OFFSET
0,2
COMMENTS
a(n-1) is, together with b(n) := A089508(n), n >= 1, the solution to a binomial problem; see A089508.
Numbers k such that 1 - 2*k + 5*k^2 is a square. - Artur Jasinski, Oct 26 2008
Also solution y of Diophantine equation x^2 + 4*y^2 = h^2 for which x = y-1. - Carmine Suriano, Jun 23 2010
REFERENCES
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 26.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1190
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).
G.f.: (1 - 2*x)/((1 - x)*(1 - 7*x + x^2)).
a(n) = F(1) + F(5) + F(9) +...+ F(4*n+1) = F(2*n)*F(2*n+3) + 1, where F(j) = Fibonacci(j).
a(n) = 7*a(n-1) - a(n-2) - 1, n >= 2. - R. J. Mathar, Nov 07 2015
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 25 do printf(`%d, `, (luc(4*n+3)+1)/5) od: # James A. Sellers, Mar 03 2003
MATHEMATICA
LinearRecurrence[{8, -8, 1}, {1, 6, 40}, 30] (* Bruno Berselli, Aug 31 2017 *)
PROG
(PARI) a(n)=([0, 1, 0; 0, 0, 1; 1, -8, 8]^n*[1; 6; 40])[1, 1] \\ Charles R Greathouse IV, Sep 28 2015
(PARI) first(n) = Vec((1-2*x)/((1-x)*(1-7*x+x^2)) + O(x^n)) \\ Iain Fox, Dec 19 2017
(Magma) [(Lucas(4*n+3) +1)/5: n in [0..30]]; // G. C. Greubel, Dec 18 2017
(Sage) [(lucas_number2(4*n+3, 1, -1) +1)/5 for n in (0..30)] # G. C. Greubel, Jul 13 2019
(GAP) List([0..30], n-> (Lucas(1, -1, 4*n+3)[2] +1)/5 ); # G. C. Greubel, Jul 13 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
R. K. Guy, Mar 01 2003
STATUS
approved