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

1,16

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

Links

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

Formula

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

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

Example

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.

Mathematica

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

Prog

(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

Crossrefs

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

Sequence in context: A101875 A081387 A261795 * A204125 A204127 A225174

Adjacent sequences: A306342 A306343 A306344 * A306346 A306347 A306348

Keyword

nonn

Author

Felix Fröhlich, Feb 08 2019

Extensions

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

Status

approved