login
a(n) = n^9 * Product_{p|n, p prime} (1 + 1/p^9).
11

%I #18 Sep 28 2024 07:39:02

%S 1,513,19684,262656,1953126,10097892,40353608,134479872,387440172,

%T 1001953638,2357947692,5170120704,10604499374,20701400904,38445332184,

%U 68853694464,118587876498,198756808236,322687697780,513000262656,794320419872,1209627165996,1801152661464,2647101800448

%N a(n) = n^9 * Product_{p|n, p prime} (1 + 1/p^9).

%C Sum of the 9th powers of the divisor complements of the squarefree divisors of n.

%H Sebastian Karlsson, <a href="/A351304/b351304.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Sum_{d|n} d^9 * mu(n/d)^2.

%F a(n) = n^9 * Sum_{d|n} mu(d)^2 / d^9.

%F Multiplicative with a(p^e) = p^(9*e) + p^(9*e-9). - _Sebastian Karlsson_, Feb 08 2022

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

%F Dirichlet g.f.: zeta(s)*zeta(s-9)/zeta(2*s).

%F Sum_{k=1..n} a(k) ~ n^10 * zeta(10) / (10 * zeta(20)) = 3273645375 * n^10 / (349222 * Pi^10).

%F Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^9/(p^18-1)) = 1.002004575331916689985388864168116922608947780516939765639888137700557... (End)

%t f[p_, e_] := p^(9*e) + p^(9*(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^9);

%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X)/(1 - p^9*X))[n], ", ")) \\ _Vaclav Kotesovec_, Feb 12 2022

%o (Python)

%o from math import prod

%o from sympy import factorint

%o def A351304(n): return prod(p**(9*e)+p**(9*(e-1)) for p,e in factorint(n).items()) # _Chai Wah Wu_, Sep 28 2024

%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), this sequence (k=9), A351305 (k=10).

%K nonn,mult

%O 1,2

%A _Wesley Ivan Hurt_, Feb 06 2022