A351300 a(n) = n^5 * Product_{p|n, p prime} (1 + 1/p^5).

1, 33, 244, 1056, 3126, 8052, 16808, 33792, 59292, 103158, 161052, 257664, 371294, 554664, 762744, 1081344, 1419858, 1956636, 2476100, 3301056, 4101152, 5314716, 6436344, 8245248, 9768750, 12252702, 14407956, 17749248, 20511150, 25170552, 28629152, 34603008, 39296688

Offset

1,2

Comments

Sum of the 5th 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^5 * mu(n/d)^2.

a(n) = n^5 * Sum_{d|n} mu(d)^2 / d^5.

Multiplicative with a(p^e) = p^(5*e) + p^(5*e-5). - Sebastian Karlsson, Feb 08 2022

From Vaclav Kotesovec, Feb 12 2022: (Start)

Dirichlet g.f.: zeta(s)*zeta(s-5)/zeta(2*s).

Sum_{k=1..n} a(k) ~ n^6 * zeta(6) / (6 * zeta(12)) = 225225 * n^6 / (1382 * Pi^6).

Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^5/(p^10-1)) = 1.03592823428850098309076014982275428113698561633329794485946580153004... (End)

a(n) = J_10(n) / J_5(n) = A069095(n) / A059378(n), where J_k is the k-th Jordan totient function. - Enrique Pérez Herrero, Nov 13 2022

Mathematica

f[p_, e_] := p^(5*e) + p^(5*(e-1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 40] (* Amiram Eldar, Feb 08 2022 *)

Prog

(PARI) a(n)=sumdiv(n, d, moebius(n/d)^2*d^5);
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X)/(1 - p^5*X))[n], ", ")) \\ Vaclav Kotesovec, Feb 12 2022

Crossrefs

Cf. A008683 (mu).

Cf. A069095, A059378.

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), this sequence (k=5), A351301 (k=6), A351302 (k=7), A351303 (k=8), A351304 (k=9), A351305 (k=10).

Sequence in context: A088703 A321561 A034679 * A017673 A001160 A294300

Adjacent sequences: A351297 A351298 A351299 * A351301 A351302 A351303

Keyword

nonn,mult

Author

Wesley Ivan Hurt, Feb 06 2022

Status

approved