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%I #35 Jul 27 2023 07:42:08
%S 0,0,0,1,0,0,0,2,1,0,0,2,0,0,0,3,0,2,0,2,0,0,0,4,1,0,2,2,0,0,0,4,0,0,
%T 0,5,0,0,0,4,0,0,0,2,2,0,0,6,1,2,0,2,0,4,0,4,0,0,0,4,0,0,2,5,0,0,0,2,
%U 0,0,0,8,0,0,2,2,0,0,0,6,3,0,0,4,0,0,0,4,0,4,0,2,0,0,0,8,0,2,2,5,0,0,0,4,0
%N Number of non-unitary divisors of n.
%C Number of zeros in row n of table A225817. - _Reinhard Zumkeller_, Jul 30 2013
%H Reinhard Zumkeller, <a href="/A048105/b048105.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = Sigma(0, n) - 2^r(n), where r() = A001221, the number of distinct primes dividing n.
%F From _Reinhard Zumkeller_, Jul 30 2013: (Start)
%F a(n) = A000005(n) - A034444(n).
%F For n > 1: a(n) = A000005(n) - 2 * A007875(n). (End)
%F Dirichlet g.f.: zeta(s)^2 - zeta(s)^2/zeta(2*s). - _Geoffrey Critzer_, Dec 10 2014
%F G.f.: Sum_{k>=1} (1 - mu(k)^2)*x^k/(1 - x^k). - _Ilya Gutkovskiy_, Apr 21 2017
%F Sum_{k=1..n} a(k) ~ (1-6/Pi^2)*n*log(n) + ((1-6/Pi^2)*(2*gamma-1)+(72*zeta'(2)/Pi^4))*n , where gamma is Euler's constant (A001620). - _Amiram Eldar_, Nov 27 2022
%e Example 1: If n is squarefree (A005117) then a(n)=0 since all divisors are unitary.
%e Example 2: n=12, d(n)=6, ud(n)=4, nud(12)=d(12)-ud(12)=2; from {1,2,3,4,6,12} {1,3,4,12} are unitary while {2,6} are not unitary divisors.
%e Example 3: n=p^k, a true prime power, d(n)=k+1, u(d)=2^r(x)=2, so nud(n)=d(p^k)-2=k+1 i.e., it can be arbitrarily large.
%p with(NumberTheory):
%p seq(SumOfDivisors(n, 0) - 2^NumberOfPrimeFactors(n, 'distinct'), n = 1..105);
%p # _Peter Luschny_, Jul 27 2023
%t Table[DivisorSigma[0, n] - 2^PrimeNu[n], {n, 1, 50}] (* _Geoffrey Critzer_, Dec 10 2014 *)
%o (Haskell)
%o a048105 n = length [d | d <- [1..n], mod n d == 0, gcd d (n `div` d) > 1]
%o -- _Reinhard Zumkeller_, Aug 17 2011
%o (PARI) a(n)=my(f=factor(n)[,2]); prod(i=1,#f,f[i]+1)-2^#f \\ _Charles R Greathouse IV_, Sep 18 2015
%Y Cf. A000005, A001221, A007875, A056170, A225817, A034444.
%Y Cf. A001620, A229099, A306016.
%K nonn
%O 1,8
%A _Labos Elemer_
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