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A112624
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If p^b(p,n) is the highest power of the prime p dividing n, then a(n) = Product_{p|n} b(p,n)!.
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11
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1, 1, 1, 2, 1, 1, 1, 6, 2, 1, 1, 2, 1, 1, 1, 24, 1, 2, 1, 2, 1, 1, 1, 6, 2, 1, 6, 2, 1, 1, 1, 120, 1, 1, 1, 4, 1, 1, 1, 6, 1, 1, 1, 2, 2, 1, 1, 24, 2, 2, 1, 2, 1, 6, 1, 6, 1, 1, 1, 2, 1, 1, 2, 720, 1, 1, 1, 2, 1, 1, 1, 12, 1, 1, 2, 2, 1, 1, 1, 24, 24, 1, 1, 2, 1, 1, 1, 6, 1, 2, 1, 2, 1, 1, 1, 120, 1, 2, 2, 4, 1
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OFFSET
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1,4
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COMMENTS
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The logarithm of the Dirichlet series with the reciprocals of this sequence as coefficients is the Dirichlet series with the characteristic function of primes A010051 as coefficients. - Mats Granvik, Apr 13 2011
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LINKS
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FORMULA
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log(a(n)) = inverse Möbius transform of log(A306694(n)).
Let f(n) = 1/a(n). Formulas from Jakimczuk (2024, pp. 12-15):
Dirichlet g.f. of f(n): Sum_{n>=1} f(n)/n^s = exp(P(s)), where P(s) is the prime zeta function.
Sum_{k=1..n} f(k) = c * n + o(n), where c = A240953.
Sum_{k=1..n} f(k)/k = c * log(n) + o(log(n)), where c = A240953. (End)
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EXAMPLE
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45 = 3^2 * 5^1. So a(45) = 2! * 1! = 2.
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MAPLE
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w := n -> op(2, ifactors(n)): a := n -> mul(factorial(w(n)[j][2]), j = 1..nops(w(n))): seq(a(n), n = 1..101); # Emeric Deutsch, May 17 2012
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MATHEMATICA
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f[n_] := Block[{fi = Last@Transpose@FactorInteger@n}, Times @@ (fi!)]; Array[f, 101] (* Robert G. Wilson v, Dec 27 2005 *)
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PROG
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(PARI) A112624(n) = { my(f = factor(n), m = 1); for (k=1, #f~, m *= f[k, 2]!; ); m; } \\ Antti Karttunen, May 28 2017
(Sage)
return mul(factorial(s[1]) for s in factor(n))
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CROSSREFS
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For row > 1: a(n) = row products of A100995(A126988), when neglecting zero elements.
Cf. A000012, A000142, A010051, A028234, A067029, A112622, A112623, A156552, A240953, A246660, A306694.
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KEYWORD
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nonn,easy,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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