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a(n) = n^5 * Product_{p|n, p prime} (1 + 1/p^5).
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%I #23 Nov 14 2022 05:35:14

%S 1,33,244,1056,3126,8052,16808,33792,59292,103158,161052,257664,

%T 371294,554664,762744,1081344,1419858,1956636,2476100,3301056,4101152,

%U 5314716,6436344,8245248,9768750,12252702,14407956,17749248,20511150,25170552,28629152,34603008,39296688

%N a(n) = n^5 * Product_{p|n, p prime} (1 + 1/p^5).

%C Sum of the 5th powers of the divisor complements of the squarefree divisors of n.

%H Sebastian Karlsson, <a href="/A351300/b351300.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Sum_{d|n} d^5 * mu(n/d)^2.

%F a(n) = n^5 * Sum_{d|n} mu(d)^2 / d^5.

%F Multiplicative with a(p^e) = p^(5*e) + p^(5*e-5). - _Sebastian Karlsson_, Feb 08 2022

%F From _Vaclav Kotesovec_, Feb 12 2022: (Start)

%F Dirichlet g.f.: zeta(s)*zeta(s-5)/zeta(2*s).

%F Sum_{k=1..n} a(k) ~ n^6 * zeta(6) / (6 * zeta(12)) = 225225 * n^6 / (1382 * Pi^6).

%F Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^5/(p^10-1)) = 1.03592823428850098309076014982275428113698561633329794485946580153004... (End)

%F a(n) = J_10(n) / J_5(n) = A069095(n) / A059378(n), where J_k is the k-th Jordan totient function. - _Enrique PĂ©rez Herrero_, Nov 13 2022

%t f[p_, e_] := p^(5*e) + p^(5*(e-1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 40] (* _Amiram Eldar_, Feb 08 2022 *)

%o (PARI) a(n)=sumdiv(n, d, moebius(n/d)^2*d^5);

%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X)/(1 - p^5*X))[n], ", ")) \\ _Vaclav Kotesovec_, Feb 12 2022

%Y Cf. A008683 (mu).

%Y Cf. A069095, A059378.

%Y Sequences of the form n^k * Product_ {p|n, p prime} (1 + 1/p^k) for k=0..10: A034444 (k=0), A001615 (k=1), A065958 (k=2), A065959 (k=3), A065960 (k=4), this sequence (k=5), A351301 (k=6), A351302 (k=7), A351303 (k=8), A351304 (k=9), A351305 (k=10).

%K nonn,mult

%O 1,2

%A _Wesley Ivan Hurt_, Feb 06 2022