A241674 - OEIS

%I #35 Jan 05 2017 02:56:41

%S 1,3,5,1,9,5,5,15,3,13,13,5,25,9,15,25,1,27,19,11,41,9,27,33,5,45,21,

%T 23,49,5,43,35,15,67,19,37,55,3,63,35,27,77,13,55,57,13,85,33,43,81,5,

%U 75,55,27,109,25,63,81,9,101,49,45,111,15,85,77,25,129,41,67,109,1,113,69,45

%N Number of factors of length 2n of the Fibonacci word (A003849) that are abelian squares.

%C A "factor" is a contiguous block.  An "abelian square" is a word of the form x x' where x' is a permutation of x.

%D Chen Fei Du, Hamoon Mousavi, Luke Schaeffer, Jeffrey Shallit, Decision Algorithms for Fibonacci-Automatic Words, III: Enumeration and Abelian Properties. See just below Theorem 11.

%H Chen Fei Du, Hamoon Mousavi, Luke Schaeffer, Jeffrey Shallit, <a href="http://arxiv.org/abs/1406.0670">Decision Algorithms for Fibonacci-Automatic Words, with Applications to Pattern Avoidance</a>, arXiv:1406.0670 [cs.FL], 2014.

%H Gabriele Fici, Filippo Mignosi, <a href="http://arxiv.org/abs/1506.03562">Words with the Maximum Number of Abelian Squares</a>, arXiv:1506.03562 [cs.DM], 2015.

%H Gabriele Fici, Filippo Mignosi, Jeffrey Shallit, <a href="https://arxiv.org/abs/1701.00948">Abelian-Square-Rich Words</a>, arXiv:1701.00948 [cs.DM], 2016. See Table 2 p. 14.

%F a(F(n)) = 2*F(n) - 1 for n >= 2, where F(n) is the n-th Fibonacci number.

%F a(n) = #{x such that x = {-i*alpha} for some 1<=i<=n and x <= {-n*alpha} if the integer part of n*alpha is even, or x >= {-n*alpha} if the integer part of n*alpha is odd}, where alpha=(sqrt(5)-1)/2. - _Gabriele Fici_, Sep 17 2015

%e a(3) = 5 because the length-6 words {010010, 001010, 010100, 100100, 001001} all occur in the Fibonacci word and are abelian squares (and no others).

%Y Cf. A003849.

%K nonn

%O 1,2

%A _Jeffrey Shallit_, Jul 10 2014