A306345 Absolute difference between the number of prime divisors and the number of composite divisors of n.

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, 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, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 5, 2, 1, 1, 5, 1, 1, 1, 3

Offset: 1

Author

Felix Fröhlich, Feb 08 2019

Keywords

• nonn

Comments

Conjecture: a(n) = 0 iff n is a term of A280076 = union of A001248 and {1}.

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

Examples

			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.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A001221, A001248, A055212, A280076.

Programs

Mathematica

Array[Abs[2 PrimeNu@ # - DivisorSigma[0, #] + 1] &, 105] (* Michael De Vlieger, Feb 17 2019 *)

PARI

a(n) = my(d=divisors(n), p=0, c=0); for(k=2, #d, if(ispseudoprime(d[k]), p++, c++)); abs(p-c)

PARI

a(n) = abs(2*omega(n) - numdiv(n) + 1); \\ Michel Marcus, Feb 12 2019

Formula

a(n) = abs(A001221(n) - A055212(n)).

a(n) = abs(2*A001221(n) - A000005(n) + 1). - Michel Marcus, Feb 12 2019

Extensions

a(1)=0 prepended by David A. Corneth, Feb 12 2019