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.
A093371
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) =1a(1) =1a(2) =2a(3) =3a(4) =6a(5) =10a(6) =20a(7) =37a(8) =74a(9) =143a(10) =286a(11) =562a(12) =1124a(13) =2230a(14) =4460a(15) =8884a(16) =17768a(17) =35465a(18) =70930a(19) =141720a(20) =283440a(21) =566600a(22) =1133200a(23) =2265843a(24) =4531686a(25) =9062261a(26) =18124522a(27) =36246826a(28) =72493652a(29) =144982872
External references
- oeis: A093371