is said to have period p ifa1a2a3a4...
For example, abcabcabcabcab has periods 3, 6, 9 and 12.ai=ai+p for each i.
If word w has length l and period p, then w is a k-power, where k=l/p. For example, eraser is a 3/2-power. An infinite word over a unary alphabet contains k-powers for arbitrarily large k. On the other hand, Thue constructed an infinite binary word avoiding any k-powers with k>2. The repetitive threshold over an n-letter alphabet is
RT(n)=inf{k: some infinite word over an n-letter alphabet avoids k-powers.}Dejean's conjecture (DC) was published in 1972, but is only now finally close to solution. She conjectured that for positive integer n>1:
Several authors have chipped away at this problem, and recently Carpi reduced its resolution to establishing finitely many cases. The current state of knowledge is as follows:
RT(n) = 7/4, n =3 7/5, n = 4 n/(n-1), n ≠ 3, 4
| Values of n | State of DC | Result by | Date |
|---|---|---|---|
| 2 | Confirmed | Thue | 1906 | 3 | Confirmed | Dejean | 1972 | 4 | Confirmed | Pansiot | 1984 | 5 ≤ n ≤ 11 | Confirmed | Moulin-Ollagnier | 1992 | 12 ≤ n ≤ 14 | Confirmed | Currie, Mohammad-Noori | 2004 | 15 ≤ n ≤ 26 | Open | ? | ? | 27 ≤ n ≤ 32 | Confirmed | Currie, Rampersad | 2008 | 33 ≤ n | Confirmed | Carpi | 2007 |
Evidently, the task now is to chip away at the gap from 15 to 29. It is quite conceivable that the methods of [4,7,8] may be sharpened to do this, eventually resolving this important conjecture.
News Flash: Dejean's conjecture has been solved! See papers by Currie and Rampersad and Rao for details.