1,3

a(n)=A001353(2n)/4. a(n) corresponds also to one-sixth the area of Fleenor-Heronian triangle with middle side A003500(n). - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 15 2002

a(n) give all (nontrivial, integer) solutions of Pell equation b(n+1)^2 - 48*a(n+1)^2 = +1 with b(n+1)=A011943(n), n>=0.

D. A. Benaron, personal communication.

E. K. Lloyd (E.K.Lloyd(AT)maths.soton.ac.uk), "The standard deviation of 1, 2, .., n, Pell's equation and rational triangles", preprint.

T. D. Noe, Table of n, a(n) for n=1..100

Tanya Khovanova, Recursive Sequences

Index entries for sequences related to Chebyshev polynomials.

Zerinvary Lajos, Sage Notebooks

a(n) = 14*a(n-1) - a(n-2). G.f.: (x^2)/(1-14*x+x^2).

a(n+1) ~ 1/24*sqrt(3)*(2 + sqrt(3))^(2*n) - Joe Keane (jgk(AT)jgk.org), May 15 2002

a(n+1) = S(n-1, 14), n>=0, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.

a(n+1) = ((7+4*sqrt(3))^n - (7-4*sqrt(3))^n)/(8*sqrt(3)).

a(n+1) = sqrt((A011943(n)^2 - 1)/48), n>=0.

Chebyshev's polynomials U(n-2, x) evaluated at x=7.

4*a(n+1) + A046184(n) = A055793(n+2) + A098301(n+1) 4*a(n+1) + A098301(n+1) + A055793(n+2) = A046184(n+1) (4*a(n+1))^2 = A098301(2n+1) (conjectures) - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Nov 02 2004

(4*a(n))^2 = A103974(n)^2 - A011922(n-1)^2. - Paul D. Hanna (pauldhanna(AT)juno.com), Mar 06 2005

a(n) = 13*(a(n-1)+a(n-2))-a(n-3), a(n) = 15*(a(n-1)-a(n-2))+a(n-3). a(n)=14*a(n-1)-a(n-2). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), May 26 2007

sage: [lucas_number1(n, 14, 1) for n in xrange(0, 20)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 25 2008

Cf. A001353, A003500.

Cf. A011945, A067900.

Cf. A103974, A011922.

Sequence in context: A091030 A055759 A086946 this_sequence A001023 A067221 A072533

Adjacent sequences: A007652 A007653 A007654 this_sequence A007656 A007657 A007658

nonn,easy

njas

Chebyshev comments from W. Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Nov 08 2002