A336327 Period of orbit of Post's tag system ({0,1},{(0,0),(1,01101)},3,100^n).

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

Offset: 1

Author

A.H.M. Smeets, Jul 17 2020

Keywords

• nonn

Comments

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)).

Links

• Emil L. Post, Formal reductions of the general combinatorial decision problem, American Journal of Mathematics, Vol. 65, No. 2 (Apr., 1943), pp. 197-215.
• Eric Weisstein's World of Mathematics, Tag System

Crossrefs

Cf. A284119, A291793, A292091, A336287.

Programs

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)

Formula

Observed: if n is even then a(n) in {0, 4, 292, 4450}, if n is odd then a(n) in {0, 6, 910}.