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A219021 Sum of cubes of first n terms of Lucas sequence U(4,1) (A001353) divided by sum of their first powers. 2

%I #16 Sep 08 2022 08:46:04

%S 1,13,172,2356,32661,454329,6325816,88099144,1227032521,17090245381,

%T 238035989412,3315412063548,46177727142301,643172746439665,

%U 8958240642814960,124772195953666576,1737852501591502353,24205162822158610557,337134426993071036956,4695676815022772628676,65402340983109050660389

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

%C 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.

%H Colin Barker, <a href="/A219021/b219021.txt">Table of n, a(n) for n = 1..875</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (19,-76,76,-19,1).

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

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

%F From _Colin Barker_, Dec 08 2015: (Start)

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

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

%F (End)

%t 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 *)

%o (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

%o (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

%Y Cf. A001353, A219020.

%K nonn,easy

%O 1,2

%A _Max Alekseyev_, Nov 09 2012

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