This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A081071 #77 Aug 03 2025 00:06:36
%S A081071 1,16,121,841,5776,39601,271441,1860496,12752041,87403801,599074576,
%T A081071 4106118241,28143753121,192900153616,1322157322201,9062201101801,
%U A081071 62113250390416,425730551631121,2918000611027441,20000273725560976
%N A081071 a(n) = Lucas(4*n+2)-2 = Lucas(2*n+1)^2.
%C A081071 Conjecture: a(n) = Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k). - _Alex Ratushnyak_, May 06 2012
%C A081071 The above conjecture is true for n >= 1. - _Nguyen Tuan Anh_, Aug 02 2025
%D A081071 Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
%H A081071 Vincenzo Librandi, <a href="/A081071/b081071.txt">Table of n, a(n) for n = 0..1000</a>
%H A081071 Pridon Davlianidze, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Problems/ElemProbMay2020.pdf">Problem B-1270</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 2 (2020), p. 179; <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Problems/FQElemProbMay2021.pdf">Four Telescopic Infinite Products</a>, Solution to Problem B-1270 by Jason L. Smith, ibid., Vol. 59, No. 2 (2021), pp. 183-184.
%H A081071 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).
%F A081071 a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
%F A081071 G.f.: -(1+8*x+x^2)/((x-1)*(x^2-7*x+1)). - _Colin Barker_, Jun 26 2012
%F A081071 From _Peter Bala_, Nov 19 2019: (Start)
%F A081071 Sum_{n >= 1} 1/(a(n) + 5) = (3*sqrt(5) - 5)/30.
%F A081071 Sum_{n >= 1} 1/(a(n) - 5) = (15 - 4*sqrt(5) )/60.
%F A081071 Sum_{n >= 1} (-1)^(n+1)/(a(n) - 5) = 1/12.
%F A081071 Sum_{n >= 1} (-1)^(n+1)/(a(n) - 25/a(n)) = (5 + 2*sqrt(5))/120. (End)
%F A081071 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
%F A081071 Product_{n>=1} (1 - 5/a(n)) = phi^2/4, where phi is the golden ratio (A001622) (Davlianidze, 2020). - _Amiram Eldar_, Dec 04 2024
%F A081071 From _Enrique Navarrete_, Mar 24 2025: (Start)
%F A081071 20 + 5*a(n) = A106729(n)^2.
%F A081071 Limit_{n->oo} a(n+1)/a(n) = (7 + 3*sqrt(5))/2. (End)
%p A081071 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
%t A081071 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 *)
%t A081071 LucasL[4*Range[0,20]+2]-2 (* _Harvey P. Dale_, Nov 25 2012 *)
%o A081071 (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
%o A081071 (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
%Y A081071 Cf. A000032 (Lucas numbers), A000045, A001622, A002878 is Lucas(2n+1), A081069.
%K A081071 nonn,easy
%O A081071 0,2
%A A081071 _R. K. Guy_, Mar 04 2003
%E A081071 More terms from _James Sellers_, Mar 05 2003