OFFSET
0,2
COMMENTS
In general, the sequence with g.f. 1/(1 - 2r*x + (r^2 + 1)*x^2) = 1/((1 - r*x)^2 + x^2) has a(n) = Sum_{k = 0..floor(n/2)} binomial(n - k, k)(r^2 - 1)^k*(2r)^(n - 2k); a(n) = Sum_{k = 0..floor((n + 1)/2)} binomial(n + 1, 2k + 1)(-1)^k*r^(n-2k).
LINKS
Robert Israel, Table of n, a(n) for n = 0..1997
Beata Bajorska-Harapińska, Barbara Smoleń, Roman Wituła, On Quaternion Equivalents for Quasi-Fibonacci Numbers, Shortly Quaternaccis, Advances in Applied Clifford Algebras (2019) Vol. 29, 54.
Index entries for linear recurrences with constant coefficients, signature (6,-10).
FORMULA
G.f.: 1/((1 - 3*x)^2 + x^2).
a(n) = Sum_{k = 0..floor(n/2)} binomial(n - k, k)(-10)^k*6^(n - 2k).
a(n) = Sum_{k = 0..floor((n + 1)/2)} binomial(n + 1, 2k + 1)(-1)^k*3^(n - 2k).
a(n) = 6*a(n - 1) - 10*a(n - 2), n >= 2. - Vincenzo Librandi, Mar 22 2011
a(n) = Im((3 + i)^(n + 1)), where i = sqrt(-1). - César Eliud Lozada, Sep 19 2012
E.g.f.: (3*sin(x) + cos(x))*exp(3*x). - Ilya Gutkovskiy, Nov 25 2016
MAPLE
f:= gfun:- rectoproc({a(n+2)=6*a(n+1)-10*a(n), a(0)=1, a(1)=6}, a(n), remember):
map(f, [$0..50]); # Robert Israel, Nov 25 2016
MATHEMATICA
CoefficientList[Series[1/(1 - 6x + 10x^2), {x, 0, 30}], x] (* or *) LinearRecurrence[{6, -10}, {1, 6}, 30] (* Harvey P. Dale, Feb 05 2015 *)
PROG
(Sage) [lucas_number1(n, 6, 10) for n in range(1, 29)] # Zerinvary Lajos, Apr 22 2009
(PARI) imag((3+I)^(n+1)); /* Joerg Arndt, Sep 20 2012 */
(PARI) x='x+O('x^100); Vec(1/((1-3*x)^2+x^2)) \\ Altug Alkan, Dec 24 2015
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Paul Barry, May 01 2005
STATUS
approved