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Search: id:A122161
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| A122161 |
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Steinbach 3 X 3 minus the identity matrix to give a new vector matrix Markov with a Steinbach characteristic polynomial of: -1 + 2 x + x^2 - x^3. |
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+0
1
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| 1, -2, 1, -4, 0, -9, -5, -23, -24, -65, -90, -196, -311, -613, -1039, -1954, -3419, -6288, -11172, -20329, -36385, -65871, -118312, -213669, -384422, -693448, -1248623, -2251097, -4054895, -7308466, -13167159, -23729196, -42755048, -77046281, -138827181, -250164695, -450772776
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Remember? 1/(1-x)=Sum[x^n,{n,0,Infinitity}] So to try with the Steinbach field: (I-A[i,j])^(-1)=Sun[A[i,j]^n,{n,0,Infinity}] It doesn't appear it shoulsd be finite? But I-A[i,j] is finite--> zero? {{1,0,0}, {{1,1,1}, {{0,-1,-1}, {0,1,0}, {1,1,0}, {-1,0,0}, {0,0,1}} - 1,0,0}}= { -1,0,1}}
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REFERENCES
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P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.
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FORMULA
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M = {{0, -1, -1}, {-1, 0, 0}, {-1, 0, 1}}; v[1] = {1, 1, 1}; v[n_] := v[n] = M.v[n - 1]; a(n) = v[n][[1]]
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MATHEMATICA
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M = {{0, -1, -1}, {-1, 0, 0}, {-1, 0, 1}}; v[1] = {1, 1, 1}; v[n_] := v[n] = M.v[n - 1]; a1 = Table[v[n][[1]], {n, 1, 50}]
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CROSSREFS
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Cf. A046854. Cf. A046854. Cf. A007700, A059455. Cf. A065941.
Sequence in context: A011017 A077954 A077979 this_sequence A067164 A140505 A117971
Adjacent sequences: A122158 A122159 A122160 this_sequence A122162 A122163 A122164
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KEYWORD
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uned,sign
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AUTHOR
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Gary Adamson and Roger Bagula (qntmpkt(AT)yahoo.com), Oct 17 2006
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