On the spacings between C-nomial coefficients

Transcription

1 On the spacings between C-nomial coefficients lorian Luca, Diego Marques,2, Pantelimon Stănică 3 Instituto de Matemáticas, Universidad Nacional Autónoma de México C.P , Morelia, Michoacán, México, 2 Departamento de Matemática, Universidade ederal do Ceará, Ceará, Brazil 3 Department of Applied Mathematics, Naval Postgraduate School, Monterey, CA , USA Abstract Let C n n 0 be the Lucas sequence C n+2 ac n+ +bc n for all n 0, where C 0 0 and C. or m let [ ] m C mc m C m + C C C be the corresponding C-nomial coefficient. When C n n is the ibonacci sequence the numbers [ m] are called ibonomials, or C n q n /q, where q > is an integer the numbers [ ] m are called q-binomial, or Gaussian coefficients, we q show that there are no nontrivial solutions to the Diophantine equation [ ] [ ] [ ] [ ] m n m n or l l with m, n,l other than the obvious ones n,l m,m. We also show that the difference [ ] [ ] m n l tends to infinity when m,,n,l are such that m/2, l n/2, m, n,l and max{m,n} tends to infinity in an effective way. Key words: Sequences 99 MSC: 2K N37 q q Corresponding author, tel addresses: fluca@matmor.unam.mx lorian Luca, diego@mat.ufc.br Diego Marques, pstanica@nps.edu Pantelimon Stănică. The first author was supported in part by Grants SEP-CONACyT and PAPIIT The second author is scholarship holder of CNPq. 3 The third author was partially supported by Air orce CVAQ D. Nussbaum and NPS RIP. Preprint submitted to Elsevier 26 May 2009

2 Introduction A famous unsolved problem in Diophantine equations is to find all pairs of binomial coefficients having the same value. That is, to find all solutions of m n. l Here, m and l n. To avoid the obvious symmetry of the Pascal triangle, we may assume that m/2 and l n/2. As of the time of this writing, the above problem has not yet been solved in its full generality. The only nontrivial solutions nown at this time are , , , 2 8 2i+2 2i+3 and 2i 2i , , i+2 2i+3 2i 2i , , 3 for i, 2,..., 2 where n is the nth ibonacci number defined by 0 0, and n+2 n+ + n for all n 0. See [5] for a proof of the fact that the above list contains all the nontrivial solutions of the Diophantine equation with l and, l {2, 3, 2, 4, 2, 6, 2, 8, 3, 4, 3,6, 4,6} and the recent paper [?] for the case, l 2, 5. Let us now loo at the sequence of ibonomial coefficients, which are defined by [ ] m 2 m m m m m + for m. These numbers are always integers as first proved by E. Lucas in [9]. Various parts of this sequence with fixed small values of appear in Sloane s On Line Encyclopedia of Integer Sequences [] see, for example, A00655, A056565, A00658, etc.. More generally, given any sequence C C n n 0 of nonzero real numbers, one can define the C-nomial coefficients as [ ] m C mc m C m +. C C C Bachmann [, p. 8], Carmichael [4, p. 40], and Jarden and Motzin [7], all showed that if C is a Lucas sequence; i.e., it has C 0 0, C and satisfies 2

3 the recurrence C n+2 ac n+ + bc n for all nonnegative integers n with some nonzero integers a and b such that the quadratic equation x 2 ax b 0 has two distinct roots and whose ratio is not a root of unity, then all the C-nomial coefficients are integers. The ibonomial coefficients are particular cases of this instance with a b. When a q + and b q, where q > is some fixed integer, the C-nomial coefficients become the so-called q-binomial coefficients given by [ ] m q qm q m +. q q In this paper, we study the analogous Diophantine equation when the binomial coefficients are replaced by C-nomial coefficients, and, more generally, we study the spacings between the C-nomial coefficients. While our results can be formulated for general C-nomial coefficients when C C n n 0 is a general Lucas sequence, we restrict our attention to the particular cases of the ibonomial coefficients, for which C n n 0, or to the case of the q-binomial coefficients, when C n q n /q for all n 0, where q > is a fixed integer. Our results are the following. Theorem None of the Diophantine equations [ ] m [ ] n l or [ ] m q [ ] n l q 3 has any positive integer solutions m/2, l n/2, m, n, l and q >. Next, let us put {[ ] m : m/2 } {f, f 2,...}, where f < f 2 < f 3 are all the elements of arranged increasingly. Note that {, 2, 3, 5, 6, 8, 3, 5, 2, 34, 40, 55,60, 89, 04,...}. This is sequence A4472 in []. Our next result shows that f N+ f N. Theorem 2 We have f N+ f N log f N /2, where the implied constant is effective. In particular, f N+ f N tends to infinity with N. 3

4 Our arguments for the proof of Theorem 2 are entirely explicit. In particular, if f N+ f N 00, then N The proof of Theorem Letting and be the two roots of the quadratic equation x 2 ax b 0 of the Lucas sequence u n n 0 with u 0 0, u of recurrence u n+2 au n+ +bu n for all n 0, we have u n n n for n 0,,.... We mae the convention that. We write 2 and we call it the discriminant of the sequence. In the particular case of the ibonacci sequence, we have + 5/2, 5/2, and 5, while in the particular case of the Lucas sequence involved in the q-binomial coefficient we have q,, and q 2. A Primitive Divisor p of the nth term u n of a Lucas sequence u n n 0 is a prime factor of u n which does not divide m n u m. It is nown that a primitive divisor p of u n exists whenever n 3 if and are real and 3 can be replaced by 7 if and are integers see, for example, [4]. The above statement is usually referred to as the Primitive Divisor Theorem see [3] for the most general version. It is also nown that such a primitive divisor p satisfies p ± mod n. We are now ready to deal with equation 3. We may assume that m n and that l. Since m n, we may assume that n > m. If l, then [ ] n l > n n m n l+ 2 l m + [ m ], n n + where we used the fact that n > m because n 3 which follows because n > m 2 2. Hence, assuming that n > m in equation 3 for the ibonomial coefficients, we deduce that l <. Thus, n > m 2 > 2l. A similar argument holds for the case of the q-binomial coefficients. Now if n 3, then by the Primitive Divisor Theorem there exists a primitive prime factor p for n. This prime will obviously divide [ ] n, since p does not divide ] l, but it cannot divide [ m, because p does not divide m. This shows that n 2 and a quic computation reveals that there are no equal ibonomial coefficients in the range 2 2l < 2 m < n 2. l 4

5 In the case of the q-binomial coefficients, the Primitive Divisor Theorem tells us that n 6. Since 2 2 < 2l m < n 6, the only possibilities are, l, m, n, 2, 4, 5,, 2, 4, 6,, 2, 5, 6. In the case, l, m, n, 2, 4, 5, the Diophantine equation 3 for q-binomial coefficients leads to q 4 q 3 q q 2 q5 q, which is equivalent to q 3 + q 4 q 2 + q 5, which is impossible because of the uniqueness of the base q expansion of a positive integer. The cases, l, m, n, 2, 4, 6 and, 2, 5, 6 of the Diophantine equation 3 for q-binomial coefficients lead to and q 4 q 3 q q 2 q6 q, or q7 + q 6 + q 2 q 8 + q 4 + q 3, q 5 q 4 q q 2 q6 q, or q9 + q 6 + q 2 q 8 + q 5 + q 4, respectively, both of which are impossible again by the uniqueness of the base q expansion of a positive integer. 3 The proof of Theorem 2 We [ eep ] the notations from the previous section, and start with some estimates m for. Let p : i i 2 i i Lemma We have [ ] m m 2 + ζ m,, p where ζ m, is a real number satisfying ζ m, < Proof We have 5

6 [ ] m m m m + 2 m+m + +m + 2 m 2 i i. i i Now observe that i i p i + It remains to estimate ζ m,. We use the inequality i p + ζ m,. e z/2 if z 0, /4, e z > + z > e 2z if z /4, 0. 5 We shall use the above inequality 4 with z / i for i +. Note that the inequality z 2+ 4 < /4 holds for all. We get that + ζ m, i < exp i i + i + +2 exp exp 2+2 / < Thus, ζ m, < By a similar calculation using the right hand side of inequalities 4, we get yielding + ζ m, exp 2 i + exp i 2 exp > 2 2+, ζ m, > 2 The desired inequality now follows from estimates 5 and

7 Lemma 2 We have where [ ] m m /2 5 /2 + ζ m,, ζ m, < 2 2m 2+. Proof We write [ ] m m m m + 2 m+m + +m + 5 /2 2 i0 m i. We now study the last product above. When, then the above product is + ζ m, m. 2m Thus, ζ m, < 2 2m 2m, so the desired inequality holds in this case. Assume now that 2. We then have, again by inequalities 4, that + ζ m, i0 m i < exp i0 2m i exp 2m + 2 exp < + 2m m 2+. In the above inequality, we used again inequality 4 with z 2/ 2m 2+ together with the fact that z 2/ 2+ 2/ 5 < /4 holds for all integers 2. Thus, ζ m, < 2 2m 2+. Using now the right hand side of inequality 4, we get + ζ m, 2 exp exp i0 2m i leading to ζ m, 2 > 2m 2+. This completes the proof of this lemma. 2 2m m 2+, 7

8 Lemma 3 The inequality holds for all > l. Proof Clearly, +/2 ll+/2 5 l/2 l+ > 8 2l+5 +/2 ll+/2 5 l/2 l+ il+ If l +, then the above expression becomes ± 2l 2 i. 2l+2 ± 2l 2 > 2l+2 + 2l+3. Assume now that l + 2. Observe that if i is odd, then i+ In particular, if l is odd, then so that il+ i il+ However, if i is even, then i+ i+2 + 2i+2 2i+4 + 2i+2 2i+4 > 4i+6 2i+3 4i+6 < 2i+3. i < 2l+3, / > 2l+3 2l+3 2 > 2l+3. i+2 + 2i+2 2i+4 + 2i+2 2i+4 4i+6 > + 2i+3 2i+5 + 2i+4, 8

9 so that, if l is even, then il+ i > + 2l+4, leading to il+ i + / > 2l+4 2l+5, which completes the proof of this lemma. It is nown that the number p appearing at?? is transcendental apply, for example, Lemma and Lemma 2 from [6] to the function fz ηz and the algebraic number q /. In particular, it cannot be of the form 5 l/2 / s for any positive integers l and s The next lemma tells us even more, namely that the number p cannot be approximated too well by numbers of the form 5 l/2 / s for positive integers l and s. Lemma 4 The inequality s 5 l/2 p > 5 3l/2 holds for all positive integers s and l with finitely many exceptions. Proof Assume that s is large and that s 5 l/2 p < 5 3l/2 holds with some positive integer l. Then also the inequality 5 l/2 p s 3s holds, where we can tae the above implied constant as 2/p 2 once s is sufficiently large. By Euler s pentagonal formula, p n3n /2 n3n+/2 n + n εn, n + ε n,2, n3n n3n+ 9

10 for some signs ε n,, ε n,2 {±}. Let N : N s be the minimal positive integer such that N3N + > 2s + 0. Then both estimates and N3N + 2s + Os /2, N + 3N + N3N + 4N + 2 c 0 s /2, c s /2 hold for large s with some positive constants c 0 and c. Thus, 5 l/2 s N n εn, + ε n,2 n3n n3n+ O +. 3s N+3N+ Multiplying both sides of the above approximation by N3N+ we get N 5 l/2 N3N+ s ε n, u n, + ε n,2 u n,2 n + s+os/2 4N c 0s /2 provided that s is sufficiently large. Here, u n, and u n,2 stand for the nonnegative exponents of the form N3N + n3n and N3N + n3n+, respectively, for all positive integers n,...,n. Put now u N3N + s > s + 0 and observe that the number N γ 5 l/2 u ε n, u n, + ε n,2 u n,2 n is an algebraic integer in K Q[ 5]. Its absolute value is, by 8, bounded above by O c 0 s. Its conjugate in K is N σγ ±5 l/2 u ε n, u n, + ε n,2 u n,2. n Since s is assumed to be large, it follows that s 5 l/2, therefore 5 l/2 u 5 l/2 u < 5 l/2 s. Since also ε n, u n, + ε n,2 u n,2 < n n 0, n N it follows that σγ. Thus, N K/Q γ γ σγ c 0 s. However, N K/Q γ is an integer. or large s, the above inequality is possible only when γ

11 We now study this last condition and show that it is impossible, which will complete the proof of the lemma. Assume that γ 0. Then we get 5 l/2 n N εn, u n, u + ε n,2 u n,2 u. 0 Conjugating in K, we get that ε5 l/2 n N εn, u n, u + ε n,2 u n,2 u, where ε {±} according to whether l is even or odd. Suppose say that ε. Then subtracting the above relations 9 and 0 and dividing both sides of the resulting relation by 5, we get εn, un, u + ε n,2 un,2 u 0. 2 n N Observe that the indices in the above relation satisfy u, u < u,2 u < u 2, u < u 2,2 u < < u N, u < u N,2 u. 3 Let M { u n,i u : n N, i, 2} {m, m 2,..., m t }, where m < m 2 < < m t : M. Observe that all members of M are nonnegative. urthermore, because of inequalities 2 for each m j M, there are at most two numbers u n,i u in the string 2 such their absolute values is m j, and if there are two then one of them is the negative of the other. Observe also that since u n,i is always even, it follows that all the numbers in M have the same parity. inally, let us observe that the first three large values of M appear only once. Indeed, the largest positive member in 2 is u N,2 u N3N + N3N + s s, whereas the first three negative ones in 2 are Thus, u, u N3N + s + 2 < s + 8, u,2 u N3N + s + 4 < s + 6, u 2, u N3N + s + 0 < s. M m t N3N + s + 2, m t N3N + s + 4 M 2, m t 2 N3N + s + 0 M 8, are the first three largest elements in M, and for each of them, there is only one element namely the corresponding negative one in the string 2 whose

12 absolute value is this given element. rom the above discussion, and using also the fact that m m m, we deduce that relation leads to the inequality M M 2 + M M M/2 Using the fact that m m / 5 + O, we get that 5 M/2 M 2 < M 0 + M M 0 + M OM 5 M 9 + OM. Thus, relation 3 leads to or M M 2 + M 8 + 2M 9 + OM, M + O M and this is false for large M; hence, for large s., 5 The case when ε is similar. In this case, we sum up relations 9 and 0 and get a relation similar to, except that the ibonacci numbers are replaced by the Lucas numbers L n n 0, where L 0 0, L and L n+2 L n+ + L n for all n 0. The general term of the Lucas sequence L n n 0 is L n n + n. A similar argument leads to inequality 4 from which the same contradiction as in the case ε is derived. This shows that γ cannot be zero and completes the proof of the lemma. The proof of Theorem 2. In parallel with the proof of this theorem, we also show that f N+ f N > 00 for N > 26. Let N be sufficiently large, say at least such that log log log f N >. Let K be tending to infinity with N. We need to show that if K tends to infinity with N sufficiently slowly, say, if K < c 2 log f N /2 with a sufficiently small positive constant c 2, then the Diophantine inequality [ ] m [ ] n l K 6 2

13 has only finitely many positive integer solutions m,, n, l with m/2, l n/2, m, n, l. We assume that n m. Observe that by Lemma, we have that for N large, f N expm 2 + O expnl l 2 + O, so that both inequalities m 2 log f N and nl l 2 log f N hold. Since m 2 m m/2 2, we get that m log f N /2. Let c 3 be the implied constant above. Assume that K c 3 log f N /2 3. Then m > K +2. or large N; hence, for large m, m has a primitive divisor which is at least as large as m K +. In the particular case that we tae K 00, we also assume that m > 00 since the remaining cases can be checed using Mathematica. Assume first that n m. [ Let ] P be [ some ] primitive prime factor of n. Then m n certainly P divides both and, therefore it divides their difference l which is K by inequality 5. Since P ± mod n, we get that P n K +. Thus, P K + 2, so the only possibility is therefore that the difference appearing in the left hand side of inequality 5 is zero. However, this is impossible for m, n, l by Theorem. Thus, we may assume that n > m. Assume next that l. Then [ ] n l [ ] m [ ] n n n n [ ] n 2 n n + n n n > n 2 > expc 3 log f N /2, 7 where we used the fact that n i i + for all i,...,. So, in particular, the above difference exceeds c 3 log f N /2 > K for N sufficiently large. Thus, we may assume that n > m 2 > 2l. In particular, n 2l + 3. Next, let us notice that m n l. Indeed, assume that this [ ] is not so. [ ] Let P m n be a primitive prime factor of m. Then P divides both and, so l again P divides their difference which is K. Since P K + 2, we get again a contradiction. Thus, m n l. 3

14 Next, we write [ ] m m 2 + ζ m, and p [ ] n l nl l2 + ζ n,l. p Put u max{nl l 2, m 2 }. By inequality 5 together with Lemma, we have m 2 nl l2 < Kp + u p ζ m, + ζ n,l < Kp + 4pu 2l+. Dividing both sides of the above inequality by u, putting λ m 2 nl l 2, and observing that u ln l ll + 3, we get that λ Kp 4p + ll+3 2l+. 8 We now distinguish two cases according to whether λ 0 or λ 0, respectively. The Case λ 0. This is the heart of the proof. We find it easier to explain it in the particular case when K 00, and to treat the general case later. The case K 00. In this case, the left hand side of the above inequality 7 is > 0.38, and we therefore get the inequality 0.38 < 23 ll l+, leading to l, 2. Since l, 2, by Lemma 2, we have where [ ] n l nl ll /2 5 l/2 + ζ n,l, ζ n,l 2 2n 3. Thus, inequality 5 together with Lemma 2 now lead to 4

15 nl l2 +ll+/2 m 2 5 l/2 p < 00 + u ll+/2 ζ 5 l/2 n,l + ζ m, p < 00 + u n p 2+ < 00 + u 2n p 2+. Observe that since n > m 2, we have that 2n provided that 3. The inequality 2n holds also when 2 since n > 00. We thus get that the inequality 2n holds always in our range, which leads to nl l2 +ll+/2 m 2 5 l/2 p u < p < 00 + u 2 2, where we also used the fact that / + 3/p < 3. We now divide both sides of the above inequality again by u getting nl l2 +ll+/2 u m 2 u 5 l/2 p < In the last inequality above, we also used the fact that u m 2. Note that the left side of the above inequality is either ll+/2 λ 5 l/2 p, or ll+/2 λ 5 l/2 p, 20 according to whether u nl l 2 or m 2. Since λ 0, in the first case we get that λ /p < /.22 < 0.5, while ll+/2 /5 l/2 / Thus, in the first case the left hand side of inequality 8 exceeds 0.2. In the second case, again since λ 0, it follows that the inequality λ ll + /2 holds in all cases except when λ, 2 and l 2. Hence, { } 2 max 5 l/2 5/2, 5 ll+/2 λ 2 5. This shows that in this second case the left hand side of inequality 8 is at least /p 2 /5 > 0.2. Thus, in both cases we have that giving < , 2 5

16 Hence, l, {, 2,, 3, 2, 3}. Next, we use again inequality 5 and Lemma 2 together with the obvious fact that max { } m /2, nl ll /2 u 5 /2 5 l/2 l 5 /2, to get that m /2 nl ll /2 5 /2 5 l/2 l u 2 < n 3 2m 5 < 00 + u 2m 7, 22 where we used the fact that n 3 2m 5 2m n 2m+2 < 2m 7, since n 2m < 4 2. Dividing again both sides of the above inequality 2 by u and using the fact that u m 2m 2, we get that m 2 u++/2 nl l2 u+ll+/2 5 /2 5 l/2 l < m 4 2m The left hand side above is either +/2 ll+/2 λ 5 /2 5 l/2 l, or +/2 λ ll+/2 5 /2 5 l/2 l, 24 according to whether u nl l 2 or m 2. In the first case, we have that +/2 /5 /2 0.8 for 2, 3, while ll+/2 λ /5 l/2 l /5 /2 < 0.5 by an argument already used. Thus, in this case the left hand side of inequality 22 is at least 0.3. In the second case, we have that ll+/2 /5 l/2 l 0.72 for l, 2, while +/2 λ { 2 max 5 /2 5, 5 } < /2 6

17 for 2, 3 and λ 0, therefore the left hand side of inequality 22 exceeds 0.9 in this case. Thus, in both cases we have leading to m < 9, which is false. 0.9 < m 4 This taes care of the case λ 0 when K 00. 2m 7, 25 The case of the general K. In the case of the general K and when λ 0, the inequality 7 shows that l Olog K /2. Then the arguments from the case K 00 show that the analog inequality 8 holds with 00 replaced by K and with the exponent 2 2 replaced by 2 + Olog K. Indeed, the only inequality to justify is the fact that the difference n grows faster than any fixed multiple of log K once N is sufficiently large. Well, assuming that this were not so, we would get that infinitely often n < c 4 log K holds with some positive constant c 4. Thus, m < n log K implying log K, and later that m < n + Olog K log K. Thus, m log K 2 log log f N 2, which combined with the fact that m log f N gives only finitely many possibilities for N. Thus, if N is sufficiently large, we get to inequality 8 with the right hand side replaced by 2+Olog K. Now inequality 8 with 00 replaced by K together with the lower bound given by Lemma 4 on the expression appearing in the right hand side in display 9 lead to an inequality of the form 5 3l/2 2+Olog K, implying l + Olog K. Since also l log K /2, we get that log K. We thus bounded both l and in terms of K. ollowing through the arguments from the case when K 00, we arrive at the analog of inequality 22 with 00 replaced by K, which implies that +/2 ll+/2± λ 5 l/2 l+ 2m+Olog K. 26 The right hand side above is not zero. Indeed, if it were, then since no power of of nonzero exponent can be an integer, we get that the exponent of in the left hand side of equation 25 is zero, and further that l+ 5 l/2. By the Primitive Divisor Theorem, the above relation is false for > 2, 7

18 and it can be checed that it does not hold for any l < 2 either. Thus, the left hand side of equation 25 is indeed nonzero. urthermore, since +/2 p5 /2 as, it follows that the only chance the above expression from the right hand side has of being small is when λ O. We now use a linear form in logarithms á la Baer [2], which states that if, 2, 3 are algebraic numbers of heights H, H 2, H 3, respectively, and b, b 2, b 3 are integers of absolute value at most B such that then Λ b b 2 2 b 3 3 0, Λ > exp c 5 logh + 2 logh logh logb + 2 for some positive constant c 5 depending on the degree of the field Q[, 2, 3 ] over Q. Recall that the height of an algebraic number is the maximum absolute value of the coefficients of its minimal polynomial over the integers. We apply this with, 2 5, 3 l+, and b + /2 ll + /2 ± λ, b 2 l, b 3. Note that H O, H 2 O, H 3 O2 and B O 2. We thus get that the left hand side of 25 is bounded below by exp c 6 2 log, where c 6 is some absolute constant. Thus, we get the inequality yielding Hence, m + Olog K 2 log log K 2 log log K, m log K 2 log log K. m log K 3 log log K log log f N 3 log log log f N. However, m > m log f N, which gives log f N log log f N 3 log log log f N, and this has only finitely many solutions N. This taes care of the case λ 0. The case λ 0. Here too we distinguish between the instance when K 00 and the general K. The case when K 00. 8

19 We use inequality 5 with K 00 together with Lemma 2 to get m /2 nl ll /2 5 /2 5 l/2 l m /2 5 /2 2m 2+ 2 nl ll /2 + 5 l/2 l 2n 2l+. We divide both sides of the above inequality by nl ll /2 / 5 l/2 l, and recalling that m 2 nl l 2 we obtain +/2 ll+/2 5 l/2 l+ <00 5l/2 l nl ll / /2 ll+/2 5 l/2 l+ 2m n 2l+. 27 Next, we estimate the terms involved in the right hand side of 26. Observe that l 5 l/2 l < + +l + <.8 ll+/2, 2i while +/2 ll+/2 5 l/2 l+ il+ i < < 2.. 2i i 2i Thus, using also the facts that l <, n l m +, and nl l 2 m 2m, we get that the right hand side of inequality 26 is < m < 84 3 < 2m 2 2m As for the left hand side of inequality 26, we use inequality 7 of Lemma 3, to arrive at < 2l+5 2m 2, 29 which leads to 2m < 2 + 2l + 6. Thus, m + l + 7. Since m 2, we get that l + 7. Hence, m 2 m l+7 2 l 2 +4l+49. Since ln l m l 2 +4l+49, we get that n l l+4+49/l, therefore n 2l+4+49/l. Thus, n l l+4+49/l. Since m n l, we get that 2 m n l l /l. Since l +, we get that 2l + 2 l /l, or l /l, leading to l 5. Thus, l and m +l , which is a contradiction. Hence, inequality 5 has no solutions with n m > 00. The case of the general K. 9

20 Arguments identical to the ones used in when K 00 lead to the analogue of inequality 28 with 2m 2 replaced by 2m 2 + Olog K, which leads to m +l+ologk. The argument used in the final step of the proof of the argument for K 00 shows first that l + Olog K, then that n 2l+Olog K, so that n l l+olog K. Since n l 2 2l+O, we get 2l+Olog K l+olog K; thus, l OlogK, therefore n Olog K. Thus, m Olog K 2 Olog log f N 2. Since also m log f N, we get that log f N log log f N 2, so again only finitely many possibilities for N. This finishes the proof of the theorem. As for the computational case when K 00 and m 00, by going through the previous bounds, a very rough estimate renders the bound n A computer program exhausted easily this entire range, revealing that f N+ f N 00, only if N 8, or N urther comments and a generalization A similar result as Theorem 2 holds when instead of the ibonomials, we arrange all the q-binomial coefficients in increasing order, or, more generally, all the C-nomial coefficients, when C C n n 0 is a Lucas sequence of integers satisfying the property that > in particular, its roots are real. In order to prove this, at least for q-binomial coefficients, one would need to prove obvious analogues of Lemmas, 2, 3 and 4. or example, the analogue of Lemma 4 would state that the inequality q s q l p > q q 3l holds for all but finitely many pairs l, s, once q is fixed, where p q. n q n It would be of interest to study mixed Diophantine equations of the type [ ] n l C [ ] m, C 2 where C and C 2 are two distinct Lucas sequences. or example, what can one say about the number of solutions of the Diophantine equation [ ] n l [ ] m, q 20

21 where l n/2, m/2 in unnowns, l, m, n once q > is a fixed integer? Does this have finitely or infinitely many solutions? When and l are fixed, then the two sides of the above equation become linear recurrent sequences of orders depending on and l with dominant roots powers of and q, respectively, where these dominant roots are multiplicatively independent. Standard results from the theory of Diophantine equations see [0], for example, will then lead to the conclusion that there are only finitely many possibilities for the pair m, n once the pair, l is fixed. We do not have an argument to the effect that the above Diophantine equation has only finitely many solutions in all four variables, l, m, n. Of a somewhat related form is the main result from [8], where it is shown that there is no non-abelian finite simple group whose order is a ibonacci number. inally, let us loo at the series N f N. 30 The fact that it is convergent follows because for each n, row n contains n + /2 ibonomial coefficients the smallest one being [ ] n n. This shows that series 29 is bounded above by n n n, and this last series is certainly convergent. What is the nature of the number 29? Is it algebraic or transcendental? We recall that it is nown that the sum of the reciprocals of the odd indexed ibonacci numbers is transcendental see [6]. Acnowledgments We than the anonymous referee for constructive criticism on a previous version of the manuscript and for pointing out a relevant reference. Wor by the third author was done in the all of 2008 while he visited Instituto de Matemáticas de la Universidad Nacional Autónoma de México. He thans this institution for its hospitality. References [] P. Bachmann, Niedere Zahlentheorie, Volume 2, Leipzig, Teubner, 90. [2] A. Baer and G. Wüstholz, Logarithmic forms and group varieties, J. Reine Angew. Math ,

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