A081079 a(n) = Lucas(4*n+2) - 3 = 5*Fibonacci(2*n)*Fibonacci(2*n+2).

0, 15, 120, 840, 5775, 39600, 271440, 1860495, 12752040, 87403800, 599074575, 4106118240, 28143753120, 192900153615, 1322157322200, 9062201101800, 62113250390415, 425730551631120, 2918000611027440, 20000273725560975, 137083915467899400, 939587134549734840

Offset

0,2

References

Hugh C. Williams, Édouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Formula

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).

From R. J. Mathar, Sep 03 2010: (Start)

a(n) = 15*A092521(n) = 5*A058038(n).

G.f.: 15*x/((1-x)*(1-7*x+x^2)). (End)

From G. C. Greubel, Jun 12 2025: (Start)

a(n) = 15*(-1)^n*A156088(n).

E.g.f.: exp(7*x/2)*( 3*cosh(3*sqrt(5)*x/2) - sqrt(5)*sinh(3*sqrt(5)*x/2) ) - 3*exp(x). (End)

Sum_{n>=1} 1/a(n) = (3 - sqrt(5))/10 = A187799 / 100. - Amiram Eldar, Dec 25 2025

Maple

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d, `, luc(4*n+2)-3) od: # James Sellers, Mar 05 2003

Mathematica

LinearRecurrence[{8, -8, 1}, {0, 15, 120}, 20] (* Jean-François Alcover, Nov 29 2023 *)

Prog

(Magma)
A081079:= func< n | Lucas(4*n+2) -3 >; // G. C. Greubel, Jun 12 2025
(SageMath)
def A081079(n): return lucas_number2(4*n+2, 1, -1) -3 # G. C. Greubel, Jun 12 2025

Crossrefs

Cf. A000032 (Lucas numbers), A000045 (Fibonacci numbers), A058038, A092521, A156088, A187799.

Sequence in context: A010967 A022580 A321950 * A138424 A279267 A357602

Adjacent sequences: A081076 A081077 A081078 * A081080 A081081 A081082

Keyword

nonn,easy

Author

R. K. Guy, Mar 04 2003

Status

approved