SOME DIVISIBILITY PROPERTIES OF BINOMIAL AND q-binomial COEFFICIENTS

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1 SOME DIVISIBILITY PROPERTIES OF BINOMIAL AND -BINOMIAL COEFFICIENTS Victor J. W. Guo an C. Krattenthaler 2 Department of Mathematics East China Normal University Shanghai People s Republic of China jwguo@math.ecnu.eu.cn WWW: 2 Fakultät für Mathematik Universität Wien Oskar-Morgenstern-Platz A-090 Vienna Austria. WWW: Abstract. We first prove that if a has a prime factor not iviing b then there are infinitely many positive integers n such that ( an+bn an is not ivisible by bn +. This confirms a recent conjecture of Z.-W. Sun. Moreover we provie some new ivisibility properties of binomial coefficients: for example we prove that ( ( 2n 3n an 2n 4n are ivisible by 6n an that ( 330n 88n is ivisible by 66n for all positive integers n. As we show the latter results are in fact conseuences of ivisibility an positivity results for uotients of -binomial coefficients by -integers generalising the positivity of -Catalan numbers. We also put forwar several relate conjectures.. Introuction The stuy of arithmetic properties of binomial coefficients has a long history. In 89 Babbage [6] prove the congruence ( 2p (mo p 2 p for primes p 3. In 862 Wolstenholme [28] showe that the above congruence hols moulo p 3 for any prime p 5. See [20] for a historical survey on Wolstenholme s theorem. 99 Mathematics Subject Classification. Primary B65; Seconary 05A0 05A30. Key wors an phrases. binomial coefficients Lucas theorem Euler s totient theorem -binomial coefficients Gaußian polynomials Catalan numbers -Catalan numbers positive polynomials. Research partially supporte by the Funamental Research Funs for the Central Universities. Research partially supporte by the Austrian Science Founation FWF grants Z30-N3 an S50- N5 the latter in the framework of the Special Research Program Algorithmic an Enumerative Combinatorics.

2 2 VICTOR J. W. GUO AND C. KRATTENTHALER Another famous congruence is (2n 0 (mo n +. n The corresponing uotients the numbers C n := n+( 2n n are calle Catalan numbers an they have many interesting combinatorial interpretations; see for example [2] an [24 pp ]. Recently Ulas an Schinzel [27] stuie ivisibility problems of Erős an Straus an of Erős an Graham. In [25 26] Sun gave some new ivisibility properties of binomial coefficients an their proucts. For example Sun prove the following result. Theorem.. [26 Theorem.] Let a b an n be positive integers. Then ( an + bn bn + 0 mo an gc(a bn +. (. Sun also propose the following conjecture. Conjecture.2. [26 Conjecture.] Let a an b be positive integers. If (bn + ( an+bn an for all sufficiently large positive integers n then each prime factor of a ivies b. In other wors if a has a prime factor not iviing b then there are infinitely many positive integers n such that (bn + ( an+bn an. Inspire by Conjecture.2 Sun [26] introuce a new function f : Z + Z + N. Namely for positive integers a an b if ( an+bn an is ivisible by bn + for all n Z + then he ( efine f(a b = 0; otherwise he let f(a b be the smallest positive integer n such that an+bn is not ivisible by bn +. Using Mathematica Sun [26] compute some values of an the function f: f(7 36 = 279 f(0 92 = 362 f( 00 = 87 f( = The present paper serves several purposes: first of all we give a proof of Conjecture.2 (see Theorem 2. below; secon we provie congruences an ivisibility results similar to the ones aresse in Theorem. an Conjecture.2 (see Theorems in Section 2; thir we show in Section 3 that among these results there is a significant number which can be lifte to the -worl; in other wors there are several such results which follow irectly from stronger ivisibility results for -polynomials. In particular Theorem. is an easy conseuence of Theorem 3.3 an Theorem 2.3 is an easy conseuence of Theorem 3.. On the other han Theorem 2.4 hints at the limitations of occurrence of these ivisibility phenomena. Sections 4 6 are evote to the proofs of our results in Sections 2 an 3. We close our paper with Section 7 by posing several open problems. 2. Results I Our first result is a more precise version of Conjecture.2. Theorem 2.. Conjecture.2 is true. Moreover if p is a prime such that p a but p b then f(a b pϕ(a+b a + b where ϕ(n is Euler s totient function.

3 SOME DIVISIBILITY PROPERTIES OF BINOMIAL AND -BINOMIAL COEFFICIENTS 3 For the proof of the above result we nee the following theorem. Theorem 2.2. Let a an b be positive integers with a > b an β an integer. Let p be a prime not iviing a. Then there are infinitely many positive integers n such that ( an ± (mo p. bn + β Our proofs of Theorems 2. an 2.2 are base on Euler s totient theorem an Lucas classical theorem on the congruence behaviour of binomial coefficients moulo prime numbers see Section 4. In [4 Corollary 2.3] the first author prove that ( 6n 0 (mo 2n. (2. 3n It is easy to see that ( ( 2n 2n = 2 n n = 4n 2 n The next theorem gives congruences similar to (2. an (2.2. ( 2n 2 0 (mo 2n. (2.2 n Theorem 2.3. Let n be a positive integer. Then ( ( 2n 2n 0 (mo 6n (2.3 3n 4n ( 30n 0 (mo (0n (5n (2.4 5n ( ( ( 60n 20n 20n 0 (mo 30n (2.5 6n 40n 45n ( 330n 0 (mo 66n. (2.6 88n We shall see that this theorem is the conseuence of a stronger result for -binomial coefficients cf. Theorem 3. in the next section. It seems that there shoul exist many more congruences like (2. (2.6. (In this irection see Conjecture 7.3. On the other han we have the following negative result. Theorem 2.4. There are no positive integers a an b such that ( an + bn 0 (mo 3n an for all n. For a possible generalisation of this theorem see Conjecture 7.2 in the last section.

4 4 VICTOR J. W. GUO AND C. KRATTENTHALER 3. Results II: -ivisibility properties Recall that the -binomial coefficients (also calle Gaußian polynomials are efine by [ ( n = k] n ( n ( if 0 k n ( k ( k ( ( nk ( nk ( 0 otherwise. We begin with the announce strengthening of Theorem 2.3. Theorem 3.. Let n be a positive integer. Then all of 2n 2n 60n 6n 3n 6n 4n 30n 6n 20n 20n 30n 40n 30n 45n 66n are polynomials in with non-negative integer coefficients. Furthermore ( 2 30n ( 0n ( 5n 5n is a polynomial in. 330n 88n (3. (3.2 For a conjectural stronger form of the last assertion in the above theorem see Conjecture 7.4 at the en of the paper. It is obvious that when a = b = the numbers ( an+bn an /(bn + (feature implicitly in Conjecture.2 an in Theorem 2. reuce to the Catalan numbers C n. There are various -analogues of the Catalan numbers. See Fürlinger an Hofbauer [0] for a survey an see [ 7 6] for the so-calle t-catalan numbers. A natural -analogue of C n is C n ( = 2n. n+ n It is well known that the -Catalan numbers C n ( are polynomials with non-negative integer coefficients (see [ 2 4 0]. Furthermore Haiman [7 (.7] prove (an it follows from Lemma 5.2 below that the polynomial bn + n bn+ n has non-negative coefficients for all b n. Another generalisation of C n ( was introuce by the first author an Zeng [5]: B nk ( := k 2n 2n 2n = n n k n k n k k k n. They note that the B nk ( s are polynomials in but i not aress the uestion whether they are polynomials with non-negative coefficients. As the next theorem shows this turns out to be the case. The theorem establishes in fact a stronger non-negativity property.

5 SOME DIVISIBILITY PROPERTIES OF BINOMIAL AND -BINOMIAL COEFFICIENTS 5 Theorem 3.2. Let n an k be non-negative integers with 0 k n. Then gc(kn 2n n n k (3.3 is a polynomial in with non-negative integer coefficients. Conseuently also B nk ( = k [ 2n n nk ] is a polynomial with non-negative coefficients. Applying the ineuality [26 (2.] we can also easily euce that C abn ( := a an + bn bn+ an is a prouct of certain cyclotomic polynomials an therefore a polynomial in. Again as it turns out all coefficients in these polynomials are non-negative. Also here we have actually a stronger result given in the theorem below. It shoul be note that it generalises Theorem. the latter being obtaine upon letting. Theorem 3.3. Let a b an n be positive integers. Then gc(anbn+ an + bn = gc(anbn+ bn+ an an+bn+ is a polynomial in with non-negative coefficients. Corollary 3.4. Let a b an n be positive integers. Then a an + bn bn+ an is a polynomial in with non-negative coefficients. The proofs of the results in this section are given in Section 5. an + bn + an 4. Proofs of Theorems 2. an 2.2 The proof of Theorem 2.2 (from which subseuently Theorem 2. is erive makes essential use of Lucas classical theorem on binomial coefficient congruences (see for example [ ]. For the convenience of the reaer we recall the theorem below. Theorem 4. (Lucas theorem. Let p be a prime an let a 0 b 0... a m b m {0... p }. Then ( a0 + a p + + a m p m m ( ai (mo p. b 0 + b p + + b m p m Proof of Theorem 2.2. Note that gc(p a =. By Euler s totient theorem (see [23] we have p ϕ(a 0 i=0 b i (mo a.

6 6 VICTOR J. W. GUO AND C. KRATTENTHALER Since a > b > 0 there exists a positive integer N such that an > bn + β > 0 hols for all n > N. Let r be a positive integer such that p rϕ(a > an an let n = (p rϕ(a /a. Then by Lucas theorem we have ( an bn + β = ( p rϕ(a bn + β m ( p ( b 0+ +b m i=0 b i (mo p where bn + β = b 0 + b p + b m p m with 0 b 0... b m p. It is clear that there are infinitely many such r an n. This completes the proof. Proof of Theorem 2.. Suppose that a an b are positive integers an p a prime such that p a but p b. We have the ecomposition ( ( an + bn an + bn = a + b ( an + bn. (4. bn + an an a an 2 It is clear that p (a+b. By the proof of Theorem 2.2 if we take n = (p rϕ(a+b /(a+b (r then ( an + bn ± (mo p an an thus ( an + bn (a + b = an ( an + bn ±(a + b 0 (mo p. (4.2 an 2 n an Combining (4. an (4.2 gives ( an + bn Z bn + an for all n = (p rϕ(a+b /(a + b (r = Namely Conjecture.2 hols an as esire. f(a b pϕ(a+b a + b 5. Proofs of Theorems an of Corollary 3.4 All the proofs in this section are similar in spirit. They all raw on a lemma from [22 Proposition 0..(iii] which extracts the essentials out of an argument of Anrews [3 Proof of Theorem 2]. (To be precise Lemma 5. below is a slight generalisation of [22 Proposition 0..(iii]. However the proof from [22] works also for this generalisation. We provie it here for the sake of completeness. Recall that a polynomial P ( = i=0 p i i in of egree is calle reciprocal if p i = p i for all i an that it is calle unimoal if there is an integer r with 0 r an 0 p 0... p r... p 0. Lemma 5.. Let P ( be a reciprocal an unimoal polynomial an m an n positive integers with m n. Furthermore assume that A( = m P ( is a polynomial in. n Then A( has non-negative coefficients.

7 SOME DIVISIBILITY PROPERTIES OF BINOMIAL AND -BINOMIAL COEFFICIENTS 7 Proof. Since P ( is unimoal the coefficient of k in ( m P ( is non-negative for 0 k eg(p /2. Conseuently the same must be true for A( = m P ( consiere n as a formal power series in. However also A( is reciprocal an its egree is at most the egree of P (. Therefore the remaining coefficients of A( must also be non-negative. Proof of Theorem 3.. In view of Lemma 5. an the well-known reciprocality an unimoality of -binomial coefficients (cf. [24 Ex ] for proving Theorem 3. it suffices to show that the expressions in (3. an (3.2 are polynomials in. We are going to accomplish this by a count of the cyclotomic polynomials which ivie numerators an enominators of these expressions respectively. We begin by showing that fact that 6n [ 2n 3n ] is a polynomial in. We recall the well-known n = n Φ ( where Φ ( enotes the -th cyclotomic polynomial in. Conseuently 2n 2n = Φ 6n 3n ( e with =2 e = χ ( (6n 2n + 3n 9n (5. where χ(s = if S is true an χ(s = 0 otherwise. This is clearly non-negative unless (6n. So let us assume that (6n which in particular means that 5. Let us write X = {3n/} where {α} := α α enotes the fractional part of α. Using this notation Euation (5. becomes e = χ ( (2X + 4X 3X. Since 0 X < we have 2X < 2. The only integers which are ivisible by in the range are 0 an. Hence we must have X = /(2 or X = ( + /(2. The former is impossible since X is a rational number which can be written with enominator. Thus the only possibility left is X = ( + /(2. For this choice it follows that 4X 3X = 2 =. (Here we use that 5. This proves that e is non-negative also in this case an completes the proof of polynomiality of 6n [ 2n 3n ]. The proof of polynomiality of to the reaer. We next turn our attention to 6n [ 2n 4n ] is completely analogous an therefore left 30n [ 60n 30n [ 60n 6n 6n ]. Again we write ] 60n = Φ ( e =2

8 8 VICTOR J. W. GUO AND C. KRATTENTHALER with e = χ ( (30n 60n + 6n 54n. (5.2 This is clearly non-negative unless (30n. We assume (30n an note that this implies = 7 or. Here we write X = {6n/}. Using this notation Euation (5.2 becomes e = χ ( (5X + 0X 9X. Since 0 X < we have (5X if an only if X is one of For the same reason as before the option X = /(5 is impossible. For the other options the corresponing value of 0X 9X is always except if X = + an = However in that case we have X = 8 which cannot be written with enominator 35 = 7. Therefore this case can actually not occur. This completes the proof that e is non-negative for all 2 an hence that 6n ] is a polynomial in. 30n [ 60n Proceeing in the same style for the proof of polynomiality of [ 20n 30n 40n ] we have to show that e = χ ( (3X + 2X 4X 8X (5.3 is non-negative for all X of the form X = x/ with 0 x < x being integral an 2. Clearly the expression in (5.3 is non-negative except possibly if (3X. With the same reasoning as before we see that the only cases to be examine are X = ( + /(3 an X = (2 + /(3 where = 7 or. If X = ( + /(3 we have 2X 4X 8X = So this will be eual to except if = 7. However in that case we have X = 8 2 which cannot be written with enominator = 7 a contraiction. Similarly if X = (2 + /(3 we have 2X 4X 8X = =. So again the exponent e in (5.3 is non-negative which establishes that is a polynomial in. For the proof of polynomiality of. 66n [ 330n 88n ] we have to show that 30n [ 20n 40n ] e = χ ( (3X + 5X 4X X (5.4 is non-negative for all X of the form X = x/ with 0 x < x being integral an 2. Clearly the expression in (5.4 is non-negative except possibly if (3X. With the same reasoning as before we see that the only cases to be examine are X = ( + /(3 an X = (2 + /(3 where = 5 = 7 or 3. If X = ( + /(3 we have 5X 4X X =

9 SOME DIVISIBILITY PROPERTIES OF BINOMIAL AND -BINOMIAL COEFFICIENTS 9 So this will be eual to except if = 7. However again this is an impossible case. Similarly if X = (2 + /(3 we have X 4X X = =. 3 So again the exponent e in (5.4 is non-negative which establishes that is a polynomial in. Turning to (3.2 to prove that show that 66n [ 330n 88n ] ( 2 ( 0n ( 5n [ 30n 5n ] is a polynomial in we must e = χ ( (2X χ ( (3X + 6X 5X (5.5 is non-negative for all X of the form X = x/ with 0 x < x being integral an 5. First of all we shoul observe that gc(0n 5n = whence the two truth functions in (5.5 cannot eual simultaneously. Therefore the expression in (5.5 is nonnegative except possibly if (2X or if (3X. With the same reasoning as before we see that the only cases to be examine are X = (+/(2 X = (+/(3 an X = (2 + /(3 where in the latter two cases the choice of = 3 is exclue. If X = ( + /(2 then 3 5 6X 5X = which always euals for 2. If X = ( + /(3 then 2 5 6X 5X = which always euals for = (sic!. Because of our assumptions we o not nee to consier the cases = 3 an = 5 so the only remaining case is = 4. However in that case X = 5 which cannot be written with enominator = 4 a contraiction. 2 Finally if X = (2 + /(3 then 2 0 6X 5X = which always euals for 2. So again the exponent e in (5.5 is non-negative in all cases which establishes that ( 2 [ ( 0n ( 5n 30n 5n ] is a polynomial in. Proof of Theorem 3.2. By Lemma 5. it suffices to establish polynomiality of (3.3. When written in terms of cyclotomic polynomials Expression (3.3 reas gc(kn 2n 2n = Φ n n k ( e with e = χ( gc(k n χ( n + =2 2n n k n + k. (5.6

10 0 VICTOR J. W. GUO AND C. KRATTENTHALER Similarly as before let us write N = {n/} an K = {k/}. Euation (5.6 becomes Using this notation e = χ( gc(k n χ( n + 2N N K N + K. (5.7 We have to istinguish several cases. If n then N = 0 an (5.7 becomes e = χ( k K. We see that this is zero (an hence non-negative regarless whether k or not. On the other han if we assume that n then (5.7 becomes e = 2N N K N + K an this is always non-negative. We have proven that (3.3 is inee a polynomial in. The statement on B nk ( follows immeiately from the previous result an the fact that gc(k n k. Finally Theorem 3.3 will follow immeiately from the following strengthening of a non-negativity result of Anrews [3 Theorem 2]. Lemma 5.2. Let a an b be positive integers. Then gc(ab a + b a+b a is a polynomial in with non-negative integer coefficients. (5.8 Proof. In view of Lemma 5. it suffices to show that the expression in (5.8 is a polynomial in. Again we start with the factorisation with gc(ab a+b a+b a + b = Φ a ( e =2 a + b a e = χ( gc(a b + b. (5.9 Next we write A = {a/} an B = {b/}. Using this notation Euation (5.9 becomes e = χ( gc(a b + A + B. This is clearly non-negative unless A = B = 0. However in that case we have a an b that is gc(a b so that e is non-negative also in this case. Proof of Theorem 3.3. Replace a by an an b by bn + in Lemma 5.2. Proof of Corollary 3.4. This follows immeiately from Theorem 3.3 an the fact that a gc(a bn + = gc(an bn +.

11 SOME DIVISIBILITY PROPERTIES OF BINOMIAL AND -BINOMIAL COEFFICIENTS 6. Proof of Theorem 2.4 The following auxiliary result on the occurrence of prime numbers congruent to 2 moulo 3 in small intervals will be crucial. Lemma 6.. If x 530 there is always at least one prime number congruent to 2 moulo 3 containe in the interval (x 20 9 x. Proof. Let θ(x; 3 2 enote the classical Chebyshev function efine by θ(x; 3 2 = log p. p prime p x p 2 (mo 3 McCurley prove the following estimates for this function (see [9 Theorems 5. an 5.3]: This implies that for x > 376 we have θ(y; 3 2 < 0.5 y y > 0 θ(y; 3 2 > 0.49 y y 376. θ ( 20 x; 3 2 θ(x; 3 2 > x 0.5 x > x >. 9 9 This means that if x > 376 there must be a prime number congruent to 2 moulo 3 strictly between x an 20 x. (To be completely accurate: the above argument only shows 9 that such a prime number exists in the half-open interval (x 20 x]. However existence 9 in the open interval (x 20 x can be easily establishe in the same manner by slightly 9 lowering the value of 20 in the above argument. 9 For the remaining range 530 x 376 one can verify the claim irectly using a computer. Proof of Theorem 2.4. Given a an b our strategy consists in fining a prime p an a positive integer n such that the p-aic valuation of ( an+bn an /(3n is negative so that 3n cannot ivie. ( an+bn an. We first verifie the possibility of fining such p an n for a b 850 using a computer. To establish the claim for the remaining values of a an b we have to istinguish several cases epening on the congruence classes of a an b moulo 3 an the relative sizes of a an b. First let (a b {(0 0 (0 ( 0 (0 2 ( (2 0} + (3Z 2. By Dirichlet s theorem [8] (see [5] we know that there are infinitely many primes congruent to 2 moulo 3. Let us take such a prime p with p > a + b an let 3n = p that is n = (p + /3. Furthermore let v p (α enote the p-aic valuation of α that is the maximal exponent e such that p e ivies α. Writing a = 3a + a 2 an b = 3b + b 2 with 0 a 2 b 2 2 by the well-known formula of Legenre [8 p. 0] for the p-aic valuation of factorials we then

12 2 VICTOR J. W. GUO AND C. KRATTENTHALER have ( ( an + bn v p = + ( (a + bn an bn 3n an p l p l p l l a2 + b 2 = + + a + b a a b2 3 + b. (6. Since in the current case we have 0 a 2 + b 2 2 an a+b < all the values of the floor 3 functions on the right-han ( sie of the above euation are zero. Conseuently the p-aic valuation of an+bn euals for our choice of p an n which in particular means that 3n ( an+bn an 3n an is not an integer. For the next case consier a pair (a b {(2 (2 2} + (3Z 2. Let us first assume that b 9 a. Since we have alreay verifie the claim for a b 850 we may assume 0 a 850. We now choose a prime p 2 (mo 3 strictly between 9 a an a. Such a 20 prime is guarantee to exist by Lemma 6. because ue to our assumption we have a > 530. Furthermore we choose n = (p + /3. We have 9 20 (a + bn 9 0 ap + 3 < 2 3 p(p + < p2 an hence with the same notation as above Euation (6. hols also in the current case. We have 3 a 2 + b 2 4 a + b 9a < 2p a = 2 < a 20 < 2 hence a2 + b 2 b2 + a + b a2 3 + a b = 0 = 0. ( Conseuently again the p-aic valuation of an+bn 3n an euals for our choice of p an ( n which in particular means that an+bn 3n an is not an integer. Next let again (a b {(2 (2 2} + (3Z 2 but 9 a < b 7 a. Since we have alreay 0 5 verifie the claim for a b 850 we may assume a 300. (If a < 300 then the above restriction imposes the boun b = 820 < 850. Here we choose a prime 5 p 2 (mo 3 strictly between 4a an 9 a. Such a prime is guarantee to exist by 5 0 Lemma 6. because ue to our assumption we have 4 a = 040 > 530. Furthermore we 5 choose n = (p + /3. We have still 3 a 2 + b 2 4 moreover 2p < 9p < 9 a a + b a < a = 2 < a 5 < 2 < 3a b 7a 7 < 2 hence p 5p 2 3 a2 + b a + b a2 3 + a b2 3 + b implying again that the p-aic valuation of n as esire. 3n = + χ(b 2 = 2 χ(b 2 = 2 = 0 ( an+bn an euals for our choice of p an The next case we iscuss is (a b {(2 } + (3Z 2 an 7 a < b 3a. Since we have 5 alreay verifie the claim for a b 850 we may assume b 850. Here we choose a prime p 2 (mo 3 strictly between 3 b an b. Such a prime is guarantee to exist by 0 3

13 SOME DIVISIBILITY PROPERTIES OF BINOMIAL AND -BINOMIAL COEFFICIENTS 3 Lemma 6. because ue to our assumption we have 3 b = 555 > 530. Furthermore we 0 choose n = (p + /3. Here we have ( ( an + bn v p = + ( (a + bn an bn 3n an p l p l p l l (a + bn an bn = + p 2 p 2 p 2 a2 + b a + b a a b2 3 + b. In this case ue to the estimations an p 2 < b p 3 we have < (a + bn = (a + bp + 3 < 2 7 bp + 3 p 2 < b p 3 < bn < (a + bn < 2p2 for p > 20 (a + bn an bn = 0 = 0. p 2 p 2 p 2 < 40 2 p(p + < 2p2 for p > 20 Moreover we have a 2 + b 2 = 3 < b < a + b 2b < 20p < 6p a = 2 < b a 3 3 9p 5b < 50 b 2 2p 63 3 = < b 0 so that 3 9 a2 + b 2 + a + b a a b2 3 + b = 2 = 0 ( implying again that the p-aic valuation of an+bn 3n an euals for our choice of p an n as esire. The last case to be iscusse is (a b {(2 } + (3Z 2 an 3a < b. Again since we have alreay verifie the claim for a b 850 we may assume b 850. Here we choose a prime p 2 (mo 3 strictly between 5 b an b. Such a prime is guarantee to exist 2 by Lemma 6. because ue to our assumption we have 5 b > 530. Furthermore we choose n = (p + /3. Since (a + bn < 4 3 bp + < p(p + < p2 for p > 44 ( for the p-aic valuation of an+bn 3n an there hols again (6.. In the current case we have furthermore a 2 + b 2 = 3 a + b 4b < 44p < a = 2 a < b < < b 2 3 9p = 3 2 < b an hence 3 5 a2 + b 2 a2 3 + a b2 + a + b b = 0 = 0 ( implying also here that the p-aic valuation of an+bn 3n an euals for our choice of p an n as esire.

14 4 VICTOR J. W. GUO AND C. KRATTENTHALER We have now covere all possible cases (in particular by symmetry in a an b we also covere the case (a b {( 2} + (3Z 2 an hence this conclues the proof of the theorem. 7. Concluing remarks an open problems In the proof of Theorem 2. assume that s is the smallest positive integer such that (a + b (p s. Then we obtain the stronger ineuality f(a b ps a + b. (7. It is easily seen that s ϕ(a + b. However such an upper boun is still likely much larger than the exact value of f(a b given by Sun [26]. For example the ineuality (7. gives f( f( f( 00 6 = 2736 = = 5960 f( = f( It seems that Theorem 2.2 can be further generalise in the following way. Conjecture 7.. Let a an b be positive integers with a > b an let α an β be integers. Furthermore let p be a prime such that gc(p a =. Then for each r = 0... p there are infinitely many positive integers n such that ( an + α r (mo p. bn + β In relation to Theorem 2.3 we propose the following two conjectures the first one generalising Theorem 2.4. Conjecture 7.2. For any o prime p there are no positive integers a > b such that ( an 0 (mo pn bn for all n. Conjecture 7.3. For any positive integer m there are positive integers a an b such that am > b an ( amn 0 (mo an bn for all n.

15 SOME DIVISIBILITY PROPERTIES OF BINOMIAL AND -BINOMIAL COEFFICIENTS 5 Note that the congruences (2. (2.6 imply that Conjecture 7.3 is true for m 5. It seems clear that for each specific prime p a proof of Conjecture 7.2 in the style of the proof of Theorem 2.4 in Section 6 can be given. On the other han a proof for arbitrary p will likely reuire a new iea. We en the paper with the following conjecture strengthening the last part of Theorem 3.. Conjecture 7.4. For all positive integers n an non-negative integers k with 0 k 25n 2 25n + 4 the coefficient of k in the polynomial ( 2 30n ( 0n ( 5n 5n is non-negative except for k = an k = 25n 2 25n+3 in which case the corresponing coefficient euals. Note Conjecture 7.2 was prove by Maji Mirzavaziri an the secon author. Conjecture 7.3 was prove by Maji Mirzavaziri. Acknowlegement We thank Pierre Dusart for very helpful iscussion concerning Lemma 6.. References [] G.E. Anrews Catalan numbers -Catalan numbers an hypergeometric series J. Combin. Theory Ser. A 44 ( [2] G.E. Anrews On the ifference of successive Gaussian polynomials J. Statist. Plann. Inference 34 ( [3] G.E. Anrews The Frieman Joichi Stanton monotonicity conjecture at primes Unusual Applications of Number Theory (M. Nathanson e. DIMACS Ser. Discrete Math. Theor. Comp. Sci. vol. 64 Amer. Math. Soc. Provience R.I pp [4] G.E. Anrews -Catalan ientities in: The Legacy of Allai Ramakrishnan in the Mathematical Sciences Springer New York 200 pp [5] T.W. Apostol Introuction to Analytic Number Theory Unergrauate Texts in Mathematics Springer-Verlag New York Heielberg 976. [6] C. Babbage Demonstration of a theorem relating to prime numbers Einburgh Philos. J. ( [7] K. S. Davis an W. A. Webb Lucas theorem for prime powers Europ. J. Combin. ( [8] P.G.L. Dirichlet Beweis es Satzes aß jee unbegrenzte arithmetische Progression eren erstes Glie un Differenz ganze Zahlen ohne gemeinschaftlichen Factor sin unenlich viele Primzahlen enthält Abhan. Aka. Wiss. Berlin 48 ( [9] N. Fine Binomial coefficients moulo a prime Amer. Math. Monthly 54 ( [0] J. Fürlinger an J. Hofbauer -Catalan numbers J. Combin. Theory Ser. A 40 ( [] A.M. Garsia an J. Haglun A proof of the t-catalan positivity conjecture Discrete Math. 256 ( [2] H.W. Goul Bell an Catalan Numbers: Research Bibliography of Two Special Number Seuences Combinatorial Research Institute Morgantown WV 978.

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