A081071 a(n) = Lucas(4*n+2)-2 = Lucas(2*n+1)^2.

1, 16, 121, 841, 5776, 39601, 271441, 1860496, 12752041, 87403801, 599074576, 4106118241, 28143753121, 192900153616, 1322157322201, 9062201101801, 62113250390416, 425730551631121, 2918000611027441, 20000273725560976

Offset: 0

Author

R. K. Guy, Mar 04 2003

Keywords

• nonn
• easy

Comments

Conjecture: a(n) = Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k). - Alex Ratushnyak, May 06 2012

The above conjecture is true for n >= 1. - Nguyen Tuan Anh, Aug 02 2025

References

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

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• 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.
• Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Crossrefs

Cf. A000032 (Lucas numbers), A000045, A001622, A002878 is Lucas(2n+1), A081069.

Programs

Magma

I:=[1, 16, 121]; [n le 3 select I[n] else 8*Self(n-1)-8*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012

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)-2) od: # James Sellers, Mar 05 2003

Mathematica

CoefficientList[Series[-(1+8*x+x^2)/((x-1)*(x^2-7*x+1)),{x,0,40}],x] (* or *) LinearRecurrence[{8,-8,1},{1,16,121},50] (* Vincenzo Librandi, Jun 26 2012 *)
LucasL[4*Range[0,20]+2]-2 (* Harvey P. Dale, Nov 25 2012 *)

PARI

x='x+O('x^30); Vec((1+8*x+x^2)/((1-x)*(x^2-7*x+1))) \\ G. C. Greubel, Dec 21 2017

Formula

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

G.f.: -(1+8*x+x^2)/((x-1)*(x^2-7*x+1)). - Colin Barker, Jun 26 2012

From Peter Bala, Nov 19 2019: (Start)

Sum_{n >= 1} 1/(a(n) + 5) = (3*sqrt(5) - 5)/30.

Sum_{n >= 1} 1/(a(n) - 5) = (15 - 4*sqrt(5) )/60.

Sum_{n >= 1} (-1)^(n+1)/(a(n) - 5) = 1/12.

Sum_{n >= 1} (-1)^(n+1)/(a(n) - 25/a(n)) = (5 + 2*sqrt(5))/120. (End)

Sum_{n>=0} 1/a(n) = (1/sqrt(5)) * Sum_{n>=1} n/F(2*n), where F(n) is the n-th Fibonacci number (A000045). - Amiram Eldar, Oct 05 2020

Product_{n>=1} (1 - 5/a(n)) = phi^2/4, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 04 2024

From Enrique Navarrete, Mar 24 2025: (Start)

20 + 5*a(n) = A106729(n)^2.

Limit_{n->oo} a(n+1)/a(n) = (7 + 3*sqrt(5))/2. (End)

Extensions

More terms from James Sellers, Mar 05 2003