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A089069
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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.
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0
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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, 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, 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
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OFFSET
| 1,3
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COMMENTS
| 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.
A new minimal Pisot recursive matrix developed with eigenvalue equation x^3-x-1=0 using the Blackmore-Kappraff "bonacci" matrix as the pattern.
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MATHEMATICA
| (* 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}]
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CROSSREFS
| Sequence in context: A163534 A030619 A016370 * A143535 A016270 A049783
Adjacent sequences: A089066 A089067 A089068 * A089070 A089071 A089072
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KEYWORD
| sign
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AUTHOR
| Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Dec 03 2003
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