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%I #31 Feb 21 2019 09:22:53
%S 0,1,1,0,1,1,1,1,0,1,1,1,1,1,1,2,1,1,1,1,1,1,1,3,0,1,1,1,1,1,1,3,1,1,
%T 1,4,1,1,1,3,1,1,1,1,1,1,1,5,0,1,1,1,1,3,1,3,1,1,1,5,1,1,1,4,1,1,1,1,
%U 1,1,1,7,1,1,1,1,1,1,1,5,2,1,1,5,1,1,1,3
%N Absolute difference between the number of prime divisors and the number of composite divisors of n.
%C Conjecture: a(n) = 0 iff n is a term of A280076 = union of A001248 and {1}.
%C Conjecture is true, since having an n with k distinct prime factors such that a(n) = 0 requires that 2k+1 can be factored into k parts > 1, and 1 is the only positive k for which this is possible. - _Charlie Neder_, Feb 12 2019
%H David A. Corneth, <a href="/A306345/b306345.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = abs(A001221(n) - A055212(n)).
%F a(n) = abs(2*A001221(n) - A000005(n) + 1). - _Michel Marcus_, Feb 12 2019
%e For n = 24: The set of divisors of 24 is {1, 2, 3, 4, 6, 8, 12, 24}. The prime divisors are {2, 3} and the composite divisors are {4, 6, 8, 12, 24}. The cardinalities of the sets are 2 and 5, respectively, and abs(2-5) = 3, so a(24) = 3.
%t Array[Abs[2 PrimeNu@ # - DivisorSigma[0, #] + 1] &, 105] (* _Michael De Vlieger_, Feb 17 2019 *)
%o (PARI) a(n) = my(d=divisors(n), p=0, c=0); for(k=2, #d, if(ispseudoprime(d[k]), p++, c++)); abs(p-c)
%o (PARI) a(n) = abs(2*omega(n) - numdiv(n) + 1); \\ _Michel Marcus_, Feb 12 2019
%Y Cf. A000005, A001221, A001248, A055212, A280076.
%K nonn
%O 1,16
%A _Felix Fröhlich_, Feb 08 2019
%E a(1)=0 prepended by _David A. Corneth_, Feb 12 2019
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