

A219020


Sum of the cubes of the first n evenindexed Fibonacci numbers divided by the sum of the first n terms.


2



1, 7, 45, 297, 2002, 13630, 93177, 638001, 4371235, 29956465, 205313076, 1407206412, 9645056785, 66107994667, 453110391657, 3105663400665, 21286529888422, 145900036590826, 1000013702089545, 6854195814790005, 46979356835860351, 322001301602738017, 2207029753248402600, 15127206968164865112
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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(n1)  33*a(n2) + 33*a(n3)  11*a(n4) + a(n5).  Vaclav Kotesovec, May 23 2013
G.f.: x*(14*x+x^2)/((1x)*(17*x+x^2)*(13*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*(14*x+x^2)/((1x)*(17*x+x^2)*(13*x+x^2)) + O(x^100)) \\ Altug Alkan, Dec 09 2015


CROSSREFS

Cf. A001906, A027941, A163198, A219021.
Sequence in context: A115194 A062274 A182556 * A280597 A301319 A143835
Adjacent sequences: A219017 A219018 A219019 * A219021 A219022 A219023


KEYWORD

nonn,easy


AUTHOR

Max Alekseyev, Nov 09 2012


STATUS

approved



