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%I #38 Jul 11 2020 07:32:16
%S 8,24,27,40,54,56,88,104,120,125,135,136,152,168,184,189,232,248,250,
%T 264,270,280,296,297,312,328,343,344,351,375,376,378,408,424,440,456,
%U 459,472,488,513,520,536,552,568,584,594,616,621,632,664,680,686,696
%N Numbers having equally many squarefree and nonsquarefree divisors; number of unitary divisors of n (A034444) = number of non-unitary divisors of n (A048105).
%C 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).
%C 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 (A086436). 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 A067295. - _Ctibor O. Zizka_, Sep 21 2008
%C 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
%C 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
%H Robert Israel, <a href="/A048109/b048109.txt">Table of n, a(n) for n = 1..10000</a>
%F Numbers k such that d(k) = 2^(omega(k)+1) or A000005(k) = 2^(A001221(k) + 1) = 2 * A034444(k).
%e 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}.
%p filter:= proc(n) local F;
%p F:= ifactors(n)[2];
%p mul(t[2]+1,t=F) = 2^(1+nops(F))
%p end proc;
%p select(filter, [$1..1000]); # _Robert Israel_, Jun 09 2015
%t Position[Table[FiniteAbelianGroupCount[n], {n, 1, 1000}],3] // Flatten (* _Geoffrey Critzer_, Jun 09 2015 *)
%o (PARI) is(n)=select(e->e>1, factor(n)[,2])==[3]~ \\ _Charles R Greathouse IV_, Jun 10 2015
%o (PARI) isok(n) = sumdiv(n, d, issquarefree(d)) == sumdiv(n, d, !issquarefree(d)); \\ _Michel Marcus_, Jun 24 2015
%Y Cf. A000005, A001221, A034444, A036537, A048106, A048107.
%K nonn
%O 1,1
%A _Labos Elemer_
%E New name based on comment by _Ivan Neretin_, Jun 19 2015
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