A389551 Cubefull numbers that are not powers of squarefree numbers.

432, 648, 864, 1728, 1944, 2000, 2592, 3456, 3888, 4000, 5000, 5184, 5488, 5832, 6912, 8000, 10125, 10368, 10976, 11664, 13824, 15552, 16000, 16875, 17496, 19208, 20000, 20736, 21296, 21952, 23328, 25000, 27648, 27783, 30375, 31104, 32000, 34992, 35152, 40000

Offset

1,1

Comments

Excludes powers k^m of squarefree numbers k, i.e., A072777.

Intersection of A359280 and A036966, where A359280 = A001694 \ A072777.

Disjoint union of A383394, A388549, and A388293 = A359280 \ A389558.

Links

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

Index entries for sequences related to powerful numbers.

Formula

From Amiram Eldar, Nov 23 2025: (Start)

Equals (A036966 \ A390540) \ {1} .

Sum_{n>=1} 1/a(n) = A065483 - A368250 + A082020 - 2 = 0.01096804024453018557... . (End)

Example

Let r = A388293, s = A383394, and t = A388549.
Table of n, a(n) for select n:
 n     a(n)
------------------------------------
 1     432 = 2^4 * 3^3        = r(1)
 2     648 = 2^3 * 3^4        = r(2)
 3     864 = 2^5 * 3^3        = r(3)
 4    1728 = 2^6 * 3^3 = 12^3 = t(1)
 5    1944 = 2^3 * 3^5        = r(4)
 6    2000 = 2^4 * 5^3        = r(5)
 7    2592 = 2^5 * 3^4        = r(6)
 8    3456 = 2^7 * 3^3        = r(7)
 9    3888 = 2^4 * 3^5        = r(8)
12    5184 = 2^6 * 5^4 = 72^2 = s(1)
17   10125 = 3^4 * 5^3        = r(13)
46   54000 = 2^4 * 3^3 * 5^3  = r(37)

Mathematica

With[{nn = 40000}, Union@ Flatten@ Table[If[! PrimePowerQ[#], If[CountDistinct[FactorInteger[#][[;; , -1]] ]>1, #, Nothing], Nothing] &[a^5 * b^4 * c^3], {c, Surd[nn, 3]}, {b, Surd[nn/(c^3), 4]}, {a, Surd[nn/(b^4 * c^3), 5] } ] ]

Prog

(Python)
from math import isqrt, gcd
from sympy import integer_nthroot, mobius, factorint
from oeis_sequences.OEISsequences import bisection, squarefreepi
def A389551(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):
        c = n+1+x+sum(squarefreepi(integer_nthroot(x, k)[0])-1 for k in range(3, x.bit_length()))
        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 02 2025

Crossrefs

Cf. A001597, A001694, A036966, A072777, A126706, A131605, A286708, A359280, A383394, A388293, A388549, A389558, A390540.

Cf. A065483, A082020, A368250.

Sequence in context: A245469 A069781 A337670 * A391593 A391980 A388293

Adjacent sequences: A389548 A389549 A389550 * A389552 A389553 A389554

Keyword

nonn,easy

Author

Michael De Vlieger, Nov 17 2025

Status

approved