%I #37 Jan 01 2024 11:45:31
%S 1,1,-4,-14,-4,76,176,-104,-1264,-1904,3776,18976,15296,-83264,
%T -258304,-17024,1515776,3133696,-2827264,-24456704,-31949824,82840576,
%U 357380096,217716736,-1708847104,-4723994624,805093376
%N Expansion of (1-x)/(1-2x+6x^2).
%C Binomial transform of [1, 0, -5, 0, 25, 0, -125, 0, 625, 0, ...]=: powers of -5 with interpolated zeros. - _Philippe Deléham_, Dec 02 2008
%H Michael De Vlieger, <a href="/A138229/b138229.txt">Table of n, a(n) for n = 0..2571</a>
%H Beata Bajorska-Harapińska, Barbara Smoleń, Roman Wituła, <a href="https://doi.org/10.1007/s00006-019-0969-9">On Quaternion Equivalents for Quasi-Fibonacci Numbers, Shortly Quaternaccis</a>, Advances in Applied Clifford Algebras (2019) Vol. 29, 54.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-6).
%F From _Philippe Deléham_, Nov 14 2008: (Start)
%F a(n) = 2*a(n-1) - 6*a(n-2), a(0)=1, a(1)=1.
%F a(n) = Sum_{k=0..n} A098158(n,k)*(-5)^(n-k). (End)
%F a(n) = Sum_{k=0..n} A124182(n,k)*(-6)^(n-k). - _Philippe Deléham_, Nov 15 2008
%F 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
%F a(n) = real part of the quaternion (1 + i + 2*j)^n. - _Peter Bala_, Mar 29 2015
%t 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 *)
%t TrigExpand@Table[6^(n/2) Cos[n ArcTan[Sqrt[5]]], {n, 0, 20}] (* or *)
%t Table[Sum[(-5)^k Binomial[n, 2 k], {k, 0, n/2}], {n, 0, 20}] (* _Vladimir Reshetnikov_, Sep 20 2016 *)
%o (Sage) [lucas_number2(n,2,6)/2 for n in range(0,28)] # _Zerinvary Lajos_, Jul 08 2008
%Y Cf. A088139.
%K easy,sign
%O 0,3
%A _Paul Barry_, Mar 06 2008