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Smallest (i.e., rightmost) Lyndon word in the Lyndon factorization of the binary representation of n (written using 1's and 2's rather than 0's and 1's, since numbers > 0 in the OEIS cannot begin with 0).
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%I #18 May 10 2014 09:50:20

%S 1,2,1,2,1,12,1,2,1,112,1,122,1,12,1,2,1,1112,1,1122,1,12,1,1222,1,

%T 112,1,122,1,12,1,2,1,11112,1,11122,1,11212,1,11222,1,112,1,12122,1,

%U 12,1,12222,1,1112,1,1122,1,12,1,1222,1,112,1,122,1,12,1,2,1,111112,1,111122,1,111212,1,111222,1,112,1,112122,1,112212,1,112222,1,1112,1,1122,1

%N Smallest (i.e., rightmost) Lyndon word in the Lyndon factorization of the binary representation of n (written using 1's and 2's rather than 0's and 1's, since numbers > 0 in the OEIS cannot begin with 0).

%C See A211095 and A211097 for further information, including Maple programs.

%F a(2k) is always 1 (i.e., 0).

%e n=25 has binary expansion 11001, which has Lyndon factorization (1)(1)(001) with three factors. The rightmost factor is 001, which we write as a(25) = 112.

%e The real sequence (written with 0's and 1's rather than 1's and 2's) is:

%e 0, 1, 0, 1, 0, 01, 0, 1, 0, 001, 0, 011, 0, 01, 0, 1, 0, 0001, 0, 0011, 0, 01, 0, 0111, 0, 001, 0, 011, 0, 01, 0, 1, 0, 00001, 0, 00011, 0, 00101, 0, 00111, 0, 001, 0, 01011, 0, 01, 0, 01111, 0, 0001, 0, 0011, 0, 01, 0, 0111, 0, 001, 0, 011, ...

%Y Cf. A211100, A211095, A211097, A211098, A211099.

%K nonn

%O 0,2

%A _N. J. A. Sloane_, Mar 31 2012