%I M2009 N0794 #56 Jan 25 2024 17:36:24
%S 2,11,123,1364,15127,167761,1860498,20633239,228826127,2537720636,
%T 28143753123,312119004989,3461452808002,38388099893011,
%U 425730551631123,4721424167835364,52361396397820127,580696784543856761,6440026026380244498,71420983074726546239
%N a(n) = 11*a(n-1) + a(n-2).
%C For odd n there is the Aurifeuillian factorization a(n) = Lucas[5n] = Lucas[n]*A[n]*B[n] = A000032[n]*A124296[n]*A124297[n], where A[n] = A124296[n] = 5*F(n)^2 - 5*F(n) + 1 and B[n] = A124297[n] = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci[n]. The largest prime divisors of a(n) for n>0 are listed in A121171[n] = {11, 41, 31, 2161, 151, 2521, 911, ...}. - _Alexander Adamchuk_, Oct 25 2006
%C For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - _Johannes W. Meijer_, Jun 12 2010
%D J. Riordan, Combinatorial Identities, Wiley, 1968, p. 139.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe, <a href="/A001946/b001946.txt">Table of n, a(n) for n = 0..200</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H <a href="/index/Rea#recur1">Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11, 1).
%F a(n) = Lucas(5n) = Fibonacci(5n-1) + Fibonacci(5n+1). - _Alexander Adamchuk_, Oct 25 2006
%F a(n) = ((11 + 5*sqrt(5))/2)^n + ((11 - 5*sqrt(5))/2)^n. - _Tanya Khovanova_, Feb 06 2007
%F Contribution from _Johannes W. Meijer_, Jun 12 2010: (Start)
%F a(2n+1) = 11*A097842(n), a(2n) = A065705(n).
%F a(3n+1) = A041226(5n), a(3n+2) = A041226(5n+3), a(3n+3) = 2* A041226(5n+4).
%F Limit(a(n+k)/a(k), k=infinity) = (A001946(n) + A049666(n)*sqrt(125))/2.
%F Limit(A001946(n)/A049666(n), n=infinity) = sqrt(125). (End)
%F From _Peter Bala_, Mar 22 2015: (Start)
%F a(n) = Fibonacci(10*n)/Fibonacci(5*n) for n >= 1.
%F a(n) = ( Fibonacci(5*n + 2*k) - F(5*n - 2*k) )/Fibonacci(2*k) for nonzero integer k.
%F a(n) = ( Fibonacci(5*n + 2*k + 1) + F(5*n - 2*k - 1) )/Fibonacci(2*k + 1) for arbitrary integer k.
%F a(n) = Sum_{k = 0..2*n} binomial(2*n,k)*Lucas(n + k). (End)
%F a(n) = [x^n] ( (1 + 11*x + sqrt(1 + 22*x + 125*x^2))/2 )^n for n >= 1. - _Peter Bala_, Jun 26 2015
%p A001946:=(-2+11*z)/(-1+11*z+z**2); # Conjectured by _Simon Plouffe_ in his 1992 dissertation
%t Table[Fibonacci[5n-1]+Fibonacci[5n+1],{n,0,30}] (* _Alexander Adamchuk_, Oct 25 2006 *)
%t LinearRecurrence[{11,1},{2,11},20] (* _Harvey P. Dale_, Jan 25 2024 *)
%o (Magma) [ Lucas(5*n) : n in [0..100]]; // _Vincenzo Librandi_, Apr 14 2011
%Y Cf. A000032, A000045, A121171, A124296, A124297.
%K easy,nonn
%O 0,1
%A _N. J. A. Sloane_, _Simon Plouffe_