A076470 Perfect powers m^k where k >= 6.

1, 64, 128, 256, 512, 729, 1024, 2048, 2187, 4096, 6561, 8192, 15625, 16384, 19683, 32768, 46656, 59049, 65536, 78125, 117649, 131072, 177147, 262144, 279936, 390625, 524288, 531441, 823543, 1000000, 1048576, 1594323, 1679616, 1771561

Offset: 1

Author

Robert G. Wilson v, Oct 14 2002

Keywords

• nonn

Comments

A necessary but not sufficient condition is that if p|n when at least p^6|n.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001597, A076467, A076468, A076469.

Different from A069493.

Cf. A002117, A013661, A013662, A013663.

Programs

Mathematica

a = {1}; Do[ If[ Apply[ GCD, Last[ Transpose[ FactorInteger[n]]]] > 4, a = Append[a, n]; Print[n]], {n, 2, 1953124}]; a

Python

from sympy import mobius, integer_nthroot
def A076470(n):
    def f(x): return int(n+9+x-(sum(integer_nthroot(x,d)[0] for d in (6,10,15))<<1)-sum(integer_nthroot(x,d)[0] for d in (8,9,12,20,25))+sum(mobius(k)*(sum(integer_nthroot(x,k*i)[0] for i in range(1,6))-5) for k in range(6,x.bit_length())))
    kmin, kmax = 1,2
    while f(kmax) >= kmax:
        kmax <<= 1
    while True:
        kmid = kmax+kmin>>1
        if f(kmid) < kmid:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    return kmax # Chai Wah Wu, Aug 14 2024

Formula

Sum_{n>=1} 1/a(n) = 5 - zeta(2) - zeta(3) - zeta(4) - zeta(5) + Sum_{k>=2} mu(k)*(5 - zeta(k) - zeta(2*k) - zeta(3*k) - zeta(4*k) - zeta(5*k)) = 1.03342597171... . - Amiram Eldar, Dec 03 2022