

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
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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 FibonacciAutomatic 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 FibonacciAutomatic 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, AbelianSquareRich 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 nth 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 length6 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



