A244098 Total number of divisors of all the ordered prime factorizations of an integer.

1, 2, 2, 3, 2, 5, 2, 4, 3, 5, 2, 9, 2, 5, 5, 5, 2, 9, 2, 9, 5, 5, 2, 14, 3, 5, 4, 9, 2, 16, 2, 6, 5, 5, 5, 19, 2, 5, 5, 14, 2, 16, 2, 9, 9, 5, 2, 20, 3, 9, 5, 9, 2, 14, 5, 14, 5, 5, 2, 35, 2, 5, 9, 7, 5, 16, 2, 9, 5, 16, 2, 34, 2, 5, 9, 9, 5, 16, 2, 20, 5, 5

Offset: 1

Author

Pierre-Louis Giscard, Jun 20 2014

Keywords

• nonn

Comments

a(n) = total number of ordered prime factorizations dividing all possible ordered prime factorizations making up n.

Example: for n = 12; a(12) = 9 because 12 = 2*2*3 = 2*3*2 = 3*2*2 the divisors of which are 1, 2, 3, 2*2, 2*3, 3*2, 2*2*3, 2*3*2, 3*2*2. This makes 9 ordered prime factorizations dividing all those making up 12.

Dirichlet convolution of A008480 with A000012.

Examples

			For n = 6; a(6) = 5 because 6 = 2*3 = 3*2, the divisors of which are 1, 2, 3, 2*3, 3*2. This makes 5 ordered prime factorizations dividing all those making up 6.
For n = 12; a(12) = 9 because 12 = 2*2*3 = 2*3*2 = 3*2*2, the divisors of which are 1, 2, 3, 2*2, 2*3, 3*2, 2*2*3, 2*3*2, 3*2*2. This makes 9 ordered prime factorizations dividing all those making up 12.
For n prime, a(n) = 2 because a prime n has a single ordered prime factorization n with divisors 1 and n. This makes two ordered prime factorizations dividing that making up n.
		

Links

• Pierre-Louis Giscard, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A000012, A008480.

Programs

Mathematica

f[s_]=Zeta[s]/(1-PrimeZetaP[s]); (* Dirichlet g.f *)
(* or *)
Clear[a, b];
a = Prepend[
   Array[Multinomial @@ Last[Transpose[FactorInteger[#]]] &, 200, 2],
   1];
b = Table[1, {u, 1, Length[a]}];
Table[Sum[If[IntegerQ[p/n], b[[n]] a[[p/n]], 0], {n, 1, p}], {p, 1,
  Length[a]}]

Formula

Dirichlet generating function: Zeta(s)/(1-P(s)) with Zeta(s) the Riemann zeta function and P(s) the prime zeta function.

G.f. A(x) satisfies: A(x) = x / (1 - x) + Sum_{k>=1} A(x^prime(k)). - Ilya Gutkovskiy, May 30 2020