%I #59 Dec 16 2023 17:34:35

%S 1,6,40,273,1870,12816,87841,602070,4126648,28284465,193864606,

%T 1328767776,9107509825,62423800998,427859097160,2932589879121,

%U 20100270056686,137769300517680,944284833567073,6472224534451830

%N a(n) = (Lucas(4*n+3) + 1)/5, or Fibonacci(2*n+1)*Fibonacci(2*n+2), or A081015(n)/5.

%C a(n-1) is, together with b(n) := A089508(n), n >= 1, the solution to a binomial problem; see A089508.

%C Numbers k such that 1 - 2*k + 5*k^2 is a square. - _Artur Jasinski_, Oct 26 2008

%C Also solution y of Diophantine equation x^2 + 4*y^2 = h^2 for which x = y-1. - _Carmine Suriano_, Jun 23 2010

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

%D A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 26.

%H G. C. Greubel, <a href="/A081016/b081016.txt">Table of n, a(n) for n = 0..1190</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).

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

%F G.f.: (1 - 2*x)/((1 - x)*(1 - 7*x + x^2)).

%F 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).

%F a(n) = 7*a(n-1) - a(n-2) - 1, n >= 2. - _R. J. Mathar_, Nov 07 2015

%p 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

%t LinearRecurrence[{8,-8,1}, {1,6,40}, 30] (* _Bruno Berselli_, Aug 31 2017 *)

%o (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

%o (PARI) first(n) = Vec((1-2*x)/((1-x)*(1-7*x+x^2)) + O(x^n)) \\ _Iain Fox_, Dec 19 2017

%o (Magma) [(Lucas(4*n+3) +1)/5: n in [0..30]]; // _G. C. Greubel_, Dec 18 2017

%o (Sage) [(lucas_number2(4*n+3,1,-1) +1)/5 for n in (0..30)] # _G. C. Greubel_, Jul 13 2019

%o (GAP) List([0..30], n-> (Lucas(1,-1,4*n+3)[2] +1)/5 ); # _G. C. Greubel_, Jul 13 2019

%Y Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081015.

%Y Partial sums of A033889. Bisection of A001654. Equals A003482 + 1.

%Y Cf. A145995, A178898.

%K nonn,easy

%O 0,2

%A _R. K. Guy_, Mar 01 2003