login
a(n) = n^10 * Product_{p|n, p prime} (1 + 1/p^10).
10

%I #14 Feb 12 2022 09:31:26

%S 1,1025,59050,1049600,9765626,60526250,282475250,1074790400,

%T 3486843450,10009766650,25937424602,61978880000,137858491850,

%U 289537131250,576660215300,1100585369600,2015993900450,3574014536250,6131066257802,10250001049600,16680163512500,26585860217050

%N a(n) = n^10 * Product_{p|n, p prime} (1 + 1/p^10).

%C Sum of the 10th powers of the divisor complements of the squarefree divisors of n.

%H Sebastian Karlsson, <a href="/A351305/b351305.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Sum_{d|n} d^10 * mu(n/d)^2.

%F a(n) = n^10 * Sum_{d|n} mu(d)^2 / d^10.

%F Multiplicative with a(p^e) = p^(10*e) + p^(10*e-10). - _Sebastian Karlsson_, Feb 08 2022

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

%F Dirichlet g.f.: zeta(s)*zeta(s-10)/zeta(2*s).

%F Sum_{k=1..n} a(k) ~ n^11 * zeta(11) / (11 * zeta(22)) = 1222532449149375 * n^11 * zeta(11) / (155366 * Pi^22)}.

%F Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^10/(p^20-1)) = 1.000993621149252443797467720671490169127513829380371486971107300011796... (End)

%t f[p_, e_] := p^(10*e) + p^(10*(e-1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 20] (* _Amiram Eldar_, Feb 08 2022 *)

%o (PARI) a(n)=sumdiv(n, d, moebius(n/d)^2*d^10);

%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X)/(1 - p^10*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), A351303 (k=8), A351304 (k=9), this sequence (k=10).

%K nonn,mult

%O 1,2

%A _Wesley Ivan Hurt_, Feb 06 2022