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%I #22 Feb 10 2021 02:20:14
%S 1,2,1,1,3,1,1,2,1,2,1,4,2,2,1,1,3,1,1,1,2,2,1,1,5,1,3,1,1,3,3,1,2,1,
%T 1,4,1,1,2,2,3,3,1,1,2,1,1,2,1,6,1,2,2,1,4,1,1,1,2,1,1,4,3,1,2,1,1,1,
%U 1,3,2,2,1,2,5,2,1,3,1,1,3,1,4,1,4,2,1
%N Excess of numbers that are not squarefree.
%C The "excess" of a number is the number of prime divisors with multiplicity (the Omega function, A001222) minus the number of distinct prime divisors (the omega function, A001221). A046660(n) gives the excess of n.
%C Since squarefree numbers have no excess, this sequence is essentially A046660 with the 0's removed.
%H G. C. Greubel, <a href="/A275699/b275699.txt">Table of n, a(n) for n = 1..5000</a>
%F a(n) = A046660(A013929(n)).
%F Asymptotic mean: lim_{m->oo} (1/m) Sum_{k=1..m} a(k) = Sum_{p prime} 1/(p*(p-1)) / (1-6/Pi^2) = A136141/A229099 = 1.9719717... - _Amiram Eldar_, Feb 10 2021
%e Since 16 = 2^4, 16 has four prime divisors, but only one distinct divisor. Hence Omega(16) - omega(16) = 4 - 1 = 3. As 16 is the fifth number that is not squarefree, its corresponding 3 is a(5) in this sequence.
%e 17 is prime and thus has no excess and no corresponding term in this sequence.
%e 18 = 2 * 3^2, Omega(18) - omega(18) = 3 - 2 = 1, thus a(6) = 1.
%t DeleteCases[Table[PrimeOmega[n] - PrimeNu[n], {n, 200}], 0] (* _Alonso del Arte_, Aug 05 2016 *)
%o (PARI) for(n=1, 200, if(bigomega(n)!=omega(n), print1(bigomega(n)-omega(n), ", ")))
%Y Cf. A001221, A001222, A013929, A046660, A136141, A229099.
%K nonn
%O 1,2
%A _Felix Fröhlich_, Aug 05 2016
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