A214733 a(n) = -a(n-1) - 3*a(n-2) with n>1, a(0)=0, a(1)=1.

0, 1, -1, -2, 5, 1, -16, 13, 35, -74, -31, 253, -160, -599, 1079, 718, -3955, 1801, 10064, -15467, -14725, 61126, -16951, -166427, 217280, 282001, -933841, 87838, 2713685, -2977199, -5163856, 14095453, 1396115, -43682474, 39494129, 91553293, -210035680

Offset: 0

Author

Roman Witula, Jul 27 2012

Keywords

• sign
• easy

Comments

The sequence a(n) is conjugate with A110523 by the following alternative relations: either ((-1 + i*sqrt(11))/2)^n = A110523(n) + a(n)*(-1 + i*sqrt(11))/2, or ((-1 - i*sqrt(11))/2)^n = A110523(n) + a(n)*(-1 - i*sqrt(11))/2 (see also comments to A110523, where these relations and many other facts on a(n) is presented).

Apart from signs, the Lucas U(P=1,Q=3)-sequence. - R. J. Mathar, Oct 24 2012

This is the Lucas U(-1, 3) sequence. (V_n(-1, 3))^2 + 11*(U_n(-1, 3))^2 = 4*Q^n = 4*3^n. For the special case where |U_n(-1, 3)| = 1, then, by the Lucas sequence identity U_2*n = U_n*V_n, we have (U_2*n(-1, 3))^2 + 11 = 4*3^n, true for n = 1, 2, 5, U_n = 1, -1, 1 and U_2*n = -1, 5, -31. E.g., (-31)^2 + 11 = 972 = 4*3^5. - Raphie Frank, Dec 09 2015

References

• Roman Witula, On Some Applications of Formulae for Unimodular Complex Numbers, Jacek Skalmierski's Press, Gliwice 2011.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..4189
• Ronald Orozco López, Deformed Differential Calculus on Generalized Fibonacci Polynomials, arXiv:2211.04450 [math.CO], 2022.
• Wikipedia, Lucas sequence
• Index entries for linear recurrences with constant coefficients, signature (-1,-3).

Crossrefs

Cf. A106852, A110523.

Programs

Magma

[n le 2 select n-1 else -Self(n-1)-3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Dec 10 2015

Mathematica

LinearRecurrence[{-1, -3}, {0, 1}, 40] (* T. D. Noe, Jul 30 2012 *)

PARI

concat(0,Vec(1/(1+x+3*x^2)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

SageMath

[(-1)^(n-1)*round(3^((n-1)/2)*chebyshev_U(n-1, 1/(2*sqrt(3)))) for n in range(41)] # G. C. Greubel, Dec 28 2023

Formula

a(n) = - a(n-1) - 3*a(n-2).

a(n) = (-1)^n*(i*sqrt(11)/11)*(((1 + i*sqrt(11))/2)^n - ((1 - i*sqrt(11))/2)^n).

G.f.: x/(1 + x + 3*x^2).

G.f.: Q(0) -1, where Q(k) = 1 - 3*x^2 - (k+2)*x + x*(k+1 + 3*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013

From G. C. Greubel, Dec 28 2023: (Start)

a(n) = (-1)^(n-1)*3^((n-1)/2)*ChebyshevU(n-1, 1/(2*sqrt(3))).

a(n) = (-1)^n * A106852(n-1).

E.g.f.: (2/sqrt(11))*exp(-x/2)*sin(sqrt(11)*x/2). (End)