

A093370


Start with any initial string of n numbers s(1), ..., s(n), with s(1) = 2, other s(i)'s = 2 or 3 (so there are 2^(n1) starting strings). The rule for extending the string is this as follows: To get s(n+1), write the string s(1)s(2)...s(n) 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. Then a(n) = number of starting strings for which k > 1.


8



0, 1, 2, 5, 10, 22, 44, 91, 182, 369, 738, 1486, 2972, 5962, 11924, 23884, 47768, 95607, 191214, 382568, 765136, 1530552, 3061104, 6122765, 12245530, 24492171, 48984342, 97970902, 195941804, 391888040
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OFFSET

1,3


LINKS

F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A SlowGrowing Sequence Defined by an Unusual Recurrence [pdf, ps].


FORMULA

a(n)/2^(n1) seems to converge to a number around 0.73.


EXAMPLE

For n=2 there are 2 starting strings, 22 and 23 and only the first has k > 1.
For n=4 there are 8 starting strings, but only 5 have k > 1, namely 2222, 2233, 2322, 2323, 2333.


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



