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A244098
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Total number of divisors of all the ordered prime factorizations of an integer.
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1
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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MATHEMATICA
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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]}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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