OFFSET
1,1
COMMENTS
Start with any initial string of n numbers s(1), ..., s(n), all = 2 or 3 (so there are 2^n starting strings). The rule for extending the string is this:
To get s(i+1), write the 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 s(1)s(2)...s(i)). Then s(i+1) = k if k >=2, but if k=1 you must stop (without writing down the 1).
a(n) = sum of final length of string, summed over all 2^n starting strings.
See A094004 for more terms. - N. J. A. Sloane, Dec 25 2012
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.
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 [pdf, ps].
B. Chaffin, J. P. Linderman, N. J. A. Sloane and Allan Wilks, On Curling Numbers of Integer Sequences, arXiv:1212.6102 [math.CO], 2012-2013.
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.
FORMULA
Equals A216813(n) + n*2^n. - N. J. A. Sloane, Sep 26 2012
A093369 is closely related.
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, May 31 2004
EXTENSIONS
a(27)-a(31) from N. J. A. Sloane, Sep 19 2012
STATUS
approved
