%I
%S 0,1,7,41,231,1289,7175,39913,221991,1234633,6866503,38188457,
%T 212387175,1181202569,6569320583,36535623529,203194800039,
%U 1130078612041,6284991883975,34954314291497,194400264968679,1081167340448777
%N A hidden Markov recursion involving the matrices: M0 = {{0, 1}, {1, 1/2}}; M = {{0, 2}, {2, 1}}; as Mh=M0.M.(M0+I); v[(n)=Mh.v(n1): first element of v.
%C Characteristic Polynomial is: 8  7 x + x^2. Binary switching of the IdentityMatrix[2] uncovers opposite signed A006131 with characteristic polynomial 4  x + x^2.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7,8).
%F M0 = {{0, 1}, {1, 1/2}}; M = {{0, 2}, {2, 1}};
%F as Mh=M0.M.(M0+I); v[(n)=Mh.v(n1):
%F a(n) first element of v(n)[[1]]/2.
%F From _R. J. Mathar_, Dec 04 2008: (Start)
%F a(n) = 7*a(n1)  8*a(n2).
%F G.f.: x/(17x+8x^2). (End)
%F a(n) = (1/17)*sqrt(17)*((7/2 + (1/2)*sqrt(17))^n  (7/2  (1/2)*sqrt(17))^n), with n >= 0.  _Paolo P. Lava_, Feb 11 2009
%t Clear[M, M0, Mh, v];
%t M0 = {{0, 1}, {1, 1/2}}; M = {{0, 2}, {2, 1}};
%t Mh = M0.(M.Inverse[IdentityMatrix[2] + M0]);
%t v[0] = {0, 1};
%t v[n_] := v[n] = Mh.v[n  1]
%t Table[ v[n][[1]]/2, {n, 0, 30}]
%o (Sage) [lucas_number1(n,7,8) for n in range(0, 22)] # _Zerinvary Lajos_, Apr 23 2009
%Y Cf. A006131.
%K nonn,easy
%O 0,3
%A _Roger L. Bagula_, Dec 01 2008
