%I #61 Sep 08 2022 08:45:05
%S 2,14,194,2702,37634,524174,7300802,101687054,1416317954,19726764302,
%T 274758382274,3826890587534,53301709843202,742397047217294,
%U 10340256951198914,144021200269567502,2005956546822746114,27939370455248878094,389145229826661547202,5420093847118012782734
%N a(n) = 14*a(n-1) - a(n-2); a(0) = 2, a(1) = 14.
%C Solves for x in x^2 - 3*y^2 = 4. [Complete nonnegative solutions are in A003500 and A052530. - _Wolfdieter Lang_, Sep 05 2021]
%C For n>0, a(n)+2 is the number of dimer tilings of a 4 X 2n Klein bottle (cf. A103999).
%C This is the Lucas sequence V(14,1). In addition to the comment above: If x = a(n) then y(n) = (a(n+1) - a(n-1))/24, n >= 1. - _Klaus Purath_, Aug 17 2021
%H G. C. Greubel, <a href="/A067902/b067902.txt">Table of n, a(n) for n = 0..850</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%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 (14,-1).
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%F G.f.: 2*(1-7*x)/(1-14*x+x^2). - _N. J. A. Sloane_, Nov 22 2006
%F a(n) = p^n + q^n, where p = 7 + 4*sqrt(3) and q = 7 - 4*sqrt(3). - _Tanya Khovanova_, Feb 06 2007
%F a(n) = 2*A011943(n+1). - _R. J. Mathar_, Sep 27 2014
%F From _Peter Bala_, Oct 16 2019: (Start)
%F Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 7 - 4*sqrt(3). This sequence gives the partial denominators in the simple continued fraction expansion of 1 + F(alpha) = 2.07140228197873197080... = 2 + 1/(14 + 1/(194 + 1/(2702 + ...))). Cf. A005248.
%F 12*Sum_{n >= 1} 1/(a(n) - 16/a(n)) = 1.
%F 16*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 12/a(n)) = 1.
%F Series acceleration formula for sum of reciprocals:
%F Sum_{n >= 1} 1/a(n) = 1/12 - 16*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 16)).
%F Sum_{n >= 1} 1/a(n) = ( (theta_3(7-4*sqrt(3)))^2 - 1 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.
%F (End)
%F From _Klaus Purath_, Aug 17 2021: (Start)
%F a(n) = (a(n-1)*a(n-2) + 2688)/a(n-3), n >= 3.
%F a(n) = (a(n-1)^2 + 192)/a(n-2), n >= 2.
%F a(2*n) = A302332(n-1) + A302332(n), n >= 1.
%F a(2*n+1) = 14*A302332(n). (End)
%F a(n) = A003500(2*n) = S(2*n,4) - S(2*n-2, 4) = 2*T(2*n,2), for n >= 0, with Chebyshev S and T. S(n, 4) = A001353(n+1) and T(n, 2) = A001075(n). - _Wolfdieter Lang_, Sep 06 2021
%e G.f. = 2 + 14*x + 194*x^2 + 2702*x^3 + 37634*x^4 + 524174*x^5 + ...
%p a := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(14) fi: 14*a(n-1)-a(n-2): end: for n from 0 to 30 do printf(`%d,`,a(n)) od:
%p seq( simplify(2*ChebyshevT(n, 7)), n=0..20); # _G. C. Greubel_, Dec 23 2019
%t a[0]=2; a[1]=14; a[n_]:= 14a[n-1] -a[n-2]; Table[a[n], {n,0,20}] (* _Robert G. Wilson v_, Jan 30 2004 *)
%t 2*ChebyshevT[Range[21] -1, 7] (* _G. C. Greubel_, Dec 23 2019 *)
%o (Sage) [lucas_number2(n,14,1) for n in range(0,20)] # _Zerinvary Lajos_, Jun 26 2008
%o (Sage) [2*chebyshev_T(n,7) for n in (0..20)] # _G. C. Greubel_, Dec 23 2019
%o (Magma) [Floor((2+Sqrt(3))^(2*n)+(1+Sqrt(3))^(-n)): n in [0..19]]; // _Vincenzo Librandi_, Mar 31 2011
%o (PARI) vector( 21, n, 2*polchebyshev(n-1, 1, 7) ) \\ _G. C. Greubel_, Dec 23 2019
%o (GAP) m:=7;; a:=[2,14];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # _G. C. Greubel_, Dec 23 2019
%Y Cf. A067900, A302332.
%Y Row 2 * 2 of array A188644.
%Y Cf. A001075, A001353, A003500, A052530.
%K nonn,easy
%O 0,1
%A _Lekraj Beedassy_, May 13 2003