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

Author

Max Alekseyev, Nov 09 2012

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A001353, A219020.

Programs

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

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 *)

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

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)