Appeared in Journal of Integer Sequences, 98.1.9.

Added March 30, 2001: Jon McCammond wrote a more efficient program (using GAP) that showed that even with the much larger haystack, in which the components Brinkhuis pairs are not related, one still gets the same kind of needles, i.e. 2^(1/17) is best possible (with this method).

Added June 12, 2001: Erratum: Uwe Grimm pointed out that the definition of Brinkhuis triple-pair, as stated, is insufficient to guarantee square-freenes-preservation of the homomorphisms, by presenting a counterexample (see Uwe Grimm's message). However, this is easily fixed. The first condition for being a Brinkhuis triple-pair is equivalent to demanding that

for every square-free word of length 2: [a,b] (there are six of them)
the four words

[U or V]_a [U or V]_b

are all square-free.

This condition needs to be replaced by the following condition.

For every square-free word of length 3: [a,b,c] (there are twelve of them)
the eight words

[U or V]_a [U or V]_b [U or V]_c ,

are square-free.

It is readily checked (by hand, or use procedure Images1 in JAN), that the proposed Brinkhuis triple-pair is indeed one, even in this new, stronger sense, hence the conclusion of the paper is upheld.

I than Uwe Grimm for his careful reading, and for spotting this inaccuracy.