A372695 Cubefull numbers that are not prime powers.

216, 432, 648, 864, 1000, 1296, 1728, 1944, 2000, 2592, 2744, 3375, 3456, 3888, 4000, 5000, 5184, 5488, 5832, 6912, 7776, 8000, 9261, 10000, 10125, 10368, 10648, 10976, 11664, 13824, 15552, 16000, 16875, 17496, 17576, 19208, 20000, 20736, 21296, 21952, 23328, 25000

Offset: 1

Author

Michael De Vlieger, May 14 2024

Keywords

• nonn

Comments

Numbers k such that rad(k)^3 | k and omega(k) > 1. In other words, numbers with at least 2 distinct prime factors whose prime power factors have exponents that exceed 2.

Proper subset of the following sequences: A001694, A036966, A126706, A286708.

Superset of A372841.

Smallest term k with omega(k) = m is k = A002110(m)^3 = A115964(m).

Examples

			Table of smallest 12 terms and instances of omega(a(n)) = m for m = 2..4
    n      a(n)
  ------------------------
    1      216 = 2^3 * 3^3
    2      432 = 2^4 * 3^3
    3      648 = 2^3 * 3^4
    4      864 = 2^5 * 3^3
    5     1000 = 2^3 * 5^3
    6     1296 = 2^4 * 3^4
    7     1728 = 2^6 * 3^3
    8     1944 = 2^3 * 3^5
    9     2000 = 2^4 * 5^3
   10     2592 = 2^5 * 3^4
   11     2744 = 2^3 * 7^3
   12     3375 = 3^3 * 5^3
  ...
   43    27000 = 2^3 * 3^3 * 5^3
  ...
  587  9261000 = 2^3 * 3^3 * 5^3 * 7^3
		

Links

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

Crossrefs

Cf. A001694, A007947, A024619, A036966, A115964, A126706, A286708, A372841.

Cf. A065483, A152441.

Programs

Mathematica

nn = 25000; Rest@ Select[Union@ Flatten@ Table[a^5 * b^4 * c^3, {c, Surd[nn, 3]}, {b, Surd[nn/(c^3), 4]}, {a, Surd[nn/(b^4 * c^3), 5]}], Not@*PrimePowerQ]

Python

from math import gcd
from sympy import primepi, integer_nthroot, factorint
def A372695(n):
    def f(x):
        c = n+1+x+sum(primepi(integer_nthroot(x, k)[0]) 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
    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
    return bisection(f,n,n) # Chai Wah Wu, Sep 12 2024

Formula

Intersection of A036966 and A024619.

Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + 1/(p^2*(p-1))) - Sum_{p prime} 1/(p^2*(p-1)) - 1 = A065483 - A152441 - 1 = 0.0188749045... . - Amiram Eldar, May 17 2024