A217208 a(n) = (conjectured) length of longest tail that can be generated by a starting string of 2's and 3's of length n before a 1 is reached, using the rule described in the Comments lines.

0, 2, 2, 4, 4, 8, 8, 58, 59, 60, 112, 112, 112, 118, 118, 118, 118, 118, 119, 119, 119, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 132, 132, 132, 132, 132, 132, 132, 132, 133, 173, 173, 173, 173

Offset: 1

Author

N. J. A. Sloane, Sep 29 2012; revised Oct 02 2012

Keywords

• nonn
• nice
• hard

Comments

Start with an initial string S of n numbers s(1), ..., s(n), all = 2 or 3. The rule for extending the string is this:

To get s(i+1), write the current string s(1)s(2)...s(i) as XY^k for words X and Y (where Y has positive length) and k is maximized, i.e. k = the maximal number of repeating blocks at the end of the sequence so far (k is the "curling number" of the string). Then set s(i+1) = k.

The "tail length" t(S) of S is defined as follows: start with S and repeatedly append the curling number (recomputing it at each step) until a 1 is reached; t(S) is the number of terms that are appended to S before a 1 is reached. If a 1 is never reached, set t(S)=oo .

The "Curling Number Conjecture" is that if one starts with any finite string and repeatedly extends it by appending the curling number k, then eventually one must reach a 1. This has not yet been proved.

The values shown for n >= 49 are only conjectures, because certain assumptions used to cut down the search have not yet been rigorously justified. However, we believe that ALL terms shown are correct.

Examples

			a(3) = 2, using the starting string 3,2,2, which extends to 3,2,2,2,3, of length 5.
a(4) = 4, using the starting string 2,3,2,3, which extends to 2,3,2,3,2,2,2,3 of length 8.
a(8) = 58: start = 23222323, end = 232223232223222322322232223232223222322322232223232223222322322332.
a(22) = 120: start = 2322322323222323223223: see A116909 for trajectory.
		

Links

• F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), #07.1.2. [pdf, ps]
• B. Chaffin, J. P. Linderman, N. J. A. Sloane and Allan Wilks, On Curling Numbers of Integer Sequences, Journal of Integer Sequences, Vol. 16 (2013), Article 13.4.3. arXiv:1212.6102
• Benjamin Chaffin and N. J. A. Sloane, The Curling Number Conjecture, preprint.
• Index entries for sequences related to curling numbers

Crossrefs

a(n) = length of n-th row of A217209.

a(n) = A094004(n) - n.

Cf. A091787, A090822, A093369, A094005, A116909, A160766, A216730, A216813.