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%I #22 Nov 26 2016 09:01:25
%S 5,5,15,35,95,245,645,1685,4415,11555,30255,79205,207365,542885,
%T 1421295,3720995,9741695,25504085,66770565,174807605,457652255,
%U 1198149155,3136795215,8212236485,21499914245,56287506245,147362604495,385800307235,1010038317215
%N Expansion of 5*(1-x-x^2)/((1+x)*(1-3*x+x^2))
%C (Start) Let A be the unit-primitive matrix (see [Jeffery])
%C A=A_(10,2)=
%C (0 0 1 0 0)
%C (0 1 0 1 0)
%C (1 0 1 0 1)
%C (0 1 0 2 0)
%C (0 0 2 0 1).
%C Then a(n)=Trace(A^n). For m=1,2,..., A^(m) can also be written
%C A^(m)=
%C [ F(m-1)^2 0 F(m)^2 0 F(m-1)*F(m) ]
%C [ 0 F(2*m-1) 0 F(2*m) 0 ]
%C [ F(m)^2 0 F(m+1)^2 0 F(m)*F(m+1) ]
%C [ 0 F(2*m) 0 F(2*m+1) 0 ]
%C [ 2*F(m-1)*F(m) 0 2*F(m)*F(m+1) 0 F(2*m+1)-F(m)*F(m+1) ],
%C where F(m-1)=A000045(n) are the Fibonacci numbers and m=n+1. Hence also a(n+1)=Trace(A^(n+1))=F(m-1)^2+F(2*m-1)+F(m+1)^2+2*F(2*m+1)-F(m)*F(m+1). (End)
%C Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers of a unit-primitive matrix A_(N,r), 0<r<Floor(N/2), and for which the closed-form expression for a(n) is derived from the eigenvalues of A_(N,r).
%H L. E. Jeffery, <a href="/wiki/User:L._Edson_Jeffery/Unit-Primitive_Matrices">Unit-primitive matrices</a>.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F G.f.: 5*(1-x-x^2)/((1+x)*(1-3*x+x^2)).
%F a(n)=2*a(n-1)+2*a(n-2)-a(n-3), n>2, a(0)=5, a(1)=5, a(2)=15.
%F a(n)=Sum_{k=1..5) ((w_k)^2-1)^n, w_k=2*cos((2*k-1)*Pi/10); hence
%F a(n)=(-1)^n+2*(1/tau^(2*n)+tau^(2*n)), tau=(1+Sqrt(5))/2=1.618033....
%F a(n)=5*A061646(n), n>=0 (offset for A061646 is -1).
%t CoefficientList[Series[5 (1-x-x^2)/((1+x)(1-3x+x^2)),{x,0,40}],x] (* or *) LinearRecurrence[{2,2,-1},{5,5,15},40] (* _Harvey P. Dale_, Nov 26 2016 *)
%Y Cf. A000045, A061646, A189315, A189317, A189318.
%K nonn
%O 0,1
%A _L. Edson Jeffery_, Apr 20 2011
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