A328408 G.f. A(x) satisfies: A(x) = A(x^2) + x * (1 + 4*x + x^2) / (1 - x)^4.

1, 9, 27, 73, 125, 243, 343, 585, 729, 1125, 1331, 1971, 2197, 3087, 3375, 4681, 4913, 6561, 6859, 9125, 9261, 11979, 12167, 15795, 15625, 19773, 19683, 25039, 24389, 30375, 29791, 37449, 35937, 44217, 42875, 53217, 50653, 61731, 59319, 73125, 68921, 83349, 79507, 97163, 91125

Offset

1,2

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

G.f.: Sum_{k>=0} x^(2^k) * (1 + 4*x^(2^k) + x^(2^(k+1))) / (1 - x^(2^k))^4.

G.f.: (1/7) * Sum_{k>=1} J_3(2*k) * x^k / (1 - x^k), where J_3() is the Jordan function (A059376).

Dirichlet g.f.: zeta(s-3) / (1 - 2^(-s)).

a(2*n) = a(n) + 8*n^3, a(2*n+1) = (2*n + 1)^3.

a(n) = Sum_{d|n} A209229(n/d) * d^3.

Product_{n>=1} (1 + x^n)^a(n) = g.f. for A023872.

Sum_{k=1..n} a(k) ~ 4*n^4/15. - Vaclav Kotesovec, Oct 15 2019

Multiplicative with a(2^e) = (8^(e+1)-1)/7, and a(p^e) = p^(3*e) for an odd prime p. - Amiram Eldar, Oct 23 2023

Mathematica

nmax = 45; CoefficientList[Series[Sum[x^(2^k) (1 + 4 x^(2^k) + x^(2^(k + 1)))/(1 - x^(2^k))^4, {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] // Rest
a[n_] := If[EvenQ[n], a[n/2] + n^3, n^3]; Table[a[n], {n, 1, 45}]
Table[DivisorSum[n, Boole[IntegerQ[Log[2, n/#]]] #^3 &], {n, 1, 45}]
f[p_, e_] :=p^(3*e); f[2, e_] := (8^(e+1)-1)/7; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 23 2023 *)

Prog

(Magma) [n eq 1 select 1 else IsOdd(n) select n^3 else Self(n div 2)+n^3 :n in [1..45]]; // Marius A. Burtea, Oct 15 2019

Crossrefs

Cf. A000578, A001511, A016755, A023872, A059376, A129527, A209229, A328407.

Sequence in context: A337628 A129957 A373479 * A198956 A110205 A319085

Adjacent sequences: A328405 A328406 A328407 * A328409 A328410 A328411

Keyword

nonn,easy,mult

Author

Ilya Gutkovskiy, Oct 14 2019

Status

approved