A372841 4-full numbers that are not prime powers.

1296, 2592, 3888, 5184, 7776, 10000, 10368, 11664, 15552, 20000, 20736, 23328, 31104, 34992, 38416, 40000, 41472, 46656, 50000, 50625, 62208, 69984, 76832, 80000, 82944, 93312, 100000, 104976, 124416, 139968, 151875, 153664, 160000, 165888, 186624, 194481, 200000

Offset: 1

Author

Michael De Vlieger, May 14 2024

Keywords

• nonn

Comments

Numbers k such that rad(k)^4 | k and omega(k) > 1. In other words, numbers with at least 2 distinct prime factors whose prime power factors have exponents that exceed 3.

Proper subset of the following sequences: A001694, A036966, A036967, A126706, A286708, A372695.

Smallest term k with omega(k) = m is k = A002110(m)^4.

Examples

			Table of smallest 12 terms:
   n      a(n)
  -----------------------
   1     1296 = 2^4 * 3^4
   2     2592 = 2^5 * 3^4
   3     3888 = 2^4 * 3^5
   4     5184 = 2^6 * 3^4
   5     7776 = 2^5 * 3^5
   6    10000 = 2^4 * 5^4
   7    10368 = 2^7 * 3^4
   8    11664 = 2^4 * 3^6
   9    15552 = 2^6 * 3^5
  10    20000 = 2^5 * 5^4
  11    20736 = 2^8 * 3^4
  12    23328 = 2^5 * 3^6
		

Links

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

Crossrefs

Cf. A001694, A007947, A024619, A036966, A036967, A126706, A286708, A372695.

Programs

Mathematica

With[{nn = 200000}, Rest@ Select[Union@ Flatten@ Table[a^7 * b^6 * c^5 * d^4, {d, Surd[nn, 4]}, {c, Surd[nn/(d^4), 5]}, {b, Surd[nn/(c^5 * d^4), 6]}, {a, Surd[nn/(b^6 * c^5 * d^4), 7]}], Not@*PrimePowerQ]]

Python

from math import gcd
from sympy import primepi, integer_nthroot, factorint
def A372841(n):
    def f(x):
        c = n+x+1+sum(primepi(integer_nthroot(x, k)[0]) for k in range(4, x.bit_length()))
        for u in range(1,integer_nthroot(x,7)[0]+1):
            if all(d<=1 for d in factorint(u).values()):
                for w in range(1,integer_nthroot(a:=x//u**7,6)[0]+1):
                    if gcd(w,u)==1 and all(d<=1 for d in factorint(w).values()):
                        for y in range(1,integer_nthroot(z:=a//w**6,5)[0]+1):
                            if gcd(w,y)==1 and gcd(u,y)==1 and all(d<=1 for d in factorint(y).values()):
                                c -= integer_nthroot(z//y**5,4)[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 10 2024

Formula

Intersection of A036967 and A024619.

Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + 1/(p^3*(p-1))) - Sum_{p prime} 1/(p^3*(p-1)) - 1 = 0.0026996042121456100761... . - Amiram Eldar, May 17 2024