How many unbordered words?

ANNALES SOCIETATIS M ATHEMATICAE POLONAE Series I: COM M ENTATION ES MATHEMATICAE XXII (1980) ROCZN1KI POLSK.IEGO TOWARZYSTWA M ATEM ATYCZNEGO

Séria I: PRACE MATEMATYCZNE XXII (1980)

D. M.

(Florianôpolis, Brasil)

How many unbordered words?

1. Introduction. Let s denote both a positive integer and also an alphabet containing exactly s letters. Let s* denote the free monoid generated by s.

When {a, /?} £ s* we say that ft is a border of a if and only if a = Xp = fig for some words X and

while 0 < \P\ < |a|, where |a| denotes the length of a.

The title question, raised independently by A. Ehrenfeucht and by V. Jacques, remains open. Precisely formulated this question asks us to characterize the function Us defined by Us(n) = |{a: a es* and |a| = n and a is unbordered}!. Less ambitiously, perhaps, specify lim Us(n)/sn.

n

It is well-known — e.g., Theorem 3.3 of [4] — that the nth Catalan number is odd if and only if n is not a power of 2. Our purpose in the present paper is to show that the function Us is interesting, by establishing that the sequence <l/s(n)/s)®=1 is in this respect like the Catalan sequence:

Th e o r e m.

Let s ^ 2. Then s2 is a factor of Us(n) if and only if n is not a power of 2.

The idea of an unbordered word is interesting in its own right. In [2], [5], and [6] links are established between unborderedness and universality for infinite sets. And, [1] argues that an infinite sequence о is periodic if and only if there is a finite upper bound on the lengths of unbordered finite segments of o .

2. P ro o fs. It is easy — e.g., via Lemma 2 of [3] — to establish the following

Le m m a.

I f a has a border, then a has a unique shortest border ft. Moreover, P itself is unbordered, and 2\P\ ^ |a|.

The expression int (x) denotes the greatest integer which does not exceed the real number x.

int (n /2 )

Pr o p o s it io n.

Let n > 1. Then Us(n) = sn— £ sn~2kUs(k). Furthermore

k = 1

U,(l) = s.

144 D. M. S i l b e r g e r

P ro o f. Choose a positive integer к ^ int (n/2). It is clear by our lemma that the number of distinct a es*, such that |a| = n and such that the length of the shortest border of a is k, is given by the expression sn~2k Us(k).

Again by the lemma we can see that the number of bordered a es* with

int (n/2)

|a| — n is given by the expression Y s"~2k Us(k). Therefore, since s* con-

k= 1

tains exactly s" distinct words oft length n, the formula for Us(n) follows. □

Co r o l l a r y.

Let m be a positive integer. Then Us(2m) = sUs(2m—l) —

— Us(m) and Us(2m+1) = sUs(2m).

m

P ro o f. By the proposition, Us(2m) = s2m— Y s2m~2k Us(k) = s(s2m~1 — m — 1

— Y s2m~1~2kUs{k)) — s2m~2mUs(m) = sUs(2m—1)—Us(m)since int ((2m—l)/2)

k= 1

= m — 1.

m

Similarly, Us(2m+1) — s2m+1 — £ s2m + l ~2k Us(k) since int ((2m-l-1)/2)

fc = 1

= m. It easily follows that Us(2m + \ ) = sUs(2m). □

P r o o f of th e o re m . Since Us( 1) = s we see that s2 it not a factor of Us(2°). Choose an arbitrary integer n > 1, and suppose for every positive integer к < n that s2 | Us (к) if and only if к is not a power of 2.

Choose a letter L e s , and for each positive integer j let ws(j) denote the number of unbordered a es* such that |a| = j and such that L is the left-most letter of a. Clearly Us(j) = sws(j). Therefore, s2 \sUs(j) for every positive integer j.

In the case that n = 2m+1 for some positive integer m, of course n is not a power of 2, and we have by the corollary that Us(n) = sUs(2m), whereupon s2 | Us(n) as is required. Hence we may suppose that n = 2m ^ 2.

Again by the corollary we have that Us(n) = sUs(2m— 1)— Us(m). Therefore, since s2 \sUs(2m— 1) we have by hypothesis that s2 \Us(n) if and only if m is not a power of 2. But n is a power of 2 if and only if m is a power of 2. □

Let f s{n) denote Us(n)/sn. Clearly f s expresses the “density” of unbordered elements in s* as a function of their lengths.

Our corollary implies that f s(2m+l) = f s(2m) = f s(2m — 1) — f s(m)/sm for every positive integer m. It follows that the sequence </s(n))^°=i is non­

increasing and has a limit. Of course 0 ^ Н т ^ (л ) < (s — l)/s. Preliminary П

guess: The limit is positive.

References

[1] A. E h r e n fe u c h t and D. M. S ilb e r g e r , Periodicity and unbordered segments of words, Discrete Mathematics 26 (1979), p. 101-109.

How many unbordered words? 145

[2] J. R. I s b e ll, On the problem o f universal terms, Bull. Acad. Polon. Sci. 14 (1966), p. 593-595.

[3] R. C. L y n d o n and M. P. S ch U tz e n b e r g e r , The equation aM = bNcp in a free group, Michigan Math. J. 9 (1962), p. 289-298.

[4] D. M. S ilb e r g e r , Occurrences of the integer (2n — 2)\/nl(n — 1)!, Comment. Math. 13 (1969), p. 91-96.

[5] — Point universal terms in a free semigroup, Diss. Ph. D. University of Washington, Seattle 1973.

[6] — When is a term point universal? Algebra Universalis (to appear).

10 - Prace Matematyczne 22.1