#A5 INTEGERS 8A (208) EXPLICIT EXAMPLES OF p-adic NUMBERS WITH PRESCRIBED IRRATIONALITY EXPONENT Yann Bugeaud IRMA, UMR 750, Université de Strasbourg et CNRS, Strasbourg, France bugeaud@math.unistra.fr Tomislav Pejković Department of Mathematics, Faculty of Science, University of Zagreb, Zagreb, Croatia pejkovic@math.hr Received: 4/6/7, Revised: /6/7, Accepted: /23/8, Published: 3/6/8 Abstract Let p be a prime number and µ > 2 a real number. We establish that the irrationality exponent of the p-adic number P + pbµic is equal to µ. This provides us with explicit examples of p-adic numbers with any prescribed irrationality exponent. To Je Shallit on his sixtieth birthday. Introduction The irrationality exponent µ( ) of an irrational real number is the supremum of the real numbers µ such that a b < b µ has infinitely many solutions in rational numbers a/b. It follows from the theory of continued fractions that µ( ) is at least 2. Furthermore, for any given real number µ > 2, it is easy to construct explicitly continued fraction expansions of real numbers µ with irrationality exponent µ. However, if one requires additional properties, like for example µ being in the middle third Cantor set, this is less straightforward. This question was solved recently in [5], also by means of the theory of continued fractions. The main goal of this paper is to consider the analogous problem for p-adic numbers (here, and throughout the paper, p denotes a prime number), that is, to construct explicit examples of p-adic numbers with prescribed irrationality exponent and only digits 0 and in their Hensel expansion.
INTEGERS: 8A (208) 2 A natural idea is to translate the argument of [5] to the p-adic setting. Unfortunately, this does not seem to work, since there is no full analogue of the theory of continued fractions in p-adic fields. Admittedly, it is possible to express a p-adic number as a continued fraction, but, unlike in the real case, one cannot read o from this expansion all the best rational approximations to. Recall that to determine the exact value of the irrationality exponent of requires, on the one hand, to bound it from below, that is, to find an infinite sequence of very good rational approximations to (this is often an easy step) and, on the other hand, to establish that only finitely many very good rational approximations to are outside this sequence (this is usually much more delicate). In most of the cases in the real setting, this delicate step requires a precise description of the continued fraction expansion of. For a reduced rational number a/b, let H(a/b) = max{ a, b } denote its height. The irrationality exponent µ( ) of an irrational p-adic number is the supremum of the real numbers µ such that a b p < H(a/b) µ has infinitely many solutions in rational numbers a/b. If a reduced rational number a/b is such that a/b p is smaller than p, then the ultrametric inequality implies a/b p = p and thus quantities like max{ a p, b p } and min{ a p, b p } would be constant which renders them useless for our work. This is why H(a/b) as defined above is used as a measure of size for a rational approximation a/b to, also in the p-adic setting. As in the real setting, we have µ( ) > 2; see for example Section 9.3 of [4]. Our main result is the following theorem. Theorem. Let p be a prime number. Let c := (c i ) i>0 be a sequence of positive integers such that c i+ > 2c i for i large enough. Then the irrationality exponent of the p-adic number is equal to c := +X p ci c := lim sup i!+ c i+ c i. In particular, for any real number µ > 2, we have µ + X p bµi c = µ. Theorem is the p-adic analogue of a result stated in the Introduction of [6].
INTEGERS: 8A (208) 3 It follows from Theorems B and 6 of [], whose proofs rest on a p-adic version of Schmidt Subspace Theorem, that all the p-adic numbers c defined in Theorem are transcendental, since their Hensel expansions have small block complexity. For every real number µ > 2, the Hausdor dimension of the set of p-adic numbers with irrationality exponent µ is equal to 2/µ; see Theorem 6.9 of [2]. This implies the existence of p-adic numbers with any prescribed irrationality exponent. As far as we are aware, explicit examples of p-adic numbers with irrationality exponent equal to µ were not known for µ < (3+ p 5)/2 (for larger values of µ, we can use the argument described in [8] in the real setting). Of course, it is clear that P + pbµi c is a good candidate for having irrationality exponent µ, but this was proved only for µ > (3 + p 5)/2. Our strategy is the following. We work in the field of power series and consider the continued fraction expansion of P + X ci and its convergents. Replacing X by p we deduce good and very good rational approximations to the p-adic number c. We are then in a position to apply Lemma 2 below to complete the proof of Theorem. Note that we do not employ as technical terms the attributes very good and good for approximations. They are used for sequences of rational approximations which give either the best possible lower bound on the irrationality exponent or a lower bound greater than two. We believe that this usage helps with the intuitive understanding of the exposition. In the following section, we establish an auxiliary lemma that will be useful for bounding from above the irrationality exponent of a (p-adic) number. In the last section we introduce continued fractions in the field of rational power series and use them to prove Theorem. We keep the notation of Theorem throughout this paper. 2. Bounding the Irrationality Exponent From Above Using the infinite family of rational approximations kx p ci, k > 0, to c, we easily get that µ( c ) > c. Let us give a detailed proof. Since lim sup i!+ c i+ /c i = c and lim i!+ c i = +, we have lim sup i!+ c i+ c i + = c.
INTEGERS: 8A (208) 4 Let > 0 and choose a subsequence (c ij ) j>0 such that i 0 >, c i0 > 2c i0, and c ij+ > c, for j > 0. c ij + Then we have Therefore, X ij H p c` = `=0 i j X `=0 p c` < p ci j +, for j > 0. c i X j p c` `=0 = p X ij ci j + < H p `=0 c i j + p c` c ij + < H X ij c` c+. p `=0 It follows that µ( c ) > c and since > 0 was arbitrary, we conclude that µ( c ) > c. This proves Theorem when c is infinite. From now on, we assume that c is finite. Establishing that µ( c ) 6 c is more delicate. The basic idea is the existence of many very good rational approximations to c, which impedes the existence of even better rational approximations. As auxiliary tools, we need the following two elementary lemmas, whose proofs rest on triangle inequalities. Lemma. Let be in Q p. Let #, 2 (, +) be real numbers and P, Q, m, n integers such that Q and n are nonzero, m/n and P/Q are reduced and distinct. If P/Q p 6 H(P/Q) # and m/n p 6 H(m/n), then H(m/n) 6 2 H(P/Q) or H(m/n) > 2 H(P/Q)#. () Proof. Using the fact that P n Qm 6 2 H(P/Q) H(m/n) and the non-archimedean nature of the p-adic absolute value, we get 2 H(P/Q) H(m/n) 6 P n Qm Qn = P p Q m n p 6 max { P/Q p, m/n p } 6 max{h(p/q) #, H(m/n) }, thus If we have /2 6 max{h(p/q) # H(m/n), H(P/Q) H(m/n) } H(P/Q) # H(m/n) 6 H(P/Q) H(m/n), i.e., H(m/n) 6 H(P/Q) #/,
INTEGERS: 8A (208) 5 then If, on the other hand, we have H(m/n) 6 2 H(P/Q). H(P/Q) # H(m/n) > H(P/Q) H(m/n), i.e., H(m/n) > H(P/Q) #/, then H(m/n) > 2 H(P/Q)#. Therefore, () always holds. Lemma 2. For 2 Q p, let (# k ) k>0 be a sequence of real numbers such that lim inf k!+ # k > and let (P k /Q k ) k>0 be a sequence of distinct rational numbers such that P k Q = H(P k/q k ) # k k p holds for k > 0. Let be a real number with > + lim sup k!+ log H(P k+ /Q k+ ) (# k ) log H(P k /Q k ). (2) Then there exist only finitely many rational numbers m/n outside the sequence (P k /Q k ) k>0 which satisfy In particular, we have n µ( ) 6 max lim sup k!+ m n p 6 H(m/n). (3) # k, + lim sup k!+ log H(P k+ /Q k+ ) o. (4) (# k ) log H(P k /Q k ) Proof. Excluding, if necessary, at most finitely many elements from the sequence (P k /Q k ) k>0, we can assume without loss of generality that for some > 0 and all k > 0, we have # k > +. Suppose that (2) is fulfilled and that there are infinitely many rational numbers m/n not in (P k /Q k ) k>0 such that (3) holds. For such a rational number m/n with H(m/n) large enough, let k 0 be the largest positive integer such that 2 H(Pk0 /Q k0 ) < H(m/n). Then Lemma implies that which gives 2 H(P k 0 /Q k0 ) # k 0 6 H(m/n) 6 2 H(Pk0+/Q k0+), 6 log 2 + log H(P k 0+/Q k0+). (# k0 ) log H(P k0 /Q k0 )
INTEGERS: 8A (208) 6 From our assumption and the fact that (H(P k /Q k )) k>0 is unbounded, the last inequality would have to hold for infinitely many positive integers k 0. This is in contradiction with (2). We stress that exact analogues of Lemmas and 2 hold in the real setting. Example. Let c > 2 be a real number. Lemma 2 can be applied to c := P + pbcic and the p-adic numbers k = P k pbcic, k > 2, to show that µ( c ) 6 c when c k+ c > + lim sup k!+ (c )c k = + c c. This implies that µ( c ) = c when c 2 3c + > 0, that is, for c > (3 + p 5)/2 = 2.68... However, besides ( k ) k>0, we can also take into consideration the following very good rational approximations to c defined for k > 2 by X p bcic + p bckc + p bck+ c bc kc + p 2(bck+ c bc k c) + k k = kx = p bcic p bck c + p. bck+ c bc k c It is not di cult to show that H( k ) = k p ck, H( k ) p ck+ c k +c k, c k p p ck+ H( k ) c, c k p p 2ck+ +c k H( k ) (2c2 c)/(c 2 c+), H( k ) < H( k ) < H( k+ ) for k large enough. Here, all the constants implicit in can be made explicit and depend only on p and c. If we now apply Lemma 2 with the union of the two sequences ( k ) k>2 and ( k ) k>2, we get after some calculation that µ( c ) 6 c for c > 2.325. Nevertheless, in order to prove that µ( c ) = c for all c > 2, we require many more good rational approximations to c. To find them, looking for repetitions and completing by periodicity is not su cient. We need an additional idea. 3. Proof of Theorem In order to construct very good rational approximations to our p-adic number, we first need to work with continued fractions in the field of power series. The field of rational power series Q((X )) is the completion of Q(X) with respect to the
INTEGERS: 8A (208) 7 non-archimedean absolute value k k defined on Q(X) by kxk = e. Instead of e, we can fix any real number greater than. If ˆ 2 Q((X )) and ˆ 6= 0, then we can write ˆ = P + k=k 0 a k X k, where k 0 2 Z, a k 2 Q and a k0 6= 0. More information on this subject can be found in [] and [7]. An element of Q(X) can be expressed as a finite continued fraction [S 0, S,..., S n ] := S 0 + S +... + S n where each partial quotient S i is a rational polynomial. This expression is unique if we require that the degree deg S i is positive for i > 0. Beside some basic facts from the theory of continued fractions (see e.g. [4]), our main tool is the so-called Folding Lemma; see [9, 2, 3, 0] and Chapter 6 of [3]. Throughout this paper, for a word! w = a, a 2,..., a h, we denote by w the word a h, a h,..., a and by w the word a h, a h,..., a. Lemma 3 (Folding Lemma). Let r, s, t be in Q[X] with t non-constant. Write r/s = [a 0, a,..., a h ], where a i 2 Q[X] for 0 6 i 6 h and denote by! w the word a,..., a h. Then, we have r s + ( )h ts 2 = [a 0,! w, t, w ] = [a 0,! w, t,,, 0, w ]. When using the Folding Lemma, we will usually immediately make the substitution [..., a, 0, b,...] = [..., a + b,...], which can be easily checked. The irrationality exponent of an irrational number does not change when multiplying it or adding to it any nonzero rational number. Thus if necessary, we can omit finitely many terms in c and multiply this number with p, so that c i+ > 2c i holds for all i > 0. This results in no loss of generality and will be assumed during the proof of Theorem. Starting from X c0 = [0, X c0, ] and successively applying Lemma 3 first with t = X c 2c0 and then with t = X c2 2c, we obtain and X c0 + X c = X c0 + ( ) 2 (X c 2c0 X 2c0 ) = [0, X c0,, X c 2c0, X c0 ] X c0 + X c + X c2 = X c0 + X c + ( ) 4 (X c2 2c X 2c ) = [0, X c0,, X c 2c0, X c0, X c2 2c,,, X c0, X c 2c0,, X c0 ].
INTEGERS: 8A (208) 8 Continuing this process indefinitely with t = X c k+ 2c k for k >, we arrive at the continued fraction expansion ˆ c = +X X ci = [a 0, a, a 2,...] = [0, X c0,, X c 2c0, X c0, X c2 2c,, X c0, X c 2c0,, X c0, X c3 2c2,, X c0 2,...] 2 Q((X )). We need to justify the convergence of the continued fraction expansion in (5) since there are partial quotients of degree 0, namely those which are equal to. The following facts are established by an easy induction and, therefore, we give only an outline of their proofs. Proposition. Let ˆ c 2 Q((X )) be the sum of the infinite series in (5). Then the following statements hold. (i) No two consecutive elements in the sequence (a i ) i>0 are equal to. (ii) For i > 0, if a i 6=, then a i is a monic integer polynomial and deg a i >. (iii) If i 3 or 8 (mod 2), then a i = X c 2c0. If i 3 2 k or 9 2 k (mod 3 2 k+ ) for some k > 2, then a i = X c k 2c k. If i does not satisfy any of these congruences, then a i 2 {, X c0, X c0, X c0 2}. (iv) Set P n /Q n = [a 0, a,..., a n ], where P n, Q n 2 Z[X] for n > 0. Then, for any n >, either deg Q n < deg Q n or deg Q n = deg Q n < deg Q n+. The polynomials P n (for n > ) and Q n are monic, thus Q n (/p) 6= 0. (5) (v) deg P n = deg Q n = nx deg a i for n > 2, i=2 nx deg a i for n >, i= deg Q n deg P n = c 0 > 0.
INTEGERS: 8A (208) 9 (vi) P 3 2k 2 = Q 3 2k 2 kx X ci, Q 3 2k 2 = X c k, for k >. (6) (vii) The sequence (P n /Q n ) n>0 converges to ˆ c in Q((X )). Sketch of the proof. Statements (i) and (ii) are easy and (iii) requires only careful attention to the indices when observing the repeated application of the Folding Lemma. The claim in (iv) follows from (i) and (ii) using the recurrence relation satisfied by P n and Q n, namely P 0 = 0, P =, P n = a n P n + P n 2, Q 0 =, Q = X c0, Q n = a n Q n + Q n 2, for n > 2 (7) The same identities are used to prove (v), while (vi) follows immediately from the Folding Lemma. The well-known identity P n+ Q n P n Q n+ = ( ) n, n > 0, (8) follows inductively from (7). Applying (8) for n consecutive indices gives P n+ = + + ( ) n. (9) Q n+ Q 0 Q Q Q 2 Q n Q n+ From (iv), we have kq i Q i k < kq i Q i+ k, so the infinite series +X ( ) i (Q i Q i+ ) converges in Q((X )) and the partial sums of this series are convergents of the continued fraction in (5). But (vi) now shows that lim n!+ P n /Q n = ˆ c so that (vii) is proved as well. Note that we have also proved (cf. (v)) that c P n = kq n Q n+ k = e 2 P n i= deg ai deg an+. (0) Q n We prove an auxiliary result on the growth of denominators of partial quotients. Recall that c i+ c = lim sup. i!+ c i
INTEGERS: 8A (208) 0 Proposition 2. The polynomials Q n defined in Proposition satisfy log n lim = 0, n!+ deg Q n () deg Q n+ deg Q n+ lim inf =, lim sup = c n!+ deg Q n n!+ deg Q n. (2) Proof. From (iv) in Proposition, we deduce that deg Q n > deg Q n 2 + for n > 2. Iterating this gives deg Q n > bn/2c for n > and () holds. The first equality in (2) is obvious if we take into account (v) from Proposition and the fact that deg a i 2 {0, } for infinitely many i. If n = 3 2 k 2 for some k >, then deg Q n+ deg Q n = c k+ c k c k = c k+ c k. (3) For an integer n > 5 not of this form, let k be the positive integer such that 3 2 k 6 n < 3 2 k+ 2. Then, by (vi), (v) and (iii) of Proposition, we get deg Q n > deg Q 3 2k = c k+ c k > c k and deg Q n+ 6 deg Q n + d, where Hence, d 6 max{c i 2c i : 0 6 i 6 k}, (c := 0). deg Q n+ 6 + d 6 + c i/c i 2 6 + c i/c i 2 6 c i, (4) deg Q n c k c k /c i 2 2c i for some i 2 {,..., k}. Since c > 2, we get c/2 6 c (3) and (4) implies the second equality in (2). and the combination of The following proposition uses the symmetric structure of the continued fraction expansion of ˆ c (obtained by means of repeated applications of the Folding Lemma) in order to express the convergents of this continued fraction expansion in terms of certain convergents with smaller indices. Proposition 3. Let n and k be positive integers such that 3 2 k 6 n < 3 2 k+ 2. Set m = 3 2 k 2 and l = 3 2 k+ 3 n. Then, P n Q n = Xck+ 2ck P m (P m Q l Q m P l ) P l X c k+ 2c kqm (P m Q l Q m P l ) Q l. (5)
INTEGERS: 8A (208) Proof. Since 0 6 l 6 3 2 k 3, we can write for l < 3 2 k 3, where This gives Setting P m Q m = (U l/v l )P l+ + P l (U l /V l )Q l+ + Q l, U l V l = [w l ], w l = a l+2,..., a m. U l V l = we deduce from the Folding Lemma that We also have Q l P m P l Q m. (6) Q l+ P m + P l+ Q m S l T l = [0, w l, X c k+ 2c k, w l ], S l T l = V l U l + ( ) l X c k+ 2c ku 2 l = Xck+ 2ck U l V l + ( ) l X c k+ 2c ku 2 l. (7) P n Q n = (T l/s l )P l+ + P l (T l /S l )Q l+ + Q l. (8) Note that the fractions that appear in (6) and (7) are reduced. Now we substitute (6) into (7) and (7) in (8) and use the fact that P l+ Q l P l Q l+ = ( ) l. Simplifying the expression, we obtain (5). Again, the fractions in (5) are reduced. If l = 3 2 k 3 = m, then S l /T l = [0, X c k+ 2c k ] and (8) gives (5) directly. Now we make the formal substitution X 7! /p and define the quantities c = +X p ci 2 Q p, A n = p deg Qn P n (/p), B n = p deg Qn Q n (/p), n > 0. The next proposition summarizes properties of c and the sequences (A n ) n>0 and (B n ) n>0. Proposition 4. For the p-adic number c and the sequences (A n ) n>0 and (B n ) n>0 defined above, the following hold: (i) The numbers A n and B n are integers satisfying p c0 A n, p c0+ - A n, B n (mod p).
INTEGERS: 8A (208) 2 (ii) For every n > 0, the integers A n and B n are relatively prime. The sequence (A n /B n ) n>0 converges in Q p to c and c A n B = p deg(qnqn+) = p 2 P n i= deg ai deg an+. (9) n p (iii) The height of A n /B n satisfies 2n 2 pdeg Qn 6 H(A n /B n ) 6 n 2 p deg Qn, n >. Proof. (i) This follows directly from (iv) and (v) of Proposition. (ii) From (8), we get for n > 0 that A n+ B n A n B n+ = ( ) n p deg Qn+deg Qn+, (20) which implies that the sequence (A n /B n ) n>0 is a Cauchy sequence and thus converges in Q p. From A 3 2k 2 c = B 3 2k 2 +X i=k+ p ci, we deduce that lim n!+ A n /B n = c. Now (9) follows from A n+ B n+ = pdeg(q0q) B 0 B p deg(qq2) B B 2 + + ( ) n pdeg(qnqn+) B n B n+, similarly as (0) was obtained from (9). The identity (20) also shows that A n and B n are always relatively prime since any common divisor would be a power of p and (i) asserts that B n is never divisible by p. Note that (9) implies that no two elements of the sequence (A n /B n ) n>0 are identical, in view of (iv) of Proposition. (iii) We have seen that A n and B n are always relatively prime. From the definition of these numbers, it is clearly su cient to prove that max{ P n (/p), Q n (/p) } belongs to the interval [/(2n 2 ), n 2 ]. We first show that P n (/p) 6 n 2. More precisely, we prove by induction on k > 0 that for 3 2 k 2 6 n < 3 2 k+ 2 the inequality P n (/p) 6 2 2k holds. This can be easily checked for k = 0. Suppose the statement holds for all k < K, where K >, and let n with 3 2 K 2 6 n < 3 2 K+ 2.
INTEGERS: 8A (208) 3 If n = 3 2 K 2, we obtain a much stronger result directly from (6) P n (/p) < p p 6 2, Q n(/p) = /p c K 6 /2. For n = 3 2 K, we have a n (/p) = p 2c K c K+, so P n (/p) 6 a n (/p)p n (/p) + P n 2 (/p) 6 2 + 2 2K 2 6 2 2K. Finally, if 3 2 K 6 n < 3 2 K+ 2, we use Proposition 3 which gives P n (/p) < p 2c K c K+ 2 2 2 2K 2 + 2 2 22K + 2 2K 2 < 2 2K. Analogously, we prove that Q n (/p) 6 n 2. Furthermore, (8) shows that thus max{ P n (/p)q n (/p), P n (/p)q n (/p) } > /2, max{ P n (/p), Q n (/p) } > 2 max{ P n (/p), Q n (/p) } > 2n 2. and We are now armed to complete the proof of Theorem. Since log c A n /B n p log H(A n /B n ) log c A n /B n p log H(A n /B n ) Proposition 2 gives lim sup n!+ log c A n /B n p log H(A n /B n ) 6 (deg Q n + deg Q n+ ) log p 3 log n + deg Q n log p > (deg Q n + deg Q n+ ) log p 2 log n + deg Q n log p = = + deg Qn+ deg Q n log n deg Q n 3 log p +, deg Qn+ + deg Q n log n 2 deg Q n log p +, log c A n /B n p = c and lim inf = 2. n!+ log H(A n /B n ) In order to apply Lemma 2 to c and (A n /B n ) n>0, we need to evaluate the right-hand side of (2). Using the results of Proposition 4, we get lim sup n!+ log H(A n+ /B n+ ) log c A n /B n p log H(A n /B n ) 6 lim sup n!+ 6 lim sup n!+ 2 log(n + ) + deg Q n+ log p (deg Q n + deg Q n+ ) log p 2 log n deg Q n log p 2 log(n+) deg Q n+ + log p 2 log n deg Q n+ + log p =,
INTEGERS: 8A (208) 4 where we used () in the last step. Bound (4) from Lemma 2 finally implies that µ( c ) 6 c and thus µ( c ) = c. This finishes the proof of Theorem. As a final remark, we show that not only the best approximations to c, but also the second best ones, can be written explicitly as series. The best approximations P 3 2k 2/Q 3 2k 2 to the associated power series ˆ c are obtained by cutting o its continued fraction expansion before the first appearance of large partial quotients, or alternatively, by deleting the terms that follow X c k = X bckc in P + X ci. These approximations of ˆ c in Q((X )) correspond to the approximations k of c 2 Q p defined in Example. The next best sequence of approximations is (P 9 2 k /Q 9 2 k ) k>2 and, as the following lemma shows, it actually corresponds to the sequence ( k ) k>2 of p-adic numbers introduced in Example. Proposition 5. For k > 2, we have k P 9 2 k X = Q 9 2 k X ci + X c k X c k c k+. (2) Proof. This is an immediate consequence of Proposition 3 where n = 9 2 k, so that m = 3 2 k 2 and l = 3 2 k 2. Since P m, Q m, P l, Q l are now explicitly given by (6), inserting their values into (5) gives (2). Acknowledgements: T.P. was supported by the Croatian Science Foundation under the project no. 6422. References [] B. Adamczewski and Y. Bugeaud, On the complexity of algebraic numbers. I. Expansions in integer bases, Ann. of Math. 65 (2007) 547 565. [2] V. I. Bernik and M. M. Dodson, Metric Diophantine Approximation on Manifolds, Cambridge Tracts in Mathematics, Cambridge University Press, Cambridge, 999. [3] J. Borwein, A. van der Poorten, J. Shallit, and W. Zudilin, Neverending Fractions: An Introduction to Continued Fractions, Cambridge University Press, 204. [4] Y. Bugeaud, Approximation by Algebraic Numbers, Cambridge Tracts in Mathematics, Cambridge University Press, Cambridge, 2004. [5] Y. Bugeaud, Diophantine approximation and Cantor sets, Math. Ann. 34 (2008), 677 684. [6] Y. Bugeaud, D. Krieger, and J. Shallit, Morphic and automatic words: maximal blocks and Diophantine approximation, Acta Arith. 49 (20), 8 99. [7] A. Lasjaunias, A survey of Diophantine approximation in fields of power series, Monatsh. Math. 30 (2000), 2 229.
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