On the degrees of irreducible factors of higher order Bernoulli polynomials


 Lester Melton
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1 ACTA ARITHMETICA LXII.4 (1992 On he degrees of irreducible facors of higher order Bernoulli polynomials by Arnold Adelberg (Grinnell, Ia. 1. Inroducion. In his paper, we generalize he curren resuls on he peisensein behavior of firs and higher order Bernoulli polynomials [4], [6 9], using he machinery of [1]. In so doing, we provide a broader framework for he known resuls, all of which are eiher immediae consequences or special cases of our more general resuls. Because of an explici formula for he coefficiens in erms of falling facorials esablished in [1], he polynomials A n (x, which we consider here are acually ranslaes of he sandard higher order Bernoulli polynomials B n (ω (x, bu all of our significan resuls apply equally well o he sandard polynomials and heir ordinary coefficiens, wih ω = n k + 1. The main resuls are summarized using sandard noaions in he research announcemen [2]. Our approach differs from he usual one in ha we make no use of congruence properies of he Bernoulli numbers, and in paricular do no use he von Saud Clausen Theorem, which is an essenial ingredien of he usual approach. Insead we characerize he behavior of he padic poles of he coefficiens in erms of he base p expansion of n. The peisensein siuaion occurs when he highes order pole is simple. The Bernoulli polynomials B n (x are defined by (cf. [11, 12] e x e 1 = n=0 B n (x n n!. The arbirary order Bernoulli polynomials B n (ω ω e x = ( e 1 n=0 B n (ω (x n n!, (x are defined by which are called higher order, or more precisely firs or higher order, if ω {1, 2,..., n}. They are monic degree n raional polynomials. In [4], Carliz showed ha if n = k(p r where 0 < k < p, hen B n (x
2 330 A. Adelberg is irreducible (always wih respec o Q unless oherwise indicaed, namely ha pb n (x is peisensein. He also showed ha if 2m + 1 = k( + 1 where p is an odd prime and 1 k p, hen B 2m+1 /x(x 1 2 (x 1 has an irreducible facor of degree 2m + 1 p. Le S(n = S(n, p be he padic digi sum of n, i.e., if n = n i p i is he base p expansion, where all digis saisfy 0 n i, hen S(n = n i. McCarhy showed in [7] ha pb 2m (x is peisensein iff S(2m =. In [6], Kimura defined a funcion N(n, p, deermined by he base p expansion of n, which he used o srenghen he resuls of Carliz and McCarhy; namely, he showed ha if S(n, p, hen B n (x has an irreducible facor of degree N(n, p. We will generalize Kimura s and McCarhy s resuls and deermine all insances of even parial peisensein behavior of he firs and higher order Bernoulli polynomials. Our main ool is an analysis of he naure of he padic poles of he coefficiens, which urns ou o be remarkably regular and uniform. In he noaion of [1], he polynomials we acually consider are A n (x, s = n!a n (x, 0, s, wih exponenial generaing funcion ( s ln(1 + (1 + x = n=0 A n (x, s n n!. These are he Narumi polynomials [12, p. 127]. I has been shown ha A n (x, s = B n (n+s+1 (x + 1, so in paricular, A n (x, n = B n (x + 1 for firs order, wih s { 1,..., n} for ω {1,..., n}. To esablish he boundaries of our sudy, i should be noed ha A n (x, 0 = (x n and A n (x, (n + 1 = (1 + x n. We will show ha he Kimura bound [6, Theorems 1, 2], alhough sharp for s = n, is far from bes for he full range s { 1,..., n}. Finally, i has been shown ha if n is odd hen 2x (n + s 1 A n (x, s and i was proved in [1] ha A n (x, s is absoluely irreducible for n even > 0, and ha he above linear facor is he only nonrivial facor for n odd. The resuls in his paper enable us o deermine he insances of peisensein behavior leading o raional irreducibiliy of his ype. 2. Preliminaries. Throughou his paper, p will be a fixed bu arbirary prime, and all numbers are assumed o be posiive inegers, unless oherwise indicaed. ν p = padic valuaion of Q, i.e. ν p (m = highes power of p m if m Z and ν p (m/n = ν p (m ν p (n, wih ν p (0 =. If b > 0, we say r has a pole of order b if ν p (r = b.
3 Higher order Bernoulli polynomials 331 We say ha f(x = n i=0 a ix n i Q[x] is peisensein down o degree d if ( ν p (a 0 = 1, ν p (a i > 0 if 0 < i < n d, ν p (a n d = 0, and ν p (a i 0 for i > n d. We refer o his as (parial peisensein behavior; f(x is said o be peisensein if i is peisensein o degree 0. The following are sandard and/or easy o demonsrae (cf. [10, Ch. 4]. (1 If f(x is peisensein down o degree d and c Z, hen so is f(x+c. (2 If f(x is peisensein down o degree d, hen f(x has an irreducible facor of degree n d. (In paricular, peisensein implies irreducible. (3 If f(x = n i=0 b i(x n i in erms of he falling facorials, he coefficiens (a i of f(x saisfy ( iff he coefficiens (b i saisfy (. We also need some sandard resuls abou ν p of facorials and binomial coefficiens, where [x] = ineger par of x and x = leas ineger x. [ ] n (a ν p (n! = p e = n S(n n, whence ν p (n! 1 (b e=1 ( n + m S(n + S(m S(n + m ν p =. n if n > 0. (In paricular, S(n + m S(n + S(m. ( n + m (c ν p = # of carries in base p addiion of n and m. n ( n (d p iff n i m i for all digis of he padic expansions m (Lucas s Theorem. I is convenien o inroduce some nonsandard noaions, namely ( ( n m + δ m n if p and m δ if m m + δ, i.e. p. m m By (d, is a parial order, and clearly m n implies m n. Obviously if m and m have nonzero digis in differen places, hen (m + m n iff m n and m n. (Observe ha n > m iff n i > m i for he larges i such ha n i m i. Also by (c, m δ iff S(m + δ = S(m + S(δ. Finally, since ( ( k + m 1 = ( 1 m, m m
4 332 A. Adelberg we have ( p iff k 1 m. m The following lemma can be proved wihou oo much difficuly. Lemma 2.1. Le n > 0 and suppose ha n. Le q = n/( = s i=r q ip i, wih q r 0. The following are equivalen: ( n (i p. q (ii q r > q r+1 q r+2... q s. (iii S(n =. P r o o f. n 1 + q = pq 1 = s i=r+1 q i p i+1 + (q r 1p r+1 + ( r p i. The equivalence of (i and (ii follows immediaely by Lucas s Theorem. n = (q = s+1 i=r+2 i=0 (q i 1 q i p i + (q r q r+1 1p r+1 + (p q r p r. I follows ha if (ii holds, hen S(n =, by elescoping he digi sum. Suppose on he oher hand ha (ii fails. If q r q r+1, hen S(n p + q r q r p q r = p q r+1 + >, while if q r > q r+1 and f is he lowes place where q f 1 < q f, he same argumen shows S(n p q f + >. R e m a r k 1. The preceding lemma shows ha if S(n = hen n/( has nonzero digis in all places below he op digi of n, down o he boom nonzero digi of n. (I has a nonzero digi in he place of he op digi iff n = (p r. We now urn o Kimura s funcion N(n, p = N(n which we will use only when S(n. In his case, N(n = he smalles such ha n and S(. This is equivalen o N(n = smalles > 0 such ha n and, which is essenially Kimura s definiion [6, Lemma 4]. Thus if n = m i=0 n ip i is he base p expansion and r is such ha r i=0 n i while r+1 i=0 n i >, hen r N(n = i=0 ( n i p i + so ha S(N(n =. The nex lemma follows easily from (c. r n i p r+1, Lemma 2.2. Suppose ha δ and δ δ/(. Then δ = 0. i=0
5 Higher order Bernoulli polynomials 333 P r o o f. Observe ha δ + δ/( = δp/(. Hence if δ 0 and f is he lowes place where he digi q f 0, hen here is a carry in place f for he base p sum. We can deduce he following lemma, characerizing N(n. Lemma 2.3. Suppose N(n n and. Then p = N(n. P r o o f. If δ = N(n, hen n 1 + = ( N(n 1 + N(n ( + δ + δ. ( n Since here is no carry in he sum of he op digi of N(n and he boom digi of δ, and oherwise he wo summands have nonzero digis in differen places, his lemma follows from he previous wo. 3. padic analysis of he higher order Bernoulli polynomials. Recall ha from [1, 3.1ii and 3.2ii], n A n (x, s = b i (s(x n i, where b i (s = w(u=i i=0 ( ( s d(u 1 (n i d(u u 1 u u 1 3 u 2,... where he sum is over all sequences (u = (u 1, u 2,... = (r 1 r 2, r 2 r 3,... of nonnegaive inegers, evenually zero, wih weigh w(u = ju j = i and arbirary degree d(u = u j. I is convenien o abbreviae he erm ( ( s d(u 1 (s d(u τ u = (n w(u d(u u 1 u u 1 3 u 2 = (n w(u... u! u, and o ake w(u as he weigh of τ u. We say ha (u or τ u is concenraed in place if u j = 0 for all j. Fix k {1,..., n} and le s =. Clearly if w(u <, hen ν p (τ u 0 since p firs occurs in he denominaor when j =, whereas if w(u = i and (u is concenraed in place, hen i and ( ( ν p (τ u = ν p (n i i i ( n! = ν p i ( (n i! + ν p i = S(n i S(n + ν p i. iff
6 334 A. Adelberg The following lemma shows ha he erms concenraed in place deermine he pole srucure. Lemma 3.1. If (u is no concenraed in place, hen τ u dominaes a erm of lower weigh, i.e. here exiss (u wih w(u < w(u and ν p (τ u ν p (τ u, whence τ u dominaes a erm of lower weigh concenraed in place. P r o o f. C a s e 1. Suppose here exiss i < p 1 wih u i > 0. Consruc (u ui by replacing u i by 0 and increasing u by 1. This clearly ui decreases degree and weigh, and since ν p (u i! 1, we ge he desired dominaion. C a s e 2. Suppose here exiss i > wih u i > 0. Consruc (u by replacing u i by 0 and increasing u by u i. This clearly preserves degree and decreases weigh. Since ν p ((u + u i! ν p (u! + ν p (u i!, he dominaion is obvious unless i = kp α 1, α > 1, where (k, p = 1. In his case, Hence and w(u w(u = u i (kp α p u i (α 1p. ν p ( u = ν p ( u + u i (α 1 ν p ((n w(u ν p ((n w(u + u i (α 1, since for any m, ν p ((m rp r. This again esablishes he dominaion. Now, if (u is no concenraed in place, repea he argumen wih (u replaced by (u. Since he weighs are decreasing, evenually we ge a erm concenraed in place dominaed by τ u. Le a {0, 1, 2,...}. We hen deduce immediaely from he preceding lemma and remarks, using sandard properies of valuaions of sums, Corollary 3.2. If is he smalles i such ha ν p (b i < a, hen (i, and if τ u is concenraed in place and has weigh, hen ν p (τ u < a, (ii ν p (τ u a for all (u wih w(u excep for he erm τ u in (i, S(n S(n (iii a if i < and i, + ν p < a and S(n i S(n + ν p i
7 (iv if i <, i, p Higher order Bernoulli polynomials 335 ( n and S(i ( > a, hen p i. i The las condiion follows from he second par of he previous one. I is no hard o furher describe and indeed characerize by a heorem which implies ha he firs pole is simple, he second higher pole has order 2, ec., and which enables us o deermine where hey occur from he base p expansion of n. All subsequen resuls of his paper follow fairly easily from his heorem. Le a {0, 1, 2,...} as above. We say ha l is a segmen of n, saring in place r, if n = m i=0 n ip i and l = s i=r n ip i. (This includes l = 0 if s < r. If r = 0, we call l a boom segmen. Theorem 3.3. If is he smalles i such ha ν p (b i < a, hen ( (i p, ( n (ii p, S(n S(n (iii = S( = 1 a, (iv ν p (b = a 1, (v he only digis of n ha and n can share are op digis of segmens of divisible by, (vi is he smalles i such ha ( n i, p, i S(i > a and p ( i. P r o o f. Assume ha S( = and ha /( /(, e.g. = N(. Le 1 =, so 0 1 <, 1, and 1 /( /(, whence by (iii of he preceding corollary, a S(n 1 S(n = S(n 1 S(n + ν p 1 S( S(n S(n S(n S(n S(n S(n + ( + ν p 1 + ν p a. ( + ν p 1
8 336 A. Adelberg Thus all he inequaliies are acually equaliies, so ( S(n S(n ν p (b = + ν p = a 1, ( ( ν p 1 = ν p, and S(n 1 = S( + S(n, i.e. n. ( I follows ha ν p = 0, since if no, and r is he lowes place where ( here is a carry base p for k 1 + /(, i.e. (k 1 r + p, ( r and if = p r (, hen 1 /( and /( agree in all r ( 1 places excep place r and = 0, so ν p 1 conradicing he( above equaion. Nex assume r ( < ν p, 0, and le = (p r. Then n, r whence (n r = 0. Therefore n has nonzero digis only in he places where /( is zero. Now, le 0 = N(. Then n 0 and n has zero digis in all he places of 0 excep possibly he op place. Nex consider 1 = N( 0. The same argumen shows n 1, and if 1 and 0 share a digi of n in place r, hen (n r = 0. Coninue he process. Since, is he sum of hese i, so n, and he heorem follows. We can deduce easily from he preceding heorem and Lemmas 2.2 and 2.3 Corollary 3.4. Le l be a boom segmen of n wih S(n l. Le k = n l. Then = N(n l is he smalles i such ha ν p (b i < 0, ν p (b = 1, and ν p (b i 1 for all i. P r o o f. Observe ha = = N(n l, following he noaion of he previous proof. If l < l is any smaller segmen of n, hen k 1 has all digis in he places of l, so p N(n l. Since p N(n l ( and p N(n l l 1 N(n l
9 Higher order Bernoulli polynomials 337 if l 1 is a boom segmen of n l N(n l by he lemmas, he resul follows immediaely. R e m a r k 2. Le j be he smalles i such ha ν p (b i < j for j = 0, 1,..., a. Then ν p (b i = j 1 if i = j by (iv, so here are no gaps in he orders of he poles. We can use (v o consruc as follows: here is a (unique sequence l 0, l 1,..., l a of segmens of n such ha l j+1 is a boom segmen of n j j λ=0 l λ and ( j+1 j+1 j = N n j l λ = N j+1. Thus he higher order poles come in he naural order. In paricular, 0 = N(n l 0 = N 0 gives he firs pole (simple, where l 0 is a boom segmen of n, and 1 = 0 + N(n 0 l 0 l 1 gives he firs higher pole (double, where l 1 is a boom segmen of n 0 l 0. Hence j = j λ=0 N λ, where S(N λ =, and he N λ give he degree differences for he successively higher order poles. The sequences ( j and (l j can be deermined recursively ogeher from n and k as follows: l j+1 is he smalles segmen l of n j saring in he place of he highes digi of j such ha p ( λ=0 N(n j j λ=0 l λ l and j+1 j = N(n j j+1 λ=0 l λ, wih iniial condiion 1 = 0. [ I follows ] ha k = 1 gives he bigges possible pole, which has order S(n, wih all l j = 0. In paricular, here are no poles if S(n <. I is no hard o show ha we can urn around he preceding consrucion, namely if is consruced as in he preceding paragraph, hen here exiss k such ha j = j λ=0 N λ is he firs i such ha ν p (b i < j for j = 0, 1,..., a, and ν p (b i a 1 for all i. Thus exhibis he complee singulariy paern for such k. I can be seen ha one way o choose k is o sar wih n l 0, subrac all he N λ for λ > 0 and also he highes place conribuion of N 0 if N 0 and N 1 share a digi of n. The case p = 2 is simples o describe since S(N = iff N is a 2power, and here is no sharing of digis. In his case, N 0, N 1,..., N a are jus successively higher powers of 2 ha occur in he base 2 expansion of n, and k = n l 0 a λ=1 N λ = n l 0 ( a 0. Now we urn o he peisensein behavior of hese polynomials. When we say ha A n (x, is peisensein down o degree d, we acually refer o pa n (x,. The condiion ha we need is a simple highes order pole, which by he above analysis is equivalen o having a (simple pole, bu no,
10 338 A. Adelberg pole of order wo, i.e. ν p (b 0 = 0, ν p (b i 0 for i < n d, ν p (b n d = 1, ν p (b i 1 all i. If S(n <, hen (S(n i S(n/( > 1 for all i, so as previously noed here is no peisensein behavior. The nex heorem, which esablishes all insances of peisensein behavior follows immediaely from he preceding discussion. Theorem 3.5. If A n (x, is peisensein down o degree d, hen d = n N(n l for some boom segmen l of he padic expansion of n wih S(n l. Conversely, if d = n N(n l, hen A n (x, (n l is peisensein down o degree d; furhermore, A n (x, is peisensein down o degree d iff he following hree condiions hold: N(n l (i p (ii p ( (iii p N(n l for all boom segmens l of n such ha l < l,, and N(n l l 1 N(n l if l 1 is any boom segmen of n l N(n l. (This condiion assumes S(n l S(l 1 (, so i is vacuous if S(n l < 2(. The case p = 2 of his heorem is paricularly easy o sae: If A n (x, is 2Eisensein down o degree d, hen d = n 2 f for some power 2 f in he base 2 expansion ( of n. Furhermore, ( A n (x, is 2Eisensein down o degree n 2 f iff 2 and 2 for all oher powers 2 g in he base 2 2 f 2 g expansion of n iff he base 2 expansion of k 1 conains all he 2 g bu does no conain 2 f iff n 2 f k 1 iff k 1 = n 2 f + m, where m is any sum of lower powers of 2 no in he base 2 expansion of n. (Observe ha k = n l if m is all hese lower powers, where l is he boom segmen of n below 2 f. The following corollaries are special cases. Take l = 0 o ge he firs hree corollaries, and l = n 0 o ge he las hree. Corollary 3.6 (Kimura. A n (x, n is peisensein down o degree n N(n. R e m a r k 3. I follows immediaely ha if S(n ( < 2(p 1, hen A n (x, is peisensein down o degree n N(n iff p N(n. Since N(n = n iff S(n =, for our resriced use of N(n, we ge Corollary 3.7 (McCarhy. A n (x, n is peisensein iff S(n =.
11 p We can sharpen his as follows: Higher order Bernoulli polynomials 339 Corollary 3.8. The following are equivalen: (i S(n =. (ii There exiss k {1,..., n} such ha A n (x, is peisensein. (iii A n (x, n is peisensein. (iv A n (x, 1 is peisensein. ( Furhermore, if hese condiions hold, A n (x, is peisensein iff n. R e m a r k 4. I follows ha if n = (p r and k n, hen A n (x, is peisensein. In paricular, if n = 2 r and k n, hen A n (x, is 2 Eisensein. In fac, i is easy o deduce ha A n (x, is peisensein for all k {1,..., n} iff n = (p r, r 0. (If n (p r and n f is he lowes nonzero digi, hen le k 1 = (p f, so k n and k 1 n. More generally, if n = m(p r wih 1 m < p, hen 0 < k (p mp r and n k > (m 1p r+1 each imply ha A n (x, is peisensein, namely n = (m 1p r+1 + (p mp r, so S(n =, and each condiion implies ha k 1 n/(. Compare his wih [8, Theorems 1, 2, 4] and [9, Theorems 1, 2]. Corollary 3.9 (where l = n 0. Suppose ha n = m + l where l > 0 and p (n l. Then he following are equivalen: (i S(m = and p m. (ii There exiss k {1,..., n} such ha A n (x, is peisensein down o degree l. (iii A n (x, m is peisensein down o degree l. Furhermore, ( if hese condiions ( hold, A n (x, is peisensein down o degree l iff p m and p N(n. P r o o f. Observe ha he condiions p (n l and p n l are equivalen o l = n 0. The resul hus follows immediaely from he heorem. Corollary 3.10 (where l = n 0 = 1. Suppose ha p n 1 and S(n 1 =. Then A n (x, (n 1 is peisensein down o degree 1 (so here is an irreducible facor of degree n 1. Also A n (x, is peisensein down o degree 1 iff p (so N(n = n i. n 1 and p n i, where i is he larges ppower < n
12 340 A. Adelberg We have no seen anyhing like he following simple corollary in he lieraure. Corollary 3.11 (where l = n 0 = 1, p = 2. If n = 2 r + 1, wih r > 0, hen A n (x, ( is peisensein ( down o degree 1 iff k is even. (This is equivalen o 2 2 r and 2. 1 These corollaries give he irreducibiliy of cerain insances of A n (x, /(2x (n k 1 for n odd. We give some addiional illusraive compuaions beyond he case n = 2 r + 1, where he Kimura bound only gives he rivial resul ha here is an irreducible facor of degree 1, since N(n = 1. Examples. (1 Le n = 13, p = 3, l = 1. Then S(12 = 2 = and p n 1, from which i follows ha A n (x, is peisensein down o degree 1 for k = 2, 3, 11, 12, whence A n (x, /(2x (n k 1 is irreducible for hese values. Since N(13 = 4, Kimura s analysis only gives an irreducible facor of degree 4, and McCarhy s higher order heorem [8, Theorem 3], wih p = 5, only gives an irreducible facor of degree 8. The preceding analysis can also be useful when n is even, e.g. (2 n = 14, p = 3, l = 2. Then A 14 (x, is 3Eisensein down o degree 2 if k = 3 or 12, so here is an irreducible facor of degree 12. On he oher hand, N(14 = 2, so Kimura s bound only gives an irreducible facor of degree 2. I is of course known ha B 14 (x is irreducible [5]. (3 Finally, we consider he excepional case n = 11, where here is an addiional irreducible quadraic facor for k = 11 (cf. [3]. Wih p = 2, N(11 = 1 and N(10 = 2, which do no help much, bu N(8 = 8, so A 11 (x, 8 has an irreducible facor of degree 8. All of he peisensein resuls for higher order Bernoulli polynomials which we have seen in he lieraure can be readily derived from Theorem 3.5 and is corollaries, usually considerably srenghened. As an example, consider he following heorem of McCarhy, wih noaions slighly changed o conform wih our usage. Theorem M [8, Theorem 3]. Le p be an odd prime, and assume ha n = r( + 1 where 0 r 1 p α. Then B n (α (x has an irreducible facor of degree n p. We acually prove a sronger if and only if heorem, which conains he preceding resul as one direcion of he special case where α < p.
13 Higher order Bernoulli polynomials 341 Theorem M. Le p be an odd prime and 1 α n. Assume ha n 1 and 2 (n 1/( p. Then B n (α (x is peisensein down o degree p iff n p α p α. p P r o o f. Le r = (n 1/(, so n = r( + 1 and (n p/( = r 1. (Thus if α < p, Theorem M follows from our Theorem 3.5; he case r = 1 is rivial since hen n = p and Theorem M is vacuous. α Le q =, so 0 qp α < p. Since rp n α, we have r q. p Bu n = (r 1p + (p r + 1 so S(n = p by our hypoheses, whence S(n < 2( since p is odd. Thus if k = n + 1 α, hen B (α N(n. n (x is peisensein down o degree n N(n iff p ( Since N(n = n p = (r 1(, we mus deermine when p, r 1 i.e. when k 1 r 1. Bu k 1 + r 1 = rp α = (r qp + (qp α and 0 r q, r 1, qp α, so he resul follows immediaely from Lucas s Theorem. References [1] A. A d e l b e r g, A finie difference approach o degenerae Bernoulli and Sirling polynomials, Discree Mah., submied. [2], Irreducible facors and padic poles of higher order Bernoulli polynomials, C. R. Mah. Rep. Acad. Sci. Canada 14 (1992, [3] J. Brillhar, On he Euler and Bernoulli polynomials, J. Reine Angew. Mah. 234 (1969, [4] L. Carliz, Noe on irreducibiliy of he Bernoulli and Euler polynomials, Duke Mah. J. 19 (1952, [5], The irreducibiliy of he Bernoulli polynomial B 14 (x, Mah. Comp. 19 (1965, [6] N. Kimura, On he degree of an irreducible facor of he Bernoulli polynomials, Aca Arih. 50 (1988, [7] P. J. M c C a r h y, Irreducibiliy of cerain Bernoulli polynomials, Amer. Mah. Monhly 68 (1961, [8], Some irreducibiliy heorems for Bernoulli polynomials of higher order, Duke Mah. J. 27 (1960, [9], Irreducibiliy of Bernoulli polynomials of higher order, Canad. J. Mah. 14 (1962, [10] I. Niven, H. Zuckerman and H. Mongomery, An Inroducion o he Theory of Numbers, 5h ed., Wiley, 1991.
14 342 A. Adelberg [11] N. E. Nörlund, Differenzenrechnung, Springer, [12] S. Roman, The Umbral Calculus, Academic Press, DEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE GRINNELL COLLEGE GRINNELL, IOWA 50112, U.S.A. Received on and in revised form on (2167
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