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A295295
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Sum of squarefree divisors of the powerful part of n: a(n) = A048250(A057521(n)).
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7
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1, 1, 1, 3, 1, 1, 1, 3, 4, 1, 1, 3, 1, 1, 1, 3, 1, 4, 1, 3, 1, 1, 1, 3, 6, 1, 4, 3, 1, 1, 1, 3, 1, 1, 1, 12, 1, 1, 1, 3, 1, 1, 1, 3, 4, 1, 1, 3, 8, 6, 1, 3, 1, 4, 1, 3, 1, 1, 1, 3, 1, 1, 4, 3, 1, 1, 1, 3, 1, 1, 1, 12, 1, 1, 6, 3, 1, 1, 1, 3, 4, 1, 1, 3, 1, 1, 1, 3, 1, 4, 1, 3, 1, 1, 1, 3, 1, 8, 4, 18, 1, 1, 1, 3, 1
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OFFSET
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1,4
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COMMENTS
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The sum of the squarefree divisors of n whose square divides n. - Amiram Eldar, Oct 13 2023
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LINKS
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FORMULA
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Multiplicative with a(p) = 1 and a(p^e) = (p+1) for e > 1.
Dirichlet g.f.: zeta(s) * zeta(2*s-1) / zeta(4*s-2).
Sum_{k=1..n} a(k) ~ (3*n/Pi^2) * (log(n) + 3*gamma - 1 - 4*zeta'(2)/zeta(2)), where gamma is Euler's constant (A001620). (End)
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MATHEMATICA
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f[p_, e_] := If[e == 1, 1, p+1] ; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2023 *)
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PROG
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(Scheme, with memoization-macro definec)
(PARI) a(n) = my(f=factor(n)); for (i=1, #f~, if (f[i, 2]==1, f[i, 1]=1)); sumdiv(factorback(f), d, d*issquarefree(d)); \\ Michel Marcus, Jan 29 2021
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CROSSREFS
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KEYWORD
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nonn,easy,mult
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AUTHOR
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STATUS
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approved
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