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%I #48 Dec 07 2019 12:18:22
%S 1,15,449,13455,403201,12082575,362074049,10850138895,325142092801,
%T 9743412645135,291977237261249,8749573705192335,262195233918508801,
%U 7857107443850071695,235451028081583642049,7055673735003659189775,211434761022028192051201
%N Chebyshev T-polynomials T(n,15) with Diophantine property.
%C Let (x_n, y_n) be n-th solution to the Pell equation x^2 = 14*y^2 + 1. Sequence gives {x_n}.
%C Numbers n such that 14*(n^2-1) is a square. - _Vincenzo Librandi_, Aug 08 2010
%C Except for the first term, positive values of x (or y) satisfying x^2 - 30xy + y^2 + 224 = 0. - _Colin Barker_, Feb 24 2014
%H Indranil Ghosh, <a href="/A068203/b068203.txt">Table of n, a(n) for n = 0..676</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H H. W. Lenstra Jr., <a href="http://www.ams.org/notices/200202/fea-lenstra.pdf">Solving the Pell Equation</a>, Notices of the AMS, Vol.49, No.2, Feb. 2002, p.182-192.
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (30,-1).
%F x_n + y_n*sqrt(14) = (x_1 + y_1*sqrt(14))^n.
%F 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
%F 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
%F 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
%F a(n) = cosh(2*n*arcsinh(sqrt(7))). - _Herbert Kociemba_, Apr 24 2008
%p Digits := 1000: q := seq(floor(evalf(((15+4*sqrt(14))^n+(15-4*sqrt(14))^n)/2)+0.1), n=1..30);
%t 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 *)
%o (Sage) [lucas_number2(n,30,1)/2 for n in range(0,15)] # _Zerinvary Lajos_, Jun 27 2008
%Y a(n)=sqrt(1 + 224*A097313(n-1)^2), n>=0. Cf. A068204.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Mar 24 2002
%E More terms from _Sascha Kurz_ and _Vladeta Jovovic_, Mar 25 2002
%E Additional term from _Colin Barker_, Feb 24 2014
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