OFFSET
0,3
COMMENTS
Binomial transform of [1, 0, -5, 0, 25, 0, -125, 0, 625, 0, ...]=: powers of -5 with interpolated zeros. - Philippe Deléham, Dec 02 2008
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..2571
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 (2,-6).
FORMULA
From Philippe Deléham, Nov 14 2008: (Start)
a(n) = 2*a(n-1) - 6*a(n-2), a(0)=1, a(1)=1.
a(n) = Sum_{k=0..n} A098158(n,k)*(-5)^(n-k). (End)
a(n) = Sum_{k=0..n} A124182(n,k)*(-6)^(n-k). - Philippe Deléham, Nov 15 2008
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k+1)/(x*(5*k+6) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n) = real part of the quaternion (1 + i + 2*j)^n. - Peter Bala, Mar 29 2015
MATHEMATICA
CoefficientList[Series[(1-x)/(1-2x+6x^2), {x, 0, 30}], x] (* or *) LinearRecurrence[ {2, -6}, {1, 1}, 30] (* Harvey P. Dale, Feb 29 2012 *)
TrigExpand@Table[6^(n/2) Cos[n ArcTan[Sqrt[5]]], {n, 0, 20}] (* or *)
Table[Sum[(-5)^k Binomial[n, 2 k], {k, 0, n/2}], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 20 2016 *)
PROG
(Sage) [lucas_number2(n, 2, 6)/2 for n in range(0, 28)] # Zerinvary Lajos, Jul 08 2008
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Paul Barry, Mar 06 2008
STATUS
approved