A067902 a(n) = 14*a(n-1) - a(n-2); a(0) = 2, a(1) = 14.

2, 14, 194, 2702, 37634, 524174, 7300802, 101687054, 1416317954, 19726764302, 274758382274, 3826890587534, 53301709843202, 742397047217294, 10340256951198914, 144021200269567502, 2005956546822746114, 27939370455248878094, 389145229826661547202, 5420093847118012782734

Offset

0,1

Comments

Solves for x in x^2 - 3*y^2 = 4. [Complete nonnegative solutions are in A003500 and A052530. - Wolfdieter Lang, Sep 05 2021]

For n>0, a(n)+2 is the number of dimer tilings of a 4 X 2n Klein bottle (cf. A103999).

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

Links

G. C. Greubel, Table of n, a(n) for n = 0..850

Tanya Khovanova, Recursive Sequences

Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)

Index entries for linear recurrences with constant coefficients, signature (14,-1).

Index entries for sequences related to Chebyshev polynomials.

Formula

G.f.: 2*(1-7*x)/(1-14*x+x^2). - N. J. A. Sloane, Nov 22 2006

a(n) = p^n + q^n, where p = 7 + 4*sqrt(3) and q = 7 - 4*sqrt(3). - Tanya Khovanova, Feb 06 2007

a(n) = 2*A011943(n+1). - R. J. Mathar, Sep 27 2014

From Peter Bala, Oct 16 2019: (Start)

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.

12*Sum_{n >= 1} 1/(a(n) - 16/a(n)) = 1.

16*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 12/a(n)) = 1.

Series acceleration formula for sum of reciprocals:

Sum_{n >= 1} 1/a(n) = 1/12 - 16*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 16)).

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.

(End)

From Klaus Purath, Aug 17 2021: (Start)

a(n) = (a(n-1)*a(n-2) + 2688)/a(n-3), n >= 3.

a(n) = (a(n-1)^2 + 192)/a(n-2), n >= 2.

a(2*n) = A302332(n-1) + A302332(n), n >= 1.

a(2*n+1) = 14*A302332(n). (End)

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

Example

G.f. = 2 + 14*x + 194*x^2 + 2702*x^3 + 37634*x^4 + 524174*x^5 + ...

Maple

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:
seq( simplify(2*ChebyshevT(n, 7)), n=0..20); # G. C. Greubel, Dec 23 2019

Mathematica

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 *)
2*ChebyshevT[Range[21] -1, 7] (* G. C. Greubel, Dec 23 2019 *)

Prog

(Sage) [lucas_number2(n, 14, 1) for n in range(0, 20)] # Zerinvary Lajos, Jun 26 2008
(Sage) [2*chebyshev_T(n, 7) for n in (0..20)] # G. C. Greubel, Dec 23 2019
(Magma) [Floor((2+Sqrt(3))^(2*n)+(1+Sqrt(3))^(-n)): n in [0..19]]; // Vincenzo Librandi, Mar 31 2011
(PARI) vector( 21, n, 2*polchebyshev(n-1, 1, 7) ) \\ G. C. Greubel, Dec 23 2019
(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

Crossrefs

Cf. A067900, A302332.

Row 2 * 2 of array A188644.

Cf. A001075, A001353, A003500, A052530.

Sequence in context: A074655 A268005 A158102 * A132611 A156327 A047796

Adjacent sequences: A067899 A067900 A067901 * A067903 A067904 A067905

Keyword

nonn,easy

Author

Lekraj Beedassy, May 13 2003

Status

approved