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URL: https://oeis.org/A080143

⇱ A080143 - OEIS

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A080143
a(n) = F(3)*F(n)*F(n+1) + F(4)*F(n+1)^2 - F(4) if n even, F(3)*F(n)*F(n+1) + F(4)*F(n+1)^2 if n odd, where F(n) is the n-th Fibonacci number (A000045).
7
0, 5, 13, 39, 102, 272, 712, 1869, 4893, 12815, 33550, 87840, 229968, 602069, 1576237, 4126647, 10803702, 28284464, 74049688, 193864605, 507544125, 1328767775, 3478759198, 9107509824, 23843770272, 62423800997, 163427632717
OFFSET
0,2
LINKS
S. Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
FORMULA
G.f.: x*(5-2*x)/((1-x^2)*(1-3*x+x^2)), see a comment on A080144 for A(4,x). - Wolfdieter Lang, Jul 30 2012
a(n) = Sum_{i=0..n} ( A000045(i+4)*A000045(i) ). - Benoit Cloitre, Jun 14 2004
a(n) = (3*A027941(n) + 7*A001654(n))/2, n >= 0. Proof: from the preceding sum given by B. Cloitre via recurrence on the first factor under the sum. See also A080097 and A059840(n+2). - Wolfdieter Lang, Jul 27 2012
a(n) = (2*Lucas(2*n + 5) + 7*(-1)^(n+1) - 15)/10. - Ehren Metcalfe, Aug 21 2017
a(n) = (2*Fibonacci(n+2)*Fibonacci(n+3) - 3 - (-1)^n)/2. - G. C. Greubel, Jul 23 2019
MATHEMATICA
CoefficientList[Series[x*(5+3*x-2*x^2)/((1-x^2)*(1-2*x-2*x^2+x^3)), {x, 0, 30}], x]
With[{F=Fibonacci}, Table[(2*F[n+2]*F[n+3] -3 -(-1)^n)/2, {n, 0, 30}]] (* G. C. Greubel, Jul 23 2019 *)
PROG
(PARI) my(x='x+O('x^30)); concat([0], Vec(x*(5+3*x-2*x^2)/((1-x^2)*(1- 2*x-2*x^2+x^3)))) \\ G. C. Greubel, Mar 05 2017
(PARI) vector(30, n, n--; f=fibonacci; (2*f(n+2)*f(n+3) -3 -(-1)^n)/2) \\ G. C. Greubel, Jul 23 2019
(Magma) F:=Fibonacci; [(2*F(n+2)*F(n+3) -3 -(-1)^n)/2: n in [0..30]]; // G. C. Greubel, Jul 23 2019
(SageMath) f=fibonacci; [(2*f(n+2)*f(n+3) -3 -(-1)^n)/2 for n in (0..30)] # G. C. Greubel, Jul 23 2019
(GAP) F:=Fibonacci;; List([0..30], n-> (2*F(n+2)*F(n+3) -3 -(-1)^n)/2); # G. C. Greubel, Jul 23 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Mario Catalani (mario.catalani(AT)unito.it), Jan 30 2003
STATUS
approved