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a(n) = (1/4)*F(n)^2 * L(n+1)^2 * F(n-1) * L(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.
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%I #12 Sep 08 2022 08:45:46

%S 0,0,28,539,9801,176175,3162160,56744792,1018249596,18271762299,

%T 327873509425,5883451505855,105574253853888,1894453118539344,

%U 33994581881622076,610008020755286075,10946149791725643705,196420688230338021807,3524626238354441796016,63246851602149831726824

%N a(n) = (1/4)*F(n)^2 * L(n+1)^2 * F(n-1) * L(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.

%C Natural bilateral extension (brackets mark index 0): ..., 9800, 540, 27, 1, -1, [0], 0, 28, 539, 9801, 176175, ... This is A163197-reversed followed by A163195. That is, A163195(-n) = A163197(n-1).

%D Stuart Clary and Paul D. Hemenway, On sums of cubes of Fibonacci numbers, Applications of Fibonacci Numbers, Vol. 5 (St. Andrews, 1992), 123-136, Kluwer Acad. Publ., 1993. See equation (3).

%H G. C. Greubel, <a href="/A163195/b163195.txt">Table of n, a(n) for n = 0..500</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (20,-35,-35,20,-1).

%F Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).

%F a(n) = (1/4)*F(n)^2 * L(n+1)^2 * F(n-1) * L(n+2).

%F a(n) = (1/20)*(F(6n+3) - 12*F(2n+1) + 10*(-1)^n).

%F a(n) = (1/4)(F(2n+1)^3 - 3*F(2n+1) + 2*(-1)^n).

%F a(n) = Sum_{k=1..n} F(2k)^3 = A163198(n) if n is even.

%F a(n) = Sum_{k=2..n} F(2k)^3 = A163199(n) if n is odd.

%F a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = 50*(-1)^n.

%F a(n) - 20 a(n-1) + 35 a(n-2) + 35 a(n-3) - 20 a(n-4) + a(n-5) = 0.

%F G.f.: (28*x^2 - 21*x^3 + x^4)/(1 - 20*x + 35*x^2 + 35*x^3 - 20*x^4 + x^5) = x^2*(28 - 21*x + x^2)/((1 + x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)).

%F a(n) - A163197(n) = (-1)^n.

%t a[n_Integer] := (1/4)*Fibonacci[n]^2 * LucasL[n+1]^2 * Fibonacci[n-1] * LucasL[n+2]

%t LinearRecurrence[{20,-35,-35,20,-1}, {0,0,28,539,9801}, 50] (* or *) Table[(Fibonacci[6n+3] - 12*Fibonacci[2n+1] + 10*(-1)^n)/20, {n,1,25}] (* _G. C. Greubel_, Dec 09 2016 *)

%o (PARI) for(n=0,30, print1((fibonacci(6*n+3) - 12*fibonacci(2*n+1) + 10*(-1)^n)/20, ", ")) \\ _G. C. Greubel_, Dec 21 2017

%o (Magma) [(1/4)*(Fibonacci(n)*Lucas(n+1))^2*(Fibonacci(n-1)*Lucas(n+2)): n in [0..30]]; // _G. C. Greubel_, Dec 21 2017

%Y Cf. A163194, A163196, A163197, A163198, A163199.

%K nonn,easy

%O 0,3

%A _Stuart Clary_, Jul 24 2009