BOCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Seria I : PRACE MATEMATYCZNE X I I I (1969) ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I : COMMENTATIONES MATHEMATICAE X I I I (1969)
D. M. Silberger (Bellingham, Wash.)
O ccurrences of the integer (2n —2)1 ln\{n — 1)!
1. Introduction. In a multiplicative set G lacking both the associative and the commutative properties, an element x may have two distinct third powers; namely, x{xx) and (xx)x. Let N (n) denote the maximum number of possibly distinct n-th powers of an element in a non-associative and non-commutative groupoid. In [3], [5], and [2 ] appear proofs of the equality
(I) N(n) = (2n—2 )l/n l(n —l ) l ,
where № is a positive integer. All three of these proofs rely upon the obvious recursive relation
' n- 1
(II) N(n) = ^ N{Tc)N{n-Tc),
k=1
and the first two of the proofs, which use generating functions, are ana
lytic in character.
In Section 2 we offer an elementary combinatorial proof, of (I), in which we sidestep relation (II) by establishing a one-to-one correspond
ence between the set of formal R-th powers of an element and the set of circular permutations of n objects of one sort with n-1 objects of another sort.
In Section 3 we prove that the integer (2n—2)![n!(n—l) l is odd if and only if n is a power of 2 .
The author is indebted to his former students P. K. Davis and T. T.
Cooper for their assistance in the work in Section 2, and also to J . B.
Isbell for his suggestions leading to increased elegance in the present paper.
2. The powers problem. Lower-case Greek letters denote words in the alphabet {x, o}. The length of the word a is written |a|, and the car
dinal number of the set A is written \A\. For each positive integer n < |a|
the symbol a(n) denotes the ti-th letter in the word a. By a- 1(y) we
mean {к: a(k) — у} for y e{x , о}. For each positive integer n the symbol JSn denotes the set {а: |а- 1(ж)| = |a_1(o)|+l = n}.
By we mean {x}, and for each integer n > 1 we mean by P n the set {dfto: deP k and ($ePn_k for some positive integer k < n } . Elements in P n are called explicit n-th powers of x. It can be seen that N (n) = \Pn\ (J).
A word aeSn is called n-appropriate if and only if |Я_1(ж)| — |A- 1(o)| ^ 1 for every non-empty left segment A of a. The set of all ^-appropriate words is written An.
We omit the easy inductive proof that Pn s Sn for every positive integer n.
T
h e o r e m2.1. P n = An for each positive integer n.
Proof. I t is clear that P x = {x} = А г. Let m be an arbitrary integer greater than 1. Suppose that the equality P k — Ak holds for every positive integer k < m . Let a be any element in P m. I t follows that а = pflo, where p eP j = A j, and where $eP m_,- = for some positive integer j < m. Let A be a non-empty left segment of a. I t is plain that [A-1 (a?) | —
— |A_1(o)| > 1 if |A| < \p\, and also that ^ - 1(ж)|— |^_ 1(o)j = 1 . Therefore, we now suppose that \p\ < |A| < \(лЩ. It follows that A = p t for some non-empty left segment r of #, and that |A_1(a?)|— |A_1(o)| = |/ а_Чж)1 +
+ \x~x{x)\— |/л—1 (o)j — |r_1(o)| = 1 + |r- 1(a?)|— I tt -1 ( o )| > 2. Furthermore, К ^ ГЧ о)! = \/Г1{х )\ - |/л_1(о)Н- |#_1(a?)|— |#-4o)| = 2. Thus we see that \a 1 (я?)| — \a 4°)! = 2 —1, and hence aeA m and P m ^ Am.
Let /3 be an arbitrary element in Am. The set {i: 0 < i < |/?| and
|A| = i for a left segment A of f such that \X~l {x)\ — |A- 1(o)| = 1} is non
empty since it contains 1, and therefore this set contains a largest ele
ment p. There exist d and у such that ft = <5y, and such that |<5| = p.
It is clear that <5eA(2,+1)/2 = P (2J+1)/2.
We claim that у = q>o, and that 9?eA(2m_p_ 1)/2 = P (2m_p_1)/2. For, if a is a left segment of y, and if 1 <|(7|< \y\, then |or 1(x)\— |ff 1(o)|
= i(<5ff)—1 (ж)J — |(<5ff)~1 (о)I —1 > 2 — 1. And, furthermore, if cp is the left segment of у such that \<p\ = \y\— 1, then I(<39?)—1 (a?)| — | ( 1 (о)| =
=_|0-1И | -| 0 -1(о)| + |^-1(Ж)|— |ęr4o)| =1+\<р-1(я)\-\<р-1(о)\>2. But I/5 1(x)\— \(3 1 (о)I = 1, and our claim is established. The theorem follows by mathematical induction.
Gummer [1] showed that |An| = (2n—2)\ln\(n—l)\. We present a simple and intuitive alternate proof of this fact. To this end we remark that there are exactly
1 (2ю—1)!
2n —l n\(n—1 )!
(1) Treat о as a binary operation symbol and P n as a set of terms generated
by о and one variable ж in a bracket free notation.
O ccurrences o f (2n — 2)\jn\(n— \)\ 93
distinguishable circular permutations of n occurrences of the letter x with n —1 occurrences of the letter o. It therefore suffices to demonstrate a one-to-one correspondence between An and the collection of all such circular permutations. This follows from our next theorem.
Let Г be any circular arrangement of occurrences of x and o. Let у be such an occurrence in Г. We call у worthy in Г if and only if every non-empty word /7, spelled counterclockwise around Г by consecutive letters the first of which is y, satisfies the inequality \fi~l {x)\ > |/?_1(o)|.
T
h e o r e m2 .2 . Let Г be any circular arrangement o f n occurrences o f о with n-\-k occurrences o f x, where n and к are any non-negative integers.
Then there occur in Г exactly к worthy letters.
Proof. Fixing an arbitrary non-negative integer к, we employ in
duction on n. The basis (n = 0) is trivial. Let m be any non-negative integer, and suppose that the theorem holds when n — m.
Let 27 be any circular arrangement of m y i -f- к occurrences of x with m +1 occurrences of o. Beading counterclockwise around 27 we encounter at least once the two-letter word xo. Isolating this word xo from 27 we obtain a circular arrangement A of m + к occurrences of x with m occurrences of o. By the inductive hypothesis there are in A exactly к worthy occurrences уг, y2, ...,У к of the letters x or o. It is now evident that y ±, y2, ---чУк are worthy also in 27, and that these are the only occurrences of letters worthy in 27. The theorem follows.
I t is an obvious corollary of 2.2 that each circular arrangement of n occurrences of x with n ~ 1 occurrences of о unwraps, from exactly one letter in the arrangement, into an n-appropriate word. The desired equal
ity (I) now follows from 2.1.
I t is another obvious corollary of 2.2 that the number of distinct paths in two-space, from the lattice point <0 , 0> to the lattice point Kn—1, n}, is (2n—2)l[n\(n—l)\, when each of these paths is subject to the following three conditions:
If the lattice point <fc, j> lies in the path, and if 0 < к < n —1, then exactly one of the lattice points <&+l, j> or <&, j + l > lies in the path.
The path is 2n —1 units long.
No point <«, 6> lies in the path if b < a.
Let G be any set. Let n and к be any positive integers. Let g1, g2, ..., gn be any n not necessarily distinct elements in G. And let * 1? *2, ..., **
be к distinct and independent, non-commutative and non-associative binary operations on G (2). Then a routine argument from 2.1 and 2.2 shows that there are exactly
kn- xN{n)
(2) I.e. <6r; *15 . . . , **> is a completely free algebra.
94 D. M. S ilb e rg e r
distinct elements, in the algebra (G) * 2, **>, associated with the ordered R-tuple (gx, g2, ..., gn}, where each such element is obtained by a judicious insertion, among the terms of the w-tuple, of n —1 operation symbols.
3. The parity of N (n). For each real number у the symbol int у denotes the greatest integer not exceeding y.
We acknowledge the assistance of H. S. Zuckerman in the proof of the following lemma:
L emma 3.1. Let n be any positive integer. Let f be the non-negative integer such that n\ = 2*p for some odd integer p. Then:
(3.1.1) I f n is a power of 2, then f = n —1.
(3.1.2) I f n is not a power of 2, then f < n —1.
Proof. It is folklore (e.g., p. 29 of [4]) that
00
/ = J^ in t (n/2l).
i=1
First, we suppose that n = 2W for some non-negative integer m.
Then it is obvious that int(w/2x) = nf2X for every positive integer i < mf and also that int(w/27) = 0 for every integer j > m. Therefore we have
m
f = У] nl%1 — n — 1.
i=1 The assertion (3.1.1) follows.
Next, we suppose that the integer n is not a power of 2. It follows that n = 2kq for some non-negative integer h, and for some odd integer q > 2 . Let s be the largest integer i such that int(^/2l) > 0. It follows that int (n/2s+1) = 0, and hence 1 > nf2S+1 = 2kq/2s+1 = 2k~s~1q > 2k~s.
Therefore, since the integer q is odd, we now see that 2k~sq is not an integer, and hence n/2s — int(w/2s) = 2k~sq — int(2 fc_sg) > 0 .
Therefore, from the choice of the integer s it is now clear that 1 < int (nj2s) < nj2s, and hence
S S
f — ^ m t(n/2l) < ^ n / 2 l = n —(n/2s) < n —1 .
i=1 i=1
The assertion (3.1.2) follows, and the lemma is proved.
C orollary 3.2. Let n be any positive integer. Let f be the non-negative integer such that n\ = 2f d for some odd integer d. Let g be the non-negative integer such that (2n)\ = 2°e for some odd integer e. Then:
(3.2.1) I f n is a power of 2, then g — 2 f—l = 0.
(3.2.2) I f n is not a power of 2, then g—2 f—l > 0.
Occurrences o f (2n — 2 ) l/n l( n — l)\
95P roo f. The lemma is trivial in the case that n = 1 = 2°. Suppose that n > 1. Then the set {г: г is a positive integer such that int(w/2l) > 0}
is both finite and non-empty, and contains a largest element k. The set {i: i is a positive integer such that in t(2^г/2г) > 0} also is both finite and non-empty, and the integer ft+1 is its largest element. We now have
00 00
0 - 2 / -1 = У (2»/2s) - 2 _yint(»/2* ) - l
г=1 г=1
k + l к
— int(n/2l~1) — 2 int(n/21) — 1
г = 1 г = 1
к к
— ^ in t{n]2v) ~ 2 m t(n/2г) —1
v —O i= 1
к
= int(w/20)— У^ тЬ(п/2г)—1
i = l oo
= n — Js] int(w/21)—1 = n —f —1.
г= 1
Therefore, when n is a power of 2, then by (3.1.1) we have f = n —l t and hence g—2 f—l — n —(n —l ) —l = 0. Likewise, on the other hand, when n is not a power of 2, then by (3.1.2) we have / < n —1, and hence g— 2 f—1 > n —{n —l ) —l — 0.
T
h e o r e m3.3. The integer N (n) is odd i f and only i f n is a power of 2.
P roof. It is easy to see that
N (n) (2n)l 1
{ n \ ) 2 2 p x ’
where p 1 is the odd integer 2n —l. Let / be the non-negative integer such that n\ = 2fp 2 for some odd integer p 2. Let g be the non-negative integer such that (2n)\ = 2°pz for some odd integer p 3. It now follows that
N(n) = 29~2f- 1p 3lp lp 1.
From this equation together with (3.2.1) we see that if n is a power of 2, then N(n) = PzlvlVi’ Therefore, since in this case the integer N (n) is a quotient of products of odd integers, N (n) itself is odd.
Likewise, we see by (3.2.2) that if n is not a power of 2, then N (n)
= ^ P s I p IP i , where t is the positive integer g — 2 f—l . Therefore, since
p lp i is an odd integer, it follows in this case that the integer N(n)
is even.
96 D. M. Si l be r g e r
References
