A219020 Sum of the cubes of the first n even-indexed Fibonacci numbers divided by the sum of the first n terms.

1, 7, 45, 297, 2002, 13630, 93177, 638001, 4371235, 29956465, 205313076, 1407206412, 9645056785, 66107994667, 453110391657, 3105663400665, 21286529888422, 145900036590826, 1000013702089545, 6854195814790005, 46979356835860351, 322001301602738017, 2207029753248402600, 15127206968164865112

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=3.

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Formula

a(n) = Sum_{k=1..n} A001906(k)^3 / Sum_{k=1..n} A001906(k).

a(n) = A163198(n) / A027941(n).

a(n) = 11*a(n-1) - 33*a(n-2) + 33*a(n-3) - 11*a(n-4) + a(n-5). - Vaclav Kotesovec, May 23 2013

G.f.: x*(1-4*x+x^2)/((1-x)*(1-7*x+x^2)*(1-3*x+x^2)). [Bruno Berselli, Jun 07 2013]

Mathematica

Table[Fibonacci[2*n+1]/4 + LucasL[4*n+2]/20 - 2/5, {n, 1, 20}] (* Vaclav Kotesovec, May 23 2013 *)
With[{f=Fibonacci[Range[2, 50, 2]]}, Accumulate[f^3]/Accumulate[f]] (* Harvey P. Dale, Feb 17 2020 *)

Prog

(PARI) Vec(x*(1-4*x+x^2)/((1-x)*(1-7*x+x^2)*(1-3*x+x^2)) + O(x^100)) \\ Altug Alkan, Dec 09 2015

Crossrefs

Cf. A001906, A027941, A163198, A219021.

Sequence in context: A062274 A182556 A363570 * A280597 A301319 A143835

Adjacent sequences: A219017 A219018 A219019 * A219021 A219022 A219023

Keyword

nonn,easy

Author

Max Alekseyev, Nov 09 2012

Status

approved