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A122173 Steinbach 6 X 6 minus the identity matrix to give a new vector matrix Markov with a characteristic polynomial of: -1 + 12 x - 34 x^2 + 30 x^3 - 6 x^4 - 3 x^5 + x^6. +0
1
1, -5, 10, -45, 110, -421, 1148, -4037, 11697, -39250, 117736, -384657, 1177235, -3787218, 11727187, -37389217, 116571621, -369712938, 1157315631, -3659226205, 11481436216, -36237006073, 113856243558, -358967583724, 1128781753801, -3556642214960, 11189229179710 (list; graph; listen)
OFFSET

1,2

COMMENT

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}}

REFERENCES

P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.

FORMULA

M = {{0, -1, -1, -1, -1, -1}, {-1, 0, -1, -1, -1, 0}, {-1, -1, 0, -1, 0, 0}, {-1, -1, -1, 1, 0, 0}, {-1, -1, 0, 0, 1, 0}, {-1, 0, 0, 0, 0, 1}}; v[1] = {1, 1, 1, 1, 1, 1}; v[n_] := v[n] = M.v[n - 1]; a(n) = v[n][[1]]

MATHEMATICA

M = {{0, -1, -1, -1, -1, -1}, {-1, 0, -1, -1, -1, 0}, {-1, -1, 0, -1, 0, 0}, {-1, -1, -1, 1, 0, 0}, {-1, -1, 0, 0, 1, 0}, {-1, 0, 0, 0, 0, 1}}; v[1] = {1, 1, 1, 1, 1, 1}; v[n_] := v[n] = M.v[n - 1]; a = Table[Floor[v[n][[1]]], {n, 1, 50}]

CROSSREFS

Cf. A046854. Cf. A046854. Cf. A007700, A059455. Cf. A065941.

Sequence in context: A001692 A038070 A136138 this_sequence A083515 A103971 A035406

Adjacent sequences: A122170 A122171 A122172 this_sequence A122174 A122175 A122176

KEYWORD

uned,sign

AUTHOR

Gary Adamson and Roger Bagula (qntmpkt(AT)yahoo.com), Oct 17 2006

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Last modified November 25 20:09 EST 2009. Contains 167514 sequences.


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