A351304 a(n) = n^9 * Product_{p|n, p prime} (1 + 1/p^9).

1, 513, 19684, 262656, 1953126, 10097892, 40353608, 134479872, 387440172, 1001953638, 2357947692, 5170120704, 10604499374, 20701400904, 38445332184, 68853694464, 118587876498, 198756808236, 322687697780, 513000262656, 794320419872, 1209627165996, 1801152661464, 2647101800448

Offset

1,2

Comments

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

Links

Sebastian Karlsson, Table of n, a(n) for n = 1..10000

Formula

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

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

Multiplicative with a(p^e) = p^(9*e) + p^(9*e-9). - Sebastian Karlsson, Feb 08 2022

From Vaclav Kotesovec, Feb 12 2022: (Start)

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

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

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

Mathematica

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 *)

Prog

(PARI) a(n)=sumdiv(n, d, moebius(n/d)^2*d^9);
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X)/(1 - p^9*X))[n], ", ")) \\ Vaclav Kotesovec, Feb 12 2022
(Python)
from math import prod
from sympy import factorint
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

Crossrefs

Cf. A008683 (mu).

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).

Sequence in context: A353942 A351272 A321565 * A017681 A013957 A294304

Adjacent sequences: A351301 A351302 A351303 * A351305 A351306 A351307

Keyword

nonn,mult

Author

Wesley Ivan Hurt, Feb 06 2022

Status

approved