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%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
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