A351305 a(n) = n^10 * Product_{p|n, p prime} (1 + 1/p^10).

1, 1025, 59050, 1049600, 9765626, 60526250, 282475250, 1074790400, 3486843450, 10009766650, 25937424602, 61978880000, 137858491850, 289537131250, 576660215300, 1100585369600, 2015993900450, 3574014536250, 6131066257802, 10250001049600, 16680163512500, 26585860217050

Offset

1,2

Comments

Sum of the 10th 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^10 * mu(n/d)^2.

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

Multiplicative with a(p^e) = p^(10*e) + p^(10*e-10). - Sebastian Karlsson, Feb 08 2022

From Vaclav Kotesovec, Feb 12 2022: (Start)

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

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

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

Mathematica

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 *)

Prog

(PARI) a(n)=sumdiv(n, d, moebius(n/d)^2*d^10);
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X)/(1 - p^10*X))[n], ", ")) \\ Vaclav Kotesovec, Feb 12 2022

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), A351304 (k=9), this sequence (k=10).

Sequence in context: A353943 A351273 A321807 * A017683 A013958 A294305

Adjacent sequences: A351302 A351303 A351304 * A351306 A351307 A351308

Keyword

nonn,mult

Author

Wesley Ivan Hurt, Feb 06 2022

Status

approved