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%I #15 Jun 13 2015 00:51:17
%S 1,3,7,21,59,171,491,1415,4073,11729,33771,97241,279993,806209,
%T 2321385,6684163,19246279,55417453,159568195,459458307,1322957467,
%U 3809304207,10968454313,31582405473,90937912211,261845282321,753953441489,2170922412257,6250921954449
%N Let M = the 4 X 4 matrix [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / -1 -1 3 2]. Set seed vector = [1 1 1 1] = first row, then take M*[1 1 1 1] = [1 1 1 3] then M * [1 1 1 3], etc. Sequence gives terms in rightmost column.
%C a(n)/a(n-1) tends to a 9-Gon diagonal.
%C 1. The other 3 columns are offsets of 1, 3, 7, 21, 59, ... starting with 1's. 2. The characteristic equation of the 4 X 4 matrix is x^4 - 2x^3 - 3x^4 + x + 1 (coefficients may be found in A066170) with roots 2.879385241..., -1, -.5320888862... and .65270364466...An alternative matrix giving the same eigenvalues (refer to A046854) relates to the 9-Gon: [1 1 1 1 / 1 1 1 0 / 1 1 0 0 / 1 0 0 0] since the eigenvalue 2.8793852...is the longest diagonal of the 9-Gon given edge = 1. Or, 2.879385... = 1/(2Cos k*Pi/9), k = 4.
%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (2,3,-1,-1).
%F G.f.: x*(1+x-2*x^2-x^3)/(1-2*x-3*x^2+x^3+x^4). [Colin Barker, Jan 31 2012]
%F a(1)=1, a(2)=3, a(3)=7, a(4)=21, a(n)=2*a(n-1)+3*a(n-2)-a(n-3)-a(n-4) [From Harvey P. Dale, Feb 17 2012]
%e a(5) = 59 since M*[1 1 1 1] then 4 iterates = [3 7 21 59]. a(5) = rightmost term.
%e a(10)/a(9) = 11729/4073 = 2.8796955...
%t Rest[CoefficientList[Series[x (1+x-2x^2-x^3)/(1-2x-3x^2+x^3+x^4),{x,0,40}],x]] (* or *) LinearRecurrence[{2,3,-1,-1},{1,3,7,21},40] (* _Harvey P. Dale_, Feb 17 2012 *)
%o (PARI) Vec((1+x-2*x^2-x^3)/(1-2*x-3*x^2+x^3+x^4)+O(x^99)) \\ _Charles R Greathouse IV_, Jan 31 2012
%Y Cf. A066170, A046854.
%K nonn,easy
%O 1,2
%A _Gary W. Adamson_, Jan 25 2004
%E More terms from Harvey P. Dale, Feb 17 2012
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