A125269 Minimal number of states required for a 2-symbol, 5-tuple Turing machine that takes n steps on an initially blank tape before halting.

1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 4, 4, 5, 5, 4, 5, 5, 5, 5, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5

Offset: 1

Author

Dustin Wehr (robert.wehr(AT)mail.mcgill.ca), Jan 16 2007, Jan 28 2007

Keywords

• nice
• nonn

Comments

If BB(n) = A060843(n), then a(BB(n)) = n and that is the last occurrence of n in this sequence. a(n) will not become monotonic; if it did, we could compute BB(n), since a(n) is computable. a(n) diverges properly, but does so very slowly. The terms with values 1,2,3 were computed by exhaustive search. The terms with value 4 were inferred from knowing that they are greater than 3 and from the observation that for all n, a(n+1) <= a(n) + 1 (an easy construction). Using exhaustive search, may be able to extend the sequence to (most of) the terms up to and a bit beyond a(107) = 4, but going much further would likely require a more sophisticated method (see A052200).

If BB(n) = A060843(n), then a(BB(n)) = n and that is the last occurrence of n in this sequence. a(n) will not become monotonic; if it did, we could compute BB(n), since a(n) is computable. a(n) diverges properly, but does so very slowly. a(n+1) <= a(n) + 1 (an easy construction). - Martin Fuller, Feb 14 2007

Links

• Martin Fuller, Table of n, a(n) for n = 1..626
• Martin Fuller, Illustration file listing a Turing machine for each n
• H. Marxen, Info on the Busy Beaver problem

Crossrefs

Cf. A060843, A052200.

Extensions

More terms from Martin Fuller, Feb 14 2007