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%I #7 Aug 13 2015 04:05:13
%S 1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,
%T 1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,
%U 1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1,1,0
%N Infinite palindromic word (a(1),a(2),a(3),...) with initial word w(1) = (1,0) and midword sequence (a(n)); see Comments.
%C Below, w* denotes the reversal of a word w, and "sequence" and "word" are interchangable. An infinite word is palindromic if it has infinitely many initial subwords w such that w = w*.
%C Many infinite palindromic words (a(1),a(2),...) are determined by an initial word w and a midword sequence (m(1),m(2),...) of palindromes, as follows: for given w of length k, take w(1) = w = (a(1),a(2),...,a(k)). Form the palindrome w(2) = w(1)m(1)w(1)* by concatenating w(1), m(1), and w(1)*. Continue inductively; i.e., w(n+1) = w(n)m(n)w(n)* for all n >= 1. Examples follow:
%C initial word midword sequence inf. palindr. word |w(n)|
%C w(1) = 10 m(i) = a(i) A260390 A083329
%C w(1) = 01 m(i) = a(i) A260393 A083329
%C w(1) = 011 m(i) = a(i) A260394 A000225
%C w(1) = 110 m(i) = a(i) A260397 A000225
%C w(1) = 101 m(i) = a(i) A035263 A000225
%C w(1) = 100 m(i) = a(i) A260444 A000225
%C w(1) = 001 m(i) = a(i) A260445 A000225
%C w(1) = 010 m(i) = a(i) A260446 A000225
%C w(1) = 0 m(i) = i A007814 A000225
%C w(1) = 123 m(i) = a(i) A260449 A000225
%C w(1) = 132 m(i) = a(i) A260450 A000225
%C w(1) = 231 m(i) = a(i) A260451 A000225
%C w(1) = 213 m(i) = a(i) A260452 A000225
%C w(1) = 321 m(i) = a(i) A260453 A000225
%C w(1) = 312 m(i) = a(i) A260454 A000225
%C w(1) = 0 (see A260455) A260455 A081254 (conjectured)
%C w(1) = 1 (see A260456) A260456 A081254 (conjectured)
%C As a sort of (obvious) converse of the above method for constructing infinite palindromic words, every such word is determined by an initial segment w(1) and a midword sequence (m(n)), where terms of the latter may be the empty word.
%H Clark Kimberling, <a href="/A260390/b260390.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = 1 - A260393(n).
%e w(1) = 10, the initial word.
%e w(2) = 10101 ( = 10+1+01, where + = concatenation)
%e w(3) = 10101010101 = w(2)+0+w(2)*
%e w(4) = w(3)+1+w(3)*
%t u[1] = {1, 0}; m[1] = {u[1][[1]]};
%t u[n_] := u[n] = Join[u[n - 1], m[n - 1], Reverse[u[n - 1]]];
%t Table[Length[u[n]], {n, 1, 20}] (* A083329 *)
%t Flatten[Position[u[8], 0]] (* A260391 *)
%t Flatten[Position[u[8], 1]] (* A260392 *)
%Y Cf. A083329, A260392, A260394.
%K nonn,easy
%O 1
%A _Clark Kimberling_, Jul 31 2015
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