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, Dec 25 2012.
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