OFFSET
0,2
COMMENTS
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r=sqrt(10). -Benoit Cloitre,Feb 14 2004
Numbers n such that 10*(n^2 - 1) is a square. -Vincenzo Librandi,Aug 08 2010
LINKS
Indranil Ghosh,Table of n, a(n) for n = 0..632
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh,Ellipse Chains and Associated Sequences,J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Tanya Khovanova,Recursive Sequences
Index entries for linear recurrences with constant coefficients,signature (38,-1).
FORMULA
a(n) = 38*a(n-1) - a(n-2), a(-1):= 19, a(0)=1.
G.f.: (1-19*x)/(1-38*x+x^2).
a(n) = T(n, 19) = (S(n, 38)-S(n-2, 38))/2 = S(n, 38)-19*S(n-1, 38) with T(n, x), resp. S(n, x), Chebyshev's polynomials of the first, resp. second, kind. SeeA053120andA049310.S(n, 38) =A078987(n).
a(n) = (ap^n + am^n)/2 with ap:= 19+6*sqrt(10) and am:= 19-6*sqrt(10).
a(n) = Sum_{k=0..floor(n/2)} ((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*19)^(n-2*k), n >= 1.
a(n) = cosh(2*arcsinh(3)*n). -Herbert Kociemba,Apr 24 2008
MATHEMATICA
LinearRecurrence[{38, -1}, {1, 19}, 15] (*Ray Chandler,Aug 11 2015 *)
PROG
(Sage) [lucas_number2(n, 38, 1)/2 for n in range(0, 16)] #Zerinvary Lajos,Nov 07 2009
(PARI) a(n) = polchebyshev(n, 1, 19); \\Michel Marcus,Jan 14 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang,Jan 10 2003
EXTENSIONS
More terms fromIndranil Ghosh,Feb 04 2017
STATUS
approved