A288419 a(n) = Sum_{d|n} d^3*A000593(n/d).

1, 9, 31, 73, 131, 279, 351, 585, 850, 1179, 1343, 2263, 2211, 3159, 4061, 4681, 4931, 7650, 6879, 9563, 10881, 12087, 12191, 18135, 16406, 19899, 22990, 25623, 24419, 36549, 29823, 37449, 41633, 44379, 45981, 62050, 50691, 61911, 68541, 76635, 68963, 97929

Offset: 1

Author

Seiichi Manyama, Jun 09 2017

Keywords

• nonn
• mult

Comments

Multiplicative because this sequence is the Dirichlet convolution of A000578 and A000593 which are both multiplicative. - Andrew Howroyd, Jul 27 2018

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000578, A027847, A183700, A288414.

Sum_{d|n} d^k*A000593(n/d): A288417 (k=0), A109386 (k=1), A288418 (k=2), this sequence (k=3), A288420 (k=4).

Programs

Mathematica

f[p_, e_] := (p^(3*e+5) - (p^2+p+1)*p^(e+1) + p + 1)/((p^3-1)*(p^2-1)); f[2, e_] := (8^(e+1)-1)/7; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 13 2022 *)

PARI

a(n)={sumdiv(n, d, (n/d)^3*sigma(d>>valuation(d,2)))} \\ Andrew Howroyd, Jul 27 2018

Formula

From Amiram Eldar, Nov 13 2022: (Start)

a(n) = A027847(n) for odd n.

Multiplicative with a(2^e) = (8^(e+1)-1)/7 and a(p^e) = (p^(3*e+5) - (p^2+p+1)*p^(e+1) + p + 1)/((p^3-1)*(p^2-1)) for p > 2.

Sum_{k=1..n} a(k) ~ c * n^4, where c = 7*Pi^4*zeta(3)/2880 = (7/32)*zeta(3)*zeta(4) = (7/32) * A183700 = 0.284596... . (End)