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A336327
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Period of orbit of Post's tag system ({0,1},{(0,0),(1,01101)},3,100^n).
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3
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0, 4, 0, 4450, 0, 4450, 0, 0, 0, 4450, 0, 0, 6, 0, 0, 4450, 0, 0, 0, 0, 910, 4450, 0, 4450, 910, 4450, 0, 4450, 0, 4450, 910, 0, 0, 4450, 0, 4450, 910, 4, 0, 4, 0, 0, 910, 0, 6, 0, 910, 0, 0, 4450, 0, 4450, 910, 4450, 0, 292, 0, 4450, 0, 0, 910, 4450, 6, 4450
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OFFSET
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1,2
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COMMENTS
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In general a tag as defined by Emil Leon Post, is given by a 4-tuple (Sigma,AF,n,w0), where Sigma is some (nonempty) set of symbols called the alphabet, AF is the associated function (sometimes also called set of production rules) AF: Sigma -> Sigma*, n is the deletion number and w0 the initial string.
From the starting sequence we obtain a new string in each step by adjoining the string associated to the prefix symbol of the string, where after the prefix n symbols are removed from the string.
The decision problem is: will the tag end up in an empty string, a(n) = 0 or not, a(n) <> 0?
a(n) is an even number. Proof: for each cycle the number of associations (productions) 0 -> 0 must equal the number of associations (productions) 1 -> 01101 applied within a cycle.
For n=85 the tag is hard to solve by a brute force method, similar to the tag for n=110 with associated function {(0,00),(1,1101)} as reported in A284119.
Also period of orbit of Post's tag system ({0,1},{(0,0),(0,11010)},3,100^(n-1)).
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LINKS
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FORMULA
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Observed: if n is even then a(n) in {0, 4, 292, 4450}, if n is odd then a(n) in {0, 6, 910}.
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PROG
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(Python)
def step(w):
i = 0
while w[0] != alfabet[i]:
i = i+1
w = w+suffix[i]
return w[n:len(w)]
alfabet, suffix, n, ws, w0, m = "01", ["0", "01101"], 3, "100", "", 0
while m >= 0:
w0, m = w0+ws, m+1
w, ww, i, a = w0, w0, 0, 0
while w != "" and a == 0:
w, i = step(w), i+1
if i%100000 == 0:
ww = w
else:
if w == ww or w == "":
if w != "":
a = i%100000
print(m, a)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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