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
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
KEYWORD
nonn,easy
AUTHOR
Max Alekseyev, Nov 09 2012
STATUS
approved