A219021 Sum of cubes of first n terms of Lucas sequence U(4,1) (A001353) divided by sum of their first powers.

1, 13, 172, 2356, 32661, 454329, 6325816, 88099144, 1227032521, 17090245381, 238035989412, 3315412063548, 46177727142301, 643172746439665, 8958240642814960, 124772195953666576, 1737852501591502353, 24205162822158610557, 337134426993071036956, 4695676815022772628676, 65402340983109050660389

Offset

1,2

Comments

For a Lucas sequence U(k,1), the sum of the cubes of the first n terms is divisible by the sum of the first n terms. This sequence corresponds to the case of k=4.

Links

Colin Barker, Table of n, a(n) for n = 1..875

Index entries for linear recurrences with constant coefficients, signature (19,-76,76,-19,1).

Formula

a(n) = Sum_{k=1..n} A001353(k)^3 / Sum_{k=1..n} A001353(k).

a(n) = Sum_{k=1..n} A001353(k)^3 / A061278(n).

From Colin Barker, Dec 08 2015: (Start)

a(n) = 19*a(n-1)-76*a(n-2)+76*a(n-3)-19*a(n-4)+a(n-5) for n>5.

G.f.: x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)*(1-4*x+x^2)).

(End)

Mathematica

CoefficientList[Series[(1 - 6 x + x^2)/((1 - x) (1 - 14 x + x^2) (1 - 4 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Dec 09 2015 *)

Prog

(PARI) Vec(x*(1-6*x+x^2)/((1-x)*(1-14*x+x^2)*(1-4*x+x^2)) + O(x^30)) \\ Colin Barker, Dec 08 2015
(Magma) I:=[1, 13, 172, 2356, 32661]; [n le 5 select I[n] else 19*Self(n-1)-76*Self(n-2)+76*Self(n-3)-19*Self(n-4)+Self(n-5): n in [1..30]]; // Vincenzo Librandi, Dec 09 2015

Crossrefs

Cf. A001353, A219020.

Sequence in context: A041314 A275293 A296585 * A065544 A362859 A096719

Adjacent sequences: A219018 A219019 A219020 * A219022 A219023 A219024

Keyword

nonn,easy

Author

Max Alekseyev, Nov 09 2012

Status

approved