A081069 a(n) = Lucas(4n)+2 = Lucas(2n)^2.

4, 9, 49, 324, 2209, 15129, 103684, 710649, 4870849, 33385284, 228826129, 1568397609, 10749957124, 73681302249, 505019158609, 3461452808004, 23725150497409, 162614600673849, 1114577054219524, 7639424778862809

Offset: 0

Author

R. K. Guy, Mar 04 2003

Keywords

• nonn
• easy

References

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

Links

• Pridon Davlianidze, Problem B-1270, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 2 (2020), p. 179; Four Telescopic Infinite Products, Solution to Problem B-1270 by Jason L. Smith, ibid., Vol. 59, No. 2 (2021), pp. 183-184.
• Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, Table 2, k=2.
• Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Crossrefs

Cf. A000032 (Lucas numbers), A001622, A005248, A056854.

Programs

Magma

[ Lucas(2*n)^2: n in [0..70] ]; // Vincenzo Librandi, Apr 16 2011

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) od: # James Sellers, Mar 05 2003
G:=(x,n)-> cos(x)^n +cos(3*x)^n: seq(simplify(2^(4*n)*G(Pi/5,2*n)^2), n=0..19) # Gary Detlefs, Dec 05 2010
t:= n-> sum(fibonacci(4*k+2),k=0..n):seq(5*t(n)+4,n=-1..18); # Gary Detlefs, Dec 06 2010

Mathematica

LucasL[4*Range[0,20]]+2 (* Harvey P. Dale, Sep 09 2012 *)

Formula

a(n) = A005248(n)^2 = A056854(n)+2.

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

a(n) = 2^(4*n)*(cos(Pi/5)^(2*n)+cos(3*Pi/5)^(2*n))^2. - Gary Detlefs, Dec 05 2010

From Gary Detlefs, Dec 06 2010: (Start)

a(n) = 7*a(n-1)-a(n-2)-10, n>1.

a(n) = 5*Sum_{k=0..n}(Fibonacci(4*k+2))+4, with offset -1. (End)

G.f.: -(9*x^2-23*x+4)/((x-1)*(x^2-7*x+1)). - Colin Barker, Jun 24 2012

Product_{n>=0} (1 + 5/a(n)) = 3*phi^2/2, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 04 2024

a(n) = Sum_{k>=0} Lucas(2*n*k)/(Lucas(2*n)^k). - Diego Rattaggi, Jan 12 2025

E.g.f.: 2*(cosh(x) + exp(7*x/2)*cosh(3*sqrt(5)*x/2) + sinh(x)). - Stefano Spezia, Jan 20 2025