A390539 Achilles numbers that are not cubefull.

72, 108, 200, 288, 392, 500, 675, 800, 968, 972, 1125, 1152, 1323, 1352, 1372, 1568, 1800, 2312, 2700, 2888, 3087, 3200, 3267, 3528, 3872, 4232, 4500, 4563, 4608, 5292, 5324, 5400, 5408, 6075, 6125, 6272, 6728, 7200, 7688, 7803, 8575, 8712, 8748, 8788, 9000, 9248

Offset

1,1

Comments

Intersection of A052486 and A362147 = A052486 \ A036966.

Links

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

Formula

From Amiram Eldar, Nov 18 2025: (Start)

Equals A325240 \ (A000290 \ A374291).

Sum_{n>=1} 1/a(n) = A082695 - A065483 - A013661 + zeta(4)*zeta(6)/zeta(12) = 0.059701351267723610429... . (End)

Example

Let s = A052486.
a(1) = s(1) = 72 = 2^3 * 3*2 is the first term, since among prime power factor exponents, there exists an exponent m < 3.
a(2) = s(2) = 108 = 2^2 * 3*3, since among prime power factor exponents, there exists an exponent m < 3.
s(6) = 432 = 2^4 * 3^3 is not a term, since 432 is the smallest Achilles number with all prime power factor exponents greater than 2.

Maple

filter:= proc(n) local F;
  F:= ifactors(n)[2][.., 2];
  min(F) = 2 and igcd(F) = 1
end proc:
select(filter, [$1..10000]); # Robert Israel, Nov 16 2025

Mathematica

With[{nn = 10^4}, Select[Rest@ Union@ Flatten@ Table[a^2*b^3, {b, Surd[nn, 3]}, {a, Sqrt[nn/b^3]}], And[Min[#] < 3, GCD @@ # == 1] &[FactorInteger[#][[;; , -1]] ] &]]

Prog

(Python)
from math import isqrt, gcd
from sympy import integer_nthroot, factorint
from oeis_sequences.OEISsequences import bisection, squarefreepi
def A390539(n):
    def f(x):
        c, l = n+x, 0
        j = z = 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)
        c += l+z-squarefreepi(integer_nthroot(x, 3)[0])-sum(isqrt(z//k**3) for k in range(1, integer_nthroot(z, 3)[0]+1) if all(d<=1 for d in factorint(k).values()))
        for w in range(1, integer_nthroot(x, 5)[0]+1):
            if all(d<=1 for d in factorint(w).values()):
                for y in range(1, integer_nthroot(z:=x//w**5, 4)[0]+1):
                    if gcd(w, y)==1 and all(d<=1 for d in factorint(y).values()):
                        c += integer_nthroot(z//y**4, 3)[0]
        return c
    return bisection(f, n, n) # Chai Wah Wu, Dec 04 2025

Crossrefs

Cf. A001694, A013929, A024619, A036966, A052486, A126706, A286708, A362147, A388293.

Cf. A000290, A013661, A065483, A082695, A325240, A374291.

Sequence in context: A390952 A378859 A114128 * A391923 A375074 A375143

Adjacent sequences: A390536 A390537 A390538 * A390540 A390541 A390542

Keyword

nonn,easy

Author

Michael De Vlieger, Nov 16 2025

Status

approved