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a(n) = n^8 * Product_{p|n, p prime} (1 + 1/p^8).
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%I #20 Nov 15 2022 17:27:11

%S 1,257,6562,65792,390626,1686434,5764802,16842752,43053282,100390882,

%T 214358882,431727104,815730722,1481554114,2563287812,4311744512,

%U 6975757442,11064693474,16983563042,25700065792,37828630724,55090232674,78310985282,110522138624,152588281250

%N a(n) = n^8 * Product_{p|n, p prime} (1 + 1/p^8).

%C Sum of the 8th powers of the divisor complements of the squarefree divisors of n.

%H Sebastian Karlsson, <a href="/A351303/b351303.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Sum_{d|n} d^8 * mu(n/d)^2.

%F a(n) = n^8 * Sum_{d|n} mu(d)^2 / d^8.

%F Multiplicative with a(p^e) = p^(8*e) + p^(8*e-8). - _Sebastian Karlsson_, Feb 08 2022

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

%F Dirichlet g.f.: zeta(s)*zeta(s-8)/zeta(2*s).

%F Sum_{k=1..n} a(k) ~ n^9 * zeta(9) / (9 * zeta(18)) = 4331032831125 * n^9 * zeta(9) / (43867 * Pi^18).

%F Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^8/(p^16-1)) = 1.004062071480173688638170669970682370243496458304179434830922739661777... (End)

%F a(n) = J_16(n)/J_8(n) = J_16(n)/A069093(n), where J_k is the k-th Jordan totient function. - _Enrique PĂ©rez Herrero_, Nov 14 2022

%t f[p_, e_] := p^(8*e) + p^(8*(e-1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 25] (* _Amiram Eldar_, Feb 08 2022 *)

%o (PARI) a(n)=sumdiv(n, d, moebius(n/d)^2*d^8);

%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X)/(1 - p^8*X))[n], ", ")) \\ _Vaclav Kotesovec_, Feb 12 2022

%Y Cf. A008683 (mu).

%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), A351300 (k=5), A351301 (k=6), A351302 (k=7), this sequence (k=8), A351304 (k=9), A351305 (k=10).

%K nonn,mult

%O 1,2

%A _Wesley Ivan Hurt_, Feb 06 2022