example, using a(23) through a(29) one obtains (A,B) = (0.047272, 1.96592) with R^2 > 0.999999. This extrapolates A006880(30) as 1.46*10^28. The exponent well may be resolving to 2. - Bill McEachen, Mar 04 2025

			16 = 2^4 = 4^2 counts double, 256 = 2^8 = 4^4 = 16^2 counts three times.

Chai Wah Wu, Table of n, a(n) for n = 1..1998 (n = 1..100 from Robert G. Wilson, n = 101..400 from Karl-Heinz Hofmann)
Karl-Heinz Hofmann, Python program.
Bill McEachen, Plot A089580 vs A006880.

Cf. A001597, A072103, A117453, A275358, A060298.

Cf. A089579 (counting multiple representations only once).

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

Python
```
# see link.
```

a(n) = Sum_{k = 1..n} A060298(k). - Karl-Heinz Hofmann, Sep 18 2023

2 more terms from Martin Renner, Oct 02 2004
More terms from T. D. Noe, Nov 16 2006
More precise name by Hugo Pfoertner, Sep 16 2023

A175066 a(1) = 1, for n >= 2: a(n) = number of ways h to write perfect powers A117453(n) as m^k (m >= 2, k >= 2).

Original entry on oeis.org

1, 2, 3, 2, 3, 2, 2, 3, 3, 2, 2, 5, 3, 2, 2, 3, 3, 2, 2, 2, 3, 2, 3, 2, 3, 4, 2, 2, 3, 2, 2, 2, 2, 5, 2, 2, 3, 2, 5, 2, 2, 2, 2, 3, 5, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 3, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 7, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 3, 2

Offset: 1

Jaroslav Krizek, Jan 23 2010

nonn

Perfect powers with first occurrence of h >= 2: 16, 64, 65536, 4096, ... (A175065)
a(n) for n>1 is the subsequence of A253642 formed by the terms which exceed 1; equivalently, a(n)+1 for n>1 is the subsequence of A175064 formed by the terms which exceed 2. Also, sum of a(n)-1 over such n that A117453(n)<10^m gives A275358(m). - Andrey Zabolotskiy, Aug 16 2016
Numbers n such that a(n) is nonprime are 1, 26, 110, ... - Altug Alkan, Aug 22 2016

			For n = 12, A117453(12) = 4096 and a(12)=5 since there are 5 ways to write 4096 as m^k: 64^2 = 16^3 = 8^4 = 4^6 = 2^12.
729=27^2=9^3=3^6 and 1024=32^2=4^5=2^10 yield a(8)=a(9)=3. - _R. J. Mathar_, Jan 24 2010

Cf. A117453.

lista(nn) = {print1(1, ", "); for (i=2, nn, if (po = ispower(i), np = sum(j=2, po, ispower(i, j)); if (np>1, print1(np, ", "));););} \\ Michel Marcus, Mar 20 2013

from math import gcd
from sympy import mobius, integer_nthroot, factorint, divisor_count, primerange
def A175066(n):
    if n == 1: return 1
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return int(n+sum(mobius(k)*(integer_nthroot(x,k)[0]-1+sum(integer_nthroot(x,p*k)[0]-1 for p in primerange((x//k).bit_length()))) for k in range(1,x.bit_length())))
    return divisor_count(gcd(*factorint(bisection(f,n,n)).values()))-1 # Chai Wah Wu, Nov 24 2024

If A117453(n) = m^k with k maximal, then a(n) = tau(k) - 1. - Charlie Neder, Mar 02 2019

Corrected and extended by R. J. Mathar, Jan 24 2010

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

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A089580 Total number of perfect powers > 1 below 10^n, counting multiple representations separately.

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A175066 a(1) = 1, for n >= 2: a(n) = number of ways h to write perfect powers A117453(n) as m^k (m >= 2, k >= 2).

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