A048109 Numbers having equally many squarefree and nonsquarefree divisors; number of unitary divisors of n (A034444) = number of non-unitary divisors of n (A048105).

8, 24, 27, 40, 54, 56, 88, 104, 120, 125, 135, 136, 152, 168, 184, 189, 232, 248, 250, 264, 270, 280, 296, 297, 312, 328, 343, 344, 351, 375, 376, 378, 408, 424, 440, 456, 459, 472, 488, 513, 520, 536, 552, 568, 584, 594, 616, 621, 632, 664, 680, 686, 696

Offset: 1

Author

Labos Elemer

Keywords

• nonn

Comments

For these terms the number of divisors should be a special power of two because ud(n) = 2^r and nud(n) = ud(n). In particular the exponent of 2 is 1+A001221(n), the number of distinct prime factors + 1. Thus this is a subsequence of A036537 where A000005(A036537(n)) = 2^s; here s = 1+A001221(n).

Let us introduce a function D(n) = sigma_0(n)/2^(alpha(1)+...+alpha(r)), sigma_0(n) number of divisors of n (A000005), prime factorization of n = p(1)^alpha(1) * ... * p(r)^alpha(r), alpha(1)+...+alpha(r) is sequence (A001222). This function splits the set of positive integers into subsets, according to the value of D(n). Squarefree numbers (A005117) has D(n)=1, other numbers are "deviated" from the squarefree ideal and have 0 < D(n) < 1. So for D(n)=1/2 we have A048109, D(n)=3/4 we have A060687. - Ctibor O. Zizka, Sep 21 2008

Integers n such that there are exactly 3 Abelian groups of order n. That is, n such that A000688(n)=3. In other words, in the prime factorization of n there is exactly one prime with exponent of 3 and the others have exponent of 1. - Geoffrey Critzer, Jun 09 2015

The asymptotic density of this sequence is (6/Pi^2) * Sum_{k>=1} 1/(prime(k)^2*(prime(k)+1)) = (1/zeta(2)) * Sum_{k>=3} (-1)^(k+1) * P(k) = 0.0741777413672596019212880156082745910562809066233004356300970463709875..., where P is the prime zeta function. - Amiram Eldar, Jul 11 2020

Examples

			n = 88 = 2*2*2*11 has 8 divisors, of which 4 are unitary divisors (because of 2 distinct prime factors) and 4 are nonunitary divisors: U={1,88,11,8} and NU = {2,44,4,22}.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A001221, A034444, A036537, A048106, A048107.

Programs

Maple

filter:= proc(n) local F;
  F:= ifactors(n)[2];
  mul(t[2]+1,t=F) = 2^(1+nops(F))
end proc;
select(filter, [$1..1000]); # Robert Israel, Jun 09 2015

Mathematica

Position[Table[FiniteAbelianGroupCount[n], {n, 1, 1000}],3] // Flatten (* Geoffrey Critzer, Jun 09 2015 *)

PARI

is(n)=select(e->e>1, factor(n)[,2])==[3]~ \\ Charles R Greathouse IV, Jun 10 2015

PARI

isok(n) = sumdiv(n, d, issquarefree(d)) == sumdiv(n, d, !issquarefree(d)); \\ Michel Marcus, Jun 24 2015

Python

from math import isqrt
from sympy import mobius, primerange
def A048109(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def g(x): return sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
    def f(x): return int(n+x+sum(sum(-g(x//p**j) if j&1 else g(x//p**j) for j in range(3,x.bit_length())) for p in primerange(isqrt(x)+1)))
    return bisection(f,n,n) # Chai Wah Wu, Feb 24 2025

Formula

Numbers k such that d(k) = 2^(omega(k)+1) or A000005(k) = 2^(A001221(k) + 1) = 2 * A034444(k).

Extensions

New name based on comment by Ivan Neretin, Jun 19 2015