A163194 a(n) = 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.

0, 0, 112, 2156, 39204, 704700, 12648640, 226979168, 4072998384, 73087049196, 1311494037700, 23533806023420, 422297015415552, 7577812474157376, 135978327526488304, 2440032083021144300, 43784599166902574820, 785682752921352087228, 14098504953417767184064, 252987406408599326907296

Offset: 0

Author

Stuart Clary, Jul 24 2009

Keywords

• nonn
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., 39200, 2160, 108, 4, -4, [0], 0, 112, 2156, 39204, 704700, ... This is A163196-reversed followed by A163194. That is, A163194(-n) = A163196(n-1).

Examples

			G.f. = 112*x^2 + 2156*x^3 + 39204*x^4 + 704700*x^5 + 12648640*x^6 + ...
		

References

• 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).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..500
• Index entries for linear recurrences with constant coefficients, signature (20,-35,-35,20,-1).

Crossrefs

Cf. A163195, A163196, A163197, A163198, A163199.

Programs

Magma

[(Fibonacci(n)*Lucas(n+1))^2*(Fibonacci(n-1)*Lucas(n+2)): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Mathematica

a[n_Integer] := Fibonacci[n]^2 LucasL[n+1]^2 Fibonacci[n-1] LucasL[n+2]
LinearRecurrence[{20, -35, -35, 20, -1}, {0, 0, 112, 2156, 39204}, 50] (* or *) Table[(1/5)*(Fibonacci[6n+3] - 12*Fibonacci[2n+1] + 10*(-1)^n), {n,0,25}] (* G. C. Greubel, Dec 09 2016 *)

PARI

for(n=0,30, print1((1/5)*(fibonacci(6*n+3) - 12*fibonacci(2*n+1) + 10*(-1)^n), ", ")) \\ G. C. Greubel, Dec 21 2017

Formula

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

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

Unfactored form: a(n) = (1/5)*(F(6n+3) - 12*F(2n+1) + 10*(-1)^n).

Unfactored form: a(n) = F(2n+1)^3 - 3*F(2n+1) + 2*(-1)^n.

Summation form: a(n) = 4*Sum_{k=1..n} F(2k)^3 = 4 A163198(n) if n is even; a(n) = 4*Sum_{k=2..n} F(2k)^3 = 4 A163199(n) if n is odd. a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = 200*(-1)^n.

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

G.f.: (112*x^2 - 84*x^3 + 4*x^4)/(1 - 20*x + 35*x^2 + 35*x^3 - 20*x^4 + x^5) = 4*x^2*(28 - 21*x + x^2)/((1 + x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)).

a(n) - A163196(n) = 4*(-1)^n.