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Total number of divisors of all the ordered prime factorizations of an integer.
1

%I #30 May 30 2020 19:14:00

%S 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,

%T 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,

%U 5,16,2,9,5,16,2,34,2,5,9,9,5,16,2,20,5,5

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

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

%C 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.

%C Dirichlet convolution of A008480 with A000012.

%H Pierre-Louis Giscard, <a href="/A244098/b244098.txt">Table of n, a(n) for n = 1..5000</a>

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

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

%e 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.

%e 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.

%e 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.

%t f[s_]=Zeta[s]/(1-PrimeZetaP[s]); (* Dirichlet g.f *)

%t (* or *)

%t Clear[a, b];

%t a = Prepend[

%t Array[Multinomial @@ Last[Transpose[FactorInteger[#]]] &, 200, 2],

%t 1];

%t b = Table[1, {u, 1, Length[a]}];

%t Table[Sum[If[IntegerQ[p/n], b[[n]] a[[p/n]], 0], {n, 1, p}], {p, 1,

%t Length[a]}]

%Y Cf. A000012, A008480.

%K nonn

%O 1,2

%A _Pierre-Louis Giscard_, Jun 20 2014