A383394 Perfect powers of Achilles numbers.

5184, 11664, 40000, 82944, 153664, 186624, 250000, 373248, 419904, 455625, 640000, 746496, 937024, 944784, 1259712, 1265625, 1327104, 1750329, 1827904, 1882384, 2458624, 3240000, 3779136, 4000000, 5345344, 6718464, 7290000, 8000000, 8340544, 9529569, 10240000

Offset

1,1

Comments

Proper subset of A131605, where A286708 is the union of A131605 and A052486. Therefore these are both powerful numbers and perfect powers, unlike Achilles numbers themselves.

Proper subset of A366854.

This sequence does not intersect A303606, also a proper subset of A131605.

Proper subset of A036967, which is a proper subset of A036966. - Michael De Vlieger, Oct 01 2025

Links

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Index entries for sequences related to powerful numbers.

Example

Table of n, a(n) for n = 1..12:
 n      a(n)
--------------------------------
 1     5184 =  72^2 = 2^6  * 3^4
 2    11664 = 108^2 = 2^4  * 3^6
 3    40000 = 200^2 = 2^6  * 5^4
 4    82944 = 288^2 = 2^10 * 3^4
 5   153664 = 392^2 = 2^6  * 7^4
 6   186624 = 432^2 = 2^8  * 3^6
 7   250000 = 500^2 = 2^4  * 5^6
 8   373248 =  72^3 = 2^9  * 3^6
 9   419904 = 648^2 = 2^6  * 3^8
10   455625 = 675^2 = 3^6  * 5^4
11   640000 = 800^2 = 2^10 * 5^4
12   746496 = 864^2 = 2^10 * 3^6

Mathematica

nn = 2^24; mm = Sqrt[nn]; i = 1; k = 2; MapIndexed[Set[S[First[#2]], #1] &, Rest@ Select[Union@ Flatten@ Table[a^2*b^3, {b, Surd[mm, 3]}, {a, Sqrt[mm/b^3]}], GCD @@ FactorInteger[#][[;; , -1]] == 1 &]]; Union@ Reap[While[j = 2; While[S[i]^j < nn, Sow[S[i]^j]; j++]; j > 2, k++; i++] ][[-1, 1]]

Prog

(Python)
from math import isqrt
from sympy import integer_nthroot, mobius
from oeis_sequences.OEISsequences import bisection, squarefreepi
def A383394(n):
    def g(x):
        c, l = squarefreepi(integer_nthroot(x, 3)[0])+sum(mobius(k)*(integer_nthroot(x, k)[0]-1) for k in range(2, x.bit_length()))-1, 0
        j = isqrt(x)
        while j>1:
            k2 = integer_nthroot(x//j**2, 3)[0]+1
            w = squarefreepi(k2-1)
            c += j*(w-l)
            l, j = w, isqrt(x//k2**3)
        return c-l
    def f(x): return n+x-sum(g(integer_nthroot(x, k)[0]) for k in range(2, x.bit_length()))
    return bisection(f, n, n) # Chai Wah Wu, Aug 11 2025

Crossrefs

Cf. A001597, A036966, A036967, A052486, A126708, A131605, A286708, A303606, A366854, A386762.

Sequence in context: A186847 A185485 A061050 * A391376 A390435 A175745

Adjacent sequences: A383391 A383392 A383393 * A383395 A383396 A383397

Keyword

nonn,easy

Author

Michael De Vlieger, Aug 01 2025

Status

approved