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^(n-1) 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.

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^(n-1) 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.

Terms

    a(0) =0a(1) =1a(2) =2a(3) =5a(4) =10a(5) =22a(6) =44a(7) =91a(8) =182a(9) =369a(10) =738a(11) =1486a(12) =2972a(13) =5962a(14) =11924a(15) =23884a(16) =47768a(17) =95607a(18) =191214a(19) =382568a(20) =765136a(21) =1530552a(22) =3061104a(23) =6122765a(24) =12245530a(25) =24492171a(26) =48984342a(27) =97970902a(28) =195941804a(29) =391888040

External references