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%I #12 Oct 16 2024 21:36:28
%S 1,1,0,2,1,0,3,1,0,0,6,1,1,0,0,9,2,2,1,0,0,18,2,4,1,0,0,0,30,4,7,1,0,
%T 1,0,0,56,5,14,1,1,1,0,0,0,99,8,25,2,1,2,1,0,0,0,186,11,48,2,2,3,2,1,
%U 0,0,0,335,18,88,3,3,6,4,1,0,0,0,0
%N Table read by antidiagonals: T(n,k) is the number of Lyndon words of length k on the alphabet {0,1} whose prefix is the bitwise complement of the binary expansion of n with n >= 1 and k >= 1.
%C T(n,k) = 0 if n is in A366195.
%C Row 1 is A059966.
%C Row 2 is A006206 for n > 1.
%C Row 3 is A065491 for n > 2.
%C Row 4 is A065417.
%C Row 6 is A349904.
%e Table begins
%e n\k| 1 2 3 4 5 6 7 8 9 10 11 12
%e ---+-------------------------------------------
%e 1 | 1, 1, 2, 3, 6, 9, 18, 30, 56, 99, 186, 335
%e 2 | 0, 1, 1, 1, 2, 2, 4, 5, 8, 11, 18, 25
%e 3 | 0, 0, 1, 2, 4, 7, 14, 25, 48, 88, 168, 310
%e 4 | 0, 0, 1, 1, 1, 1, 2, 2, 3, 4, 6, 7
%e 5 | 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 12, 18
%e 6 | 0, 0, 1, 1, 2, 3, 6, 10, 18, 31, 56, 96
%e 7 | 0, 0, 0, 1, 2, 4, 8, 15, 30, 57, 112, 214
%e 8 | 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 3, 3
%e 9 | 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 3, 4
%e 10 | 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 12, 18
%e 11 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 12 | 0, 0, 0, 1, 1, 2, 3, 5, 9, 15, 26, 43
%e T(6,5) = 2 because 6 is 110 in base 2, its bitwise complement is 001, and there are T(6,5) = 2 length-5 Lyndon words that begin with 001: 00101 and 00111.
%Y Cf. A059966, A006206, A065491, A065417, A349904.
%Y Cf. A365746.
%K nonn,base,tabl
%O 1,4
%A _Peter Kagey_, Oct 04 2024