%I A089069
%S A089069 0,0,2,2,1,0,0,0,1,1,1,0,0,0,1,1,1,0,0,0,1,1,1,0,0,0,1,1,1,0,0,0,1,1,1,
0,0,
%T A089069 0,1,1,1,0,0,0,1,1,0,1,0,0,2,2,1,0,0,0,1,2,1,0,0,0,1,1,0,1,3,2,1,0,0,1,
3,2,1,
%U A089069 0,0,1,3,2,1,0,0,1,3,2,1,0,0,1,3,2,1,0,0,1,3,2,1,0,0,1,3,2,1,0,0,1,3,2,
1,0,0,1,3
%V A089069 0,0,2,-2,-1,0,0,0,1,-1,-1,0,0,0,1,-1,-1,0,0,0,1,-1,-1,0,0,0,1,-1,-1,0,
0,0,1,-1,-1,0,0,
%W A089069 0,1,-1,-1,0,0,0,1,-1,0,1,0,0,2,-2,-1,0,0,0,1,-2,-1,0,0,0,1,-1,0,1,3,-2,
1,0,0,1,3,-2,1,
%X A089069 0,0,1,3,-2,1,0,0,1,3,-2,1,0,0,1,3,-2,1,0,0,1,3,-2,1,0,0,1,3,-2,1,0,0,
1,3,-2,1,0,0,1,3
%N A089069 Let m0 be the matrix {{0,1,0},{0,1,1},{1,0,-q}}, m1={{0,1,0},{0,0,1},
{1,1,q1}}; if n even a(n) = (3,3)-element of m[n-1]*m0, if n odd
a(n) = (3,3)-element of m[n-1]*m1.
%C A089069 Alternating even-odd matrix recursive sequence using the x^3-x^2-x-1=0
Pisot balanced by the x^3-x-1=0 minimal Pisot.
%C A089069 A new minimal Pisot recursive matrix developed with eigenvalue equation
x^3-x-1=0 using the Blackmore-Kappraff "bonacci" matrix as the pattern.
%t A089069 (* Adamson's matrix functions alternating x^3 Pisot and minimal Pisot*)
digits=200 Solve[x^3-x-1==0, x] k=positive root of minimal Pisot
q=N[k^2-1/k, 20] m0={{0, 1, 0}, {0, 1, 1}, {1, 0, -q}} NSolve[x^3-x^2-x-1==0,
x] k1=1.83928675521416113 q1=k1^2-k1-1/k1 m1={{0, 1, 0}, {0, 0, 1},
{1, 1, q1}} m[n_Integer?Positive] := If[Mod[n, 2]==0, m[n-1].m0,
m[n-1].m1] m[0] = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}} a=Table[Floor[m[n][[3,
3]]], {n, 1, digits}]
%Y A089069 Sequence in context: A163534 A030619 A016370 this_sequence A143535 A016270
A049783
%Y A089069 Adjacent sequences: A089066 A089067 A089068 this_sequence A089070 A089071
A089072
%K A089069 sign
%O A089069 1,3
%A A089069 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Dec 03 2003
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