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COMMENTS
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The head starts in state 1 on the leftmost cell (cell 1.) The machine halts if the head moves to the left of the first cell. a(n) gives the maximum halting time for any of the 3^n initial configurations of the first n cells.
Because the rule (rule 596440) is universal, the halting problem is undecidable for arbitrary tapes. However, it is not known whether universal computation can be achieved with a finite number of nonzero starting cells. Thus, this function might be computable.
Since the head starts in state one, it will immediately move left and halt unless the first cell starts out with a 0.
For initial conditions with n<=11, the machine always halts (the Mathematica code given assumes that the machine halts for all finite configurations). Whether this remains true is an open question.
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MATHEMATICA
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r = {{1, 2} -> {1, 1, -1}, {1, 1} -> {1, 2, -1}, {1, 0} -> {2, 1, 1}, {2,
2} -> {1, 0, 1}, {2, 1} -> {2, 2, 1}, {2, 0} -> {1, 2, -1}}; len[n_, k_] := Length[NestWhileList[TuringMachine[r, #] &, {{1, 2}, {Prepend[IntegerDigits[k, 3, n], 3], 0}}, UnsameQ, All]] - 2; a[n_] := Max[Table[len[n, k], {k, 0, 3^(n - 1) - 1}]]; Table[a[n], {n, 1, 8}]
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