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A241674 Number of factors of length 2n of the Fibonacci word (A003849) that are abelian squares. 0
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, 23, 49, 5, 43, 35, 15, 67, 19, 37, 55, 3, 63, 35, 27, 77, 13, 55, 57, 13, 85, 33, 43, 81, 5, 75, 55, 27, 109, 25, 63, 81, 9, 101, 49, 45, 111, 15, 85, 77, 25, 129, 41, 67, 109, 1, 113, 69, 45 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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

REFERENCES

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.

LINKS

Table of n, a(n) for n=1..75.

Chen Fei Du, Hamoon Mousavi, Luke Schaeffer, Jeffrey Shallit, Decision Algorithms for Fibonacci-Automatic Words, with Applications to Pattern Avoidance, arXiv:1406.0670 [cs.FL], 2014.

Gabriele Fici, Filippo Mignosi, Words with the Maximum Number of Abelian Squares, arXiv:1506.03562 [cs.DM], 2015.

Gabriele Fici, Filippo Mignosi, Jeffrey Shallit, Abelian-Square-Rich Words, arXiv:1701.00948 [cs.DM], 2016. See Table 2 p. 14.

FORMULA

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

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

EXAMPLE

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).

CROSSREFS

Cf. A003849.

Sequence in context: A112411 A283838 A228146 * A021970 A115335 A214062

Adjacent sequences:  A241671 A241672 A241673 * A241675 A241676 A241677

KEYWORD

nonn

AUTHOR

Jeffrey Shallit, Jul 10 2014

STATUS

approved

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Last modified June 20 11:38 EDT 2019. Contains 324234 sequences. (Running on oeis4.)