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Search: id:A133381
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| A133381 |
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Vector Matrix Markov designed so that the matrix row sums are all zero: characteristic polynomial: -3149685x - 88636 x^2 + 1037 x^3 + x^4. |
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+0
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| 0, -1, 5, -90986, 91645977, -103085764820, 114736493696302, -127830197854583449, 142405011343301378985, -158642970366894551628139, 176732345999164897395531038, -196884390194421237873409045085, 219334285467476751254430593851098
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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It is certainly an off the wall model where the Turing symbol number and tape reading are used as combinatorial "information" for the Markov matrix. The d information is Boson like, and the h information is Fermion like, making the alpha information Matter/ Antimatter like in a SU(7) or A_6 model of the Chaitan "universe". Root structure is: { -1114.03, -27.1418, 0, 104.168}
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FORMULA
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a0=Binomial[46,2]+Binomial[4,2]-Binomial[49,2]; b0=Binomial[46,3]+Binomial[4,3]-Binomial[49,3]; M = {{1, 1, -1, -1}, {49, 1, -46, -4}, {Binomial[49, 2], a0, -Binomial[46, 2], -Binomial[4, 2]}, {Binomial[49, 3], b0, -Binomial[46, 3], -Binomial[4, 3]}}; v(n)=M*v(n-1); a(n) = v(n)[[1]]
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EXAMPLE
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system equations:O=omega; A=alpha
O+d=A+h
49*O+d=46*A+4*d
Binomial[49,2]*O+a0*d=Binomial[46,2]*A+Binomial[4,2]*h
Binomial[49,3]*O+b0*d=Binomial[46,3]*A+Binomial[4,3]*h
a0,b0 adjusted to give zero row sum or constant information in a 3,2
universal Turing machine system.
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MATHEMATICA
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a0 = Binomial[46, 2] + Binomial[4, 2] - Binomial[49, 2]; b0 = Binomial[46, 3] + Binomial[4, 3] - Binomial[49, 3]; M = {{1, 1, -1, -1}, {49, 1, -46, -4}, {Binomial[49, 2], a0, -Binomial[46, 2], -Binomial[4, 2]}, {Binomial[49, 3], b0, -Binomial[46, 3], -Binomial[4, 3]}}; v[0] = {0, 0, 0, 1}; v[n_] := v[n] = M.v[n - 1]; a = Table[v[n][[1]], {n, 0, 20}]
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CROSSREFS
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Sequence in context: A050816 A145232 A123591 this_sequence A038027 A057679 A123751
Adjacent sequences: A133378 A133379 A133380 this_sequence A133382 A133383 A133384
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KEYWORD
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uned,sign
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AUTHOR
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Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Oct 28 2007
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