1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 2, 2, 1, 0, 1, 2, 3, 2, 1, 0, 1, 2, 4, 5, 2, 1, 0, 1, 2, 4, 4, 8, 2, 1, 0, 1, 2, 4, 5, 4, 13, 2, 1, 0, 1, 2, 4, 6, 7, 4, 21, 2, 1, 0, 1, 2, 4, 7, 10, 11, 4, 34, 2, 1, 0, 1, 2, 4, 8, 13, 16, 16, 4, 55, 2, 1, 0, 1, 2, 4, 8, 8, 24
Offset: 0
Table begins:
n\k | 0 1 2 3 4 5 6 7 8 9 10 11
-----+----------------------------------------------------
0 | 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
1 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
2 | 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ...
3 | 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...
4 | 1, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, ...
5 | 1, 2, 4, 5, 7, 11, 16, 23, 34, 50, 73, 107, ...
6 | 1, 2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, ...
7 | 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, ...
8 | 1, 2, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, ...
9 | 1, 2, 4, 8, 9, 11, 15, 23, 32, 43, 58, 81, ...
For (n,k) = (3,4), we see that T(3,4) = 8 because there are 8 binary strings of length k = 4 where all length A070939(3) = 2 substrings are lexicographically earlier than "11" (the binary expansion of n = 3): 0000, 0001, 0010, 0100, 0101, 1000, 1001, and 1010.
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