OFFSET
0,2
COMMENTS
Let (x_n, y_n) be n-th solution to the Pell equation x^2 = 14*y^2 + 1. Sequence gives {x_n}.
Numbers n such that 14*(n^2-1) is a square. - Vincenzo Librandi, Aug 08 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 30xy + y^2 + 224 = 0. - Colin Barker, Feb 24 2014
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..676
Tanya Khovanova, Recursive Sequences
H. W. Lenstra Jr., Solving the Pell Equation, Notices of the AMS, Vol.49, No.2, Feb. 2002, p.182-192.
Index entries for linear recurrences with constant coefficients, signature (30,-1).
FORMULA
x_n + y_n*sqrt(14) = (x_1 + y_1*sqrt(14))^n.
a(n) = (-15/2-2*sqrt(14))*(-1/(-15-4*sqrt(14)))^n/(-15-4*sqrt(14))+(2*sqrt(14)-15/2)*(-1/(-15+4*sqrt(14)))^n/(-15+4*sqrt(14)). Recurrence: a(n) = 30*a(n-1)-a(n-2). G.f.: (1-15*x)/(1-30*x+x^2). - Vladeta Jovovic, Mar 25 2002
a(n) = T(n, 15)= (S(n, 30)-S(n-2, 30))/2 = S(n, 30)-15*S(n-1, 30) with T(n, x), resp. S(n, x), Chebyshev's polynomials of the first, resp.second, kind. See A053120 and A049310. S(n, 30)=A097313(n). - Wolfdieter Lang, Aug 31 2004
a(n) = sum(((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*15)^(n-2*k), k=0..floor(n/2)), n>=1. - Wolfdieter Lang, Aug 31 2004
a(n) = cosh(2*n*arcsinh(sqrt(7))). - Herbert Kociemba, Apr 24 2008
MAPLE
Digits := 1000: q := seq(floor(evalf(((15+4*sqrt(14))^n+(15-4*sqrt(14))^n)/2)+0.1), n=1..30);
MATHEMATICA
a[0] = 1; a[1] = 15; a[n_] := 30a[n-1] - a[n-2]; Table[a[n], {n, 0, 16}] (* or *) LinearRecurrence[{30, -1}, {1, 15}, 17] (* Indranil Ghosh, Feb 18 2017 *)
PROG
(Sage) [lucas_number2(n, 30, 1)/2 for n in range(0, 15)] # Zerinvary Lajos, Jun 27 2008
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Mar 24 2002
EXTENSIONS
More terms from Sascha Kurz and Vladeta Jovovic, Mar 25 2002
Additional term from Colin Barker, Feb 24 2014
STATUS
approved