On the number of ways of writing t as a product of factorials

Transcription

1 On the number of ways of writing t as a product of factorials Daniel M. Kane December 3, 005 Abstract Let N 0 denote the set of non-negative integers. In this paper we prove that lim sup n, m N 0 : n!m! t 6. Introduction Let N 0 denote the set of non-negative integers. In this paper we will prove that lim sup { n, m N 0 : n!m! t } 6. We use three techniques to prove this result. First, it is not difficult to generate an infinite set of t each of which has at least 6 representations as a product of factorials thus establishing the lower bound. We then use considerations of the number of times two divides t in order to show that all of the solutions must be near each other. Lastly we use some analytic techniques analogous to those in []. The following three conjectures also seem likely. Conjecture. Conjecture. Conjecture 3. max { n, m N 0 : n!m! t } 6. lim sup { n, m N : n!m! t } 4. max { n, m N : n!m! t } 4. It is true that conjecture 3 would imply both other conjectures, and that any of these conjectures is stronger than our main theorem.

2 The lower bound Notice that for any integer n >, we have that n!! 0! n!!! n!! n! n!! n!! n! n!!! n!! 0!. Therefore, we have that lim sup 3 The first technique { n, m N 0 : n!m! t } 6. For a positive integer n, let en denote the largest k so that k divides n. Notice that en! log n i log n i n i n + Olog n. log n n i i O Therefore we have that if n!m! t, then en! + em! et, and therefore, n + m + Olog n + log m et. Since n/ n/ < n! < t, we have that n < log t for sufficiently large t. Therefore, for sufficiently large t, n + m + O et. Hence if n!m! n!m! t, then n + m n + m + O. This fact provides an elementary proof that for fixed t the number of solutions to n!m! t is O because by convexity of the log of the factorial function, at most two solutions to n!m! t have a given sum of n + m, and this sum cannot vary by more than O. 4 The second technique Our second technique is similar to that used in []. We begin with the following lemma: Lemma. If F x : R R is an infinitely differentiable function and if F x 0 for x x, x,..., x n+ where x < x <... < x n+, then F n y 0 for some y x, x n+. Proof. We proceed by induction on n. The case of n is Rolle s Theorem. Given the statement of Lemma. for n, if there exists such an F with n + zeroes, x < x <... < x n+, then by Rolle s theorem, there exist points y i x i, x i+ i n so that F y i 0. Then since F has at least n roots, by the induction hypothesis there exists a y with x < y < y < y n < x n+, and F n y F n y 0.

3 We now state a lemma that helps us to count the number of integer points on smooth curves. Lemma. Let f : R R be a C k function. Suppose that for x a, b, that 0 < k k! x k fx < α. Then if we have a < x 0 < x <... < x k < b where x i Z and fx i Z for all 0 i k, then x k x 0 α kk+. Proof. Let gx k fx i i0 0 j k i j x x j x i x j be the polynomial of degree k that interpolates f at the x i. Let hx fx gx. Then hx i 0. Hence by Lemma we have that for some a < x 0 < y < x k < b that k x k hy 0. Or that Therefore, k k! x k fx xy k k! x k gx xy k s k! x k fx k fx i i0 xy 0 j k j i x i x j. is an integer multiple of M 0 i<j k x j x i. Therefore, either s 0 or else s M. But by assumption, 0 < s < α. Therefore, α s M. Hence α x k x 0 kk+, and hence we have that α kk+ xk x 0 as desired. We will also make use of a generalization of Stirling s formula which states that: logγz + z + logz z + logπ + Oz uniformly for Rz > 0. This follows readily from the m case of log Γz + logπ + z + logz z + m j B j j jz j m 0 B m x [x] dx. x + z m where B j and B m are the Bernoulli numbers and Bernoulli polynomials see []. 3

4 5 The Strategy We have yet to prove that for sufficiently large t { n, m N 0 : n!m! t } 6. It is sufficient to show that { n, m N 0 : n m, n!m! t } 3 for all sufficiently large t. We will split solutions of this form into three overlapping cases:. m < exp. m > exp, n m > log t 5/36 3. n m < log t 6/36 Furthermore, we will show by our results from sections 3 and 4, that for all sufficiently large t, that all integer solutions to n!m! t lie in one of these regions. Define the function f : R R implicitly by Γfx + Γx + t. It is clear that f x gx gfx where gx x log Γx + logx + Ox by our strong form of Stirling s formula. So f log x + O x logfx. So if we have two pairs of solutions n, m, n, m to n m, n!m! t, where m > m, then O > n + m n + m m m + f xdx. We need to show that if there are solutions with m too big for region, there are none with m too small for region, and that if there are solutions with m too small for region 3, there are none with m too big for region. We can show the first of these by verifying that for sufficiently large t exp exp exp exp + f xdx > log t 4 + O dx log t exp + O. This shows that for sufficiently large t, it is impossible to have two solutions, one of which has m too small to be in region, and the other of which has m too large to be in region since this would imply that n +m n +m. 4

5 Now if x and x are the numbers so that fx x log t 5/36 and fx x log t 6/36, we notice that since the log of the gamma-function is convex that x x > 3 log t6/36. Thus we verify the second of these statements by noticing that x + f xdx > x 3 log t6/36 + f x > 3 log logfx /x + Ox t6/36 > logfx Ω log t 6/36 fx x /x logfx log t 5/36 Ω. Recall that at Ωbt means that there exists a constant c > 0 so that for all sufficiently large t, at > cbt, and that at Θbt means that there exist c > 0 and c > 0 so that for all sufficiently large t, c at > bt > c at. In section 6, we will cover the case where there are solutions in the first region. In section 7, we will cover the case where there are solutions in the second region. In section 8, we will cover the case where there are solutions in the third region. 6 The First Region In this section, we will prove that for sufficiently large t, that there are at most solutions to n!m! t with 0 < m exp. Notice that n + x! e en + x! en! n! n + x n i i log x i i x + O + max ec log x i n<c n+x x + max ec + Olog x. n<c n+x Therefore, if we have any two such solutions, n!m! n!m! t, with n > n then e e. Therefore, n! n! m! m! n n + Ologn n + max ec m m + Ologm m. n <c n 5

6 Which implies that max ec > m m n n + Ologm m. n <c n Notice that if n!m! n!m!, then n! n m!! m, and therefore, m! m > log n n n log m. Now, n log n > log t exp. Therefore, n Ω log t, so log n Ω. Hence m m Ω n n. Therefore, we have that max ec > m m + O / + Ologm m n <c n Ωm m Ω. Therefore, if we have three solutions in region, n, m, n, m, n 3, m 3 with 0 < m < m < m 3, then we have that there exist n 3 < c n < c n with ec i Ω. Therefore, since minex, ey ex y, we have that ec c Ω. Therefore, n n 3 > c c > expω. But we notice that this and previous inequalities imply that m 3 + n 3 m + n Ω expω. Since this cannot be O, we have that for sufficiently large t, there are at most solutions with m 0 in region. Hence there are at most 3 solutions in region. 7 The Second Region In this section we will show that there are at most solutions with m > exp and n m > log t 5/36. Recall that f : R R so that Γfx + Γx + t. This is defined in the range we are interested in, because the gamma-function is increasing. Let gx log Γx +. So gfx + gx log t. Differentiating implicitly, we get that Therefore, f x g x g fx. f x g x g fx + f x g xg fx g fx g xg fx + g x g fx g fx 3. By differentiating our strong form of Stirling s formula, we find that for fx > x f x log fx x + log x fx + Ox log fx + Ofx 3. 6

7 Therefore, for all sufficiently large t, for x > exp and fx x > log t 5/36 we have that 0 < f x < O. xlog fx Assume for sake of contradiction that we have three solutions to n!m! t in region n i!m i! t for i 3 where m i < m i+. Then we have that m i is an integer and that fm i is an integer. Since between m and m 3 we have that 0 < f x < O. m Therefore, by Lemma, m 3 m > Ωm /3. But we also have that m 3 + n 3 m + n O. Therefore m+ωm /3 m log fx log x dx O. log fx Now we have that for x in the range we are concerned with that log fx log x fx x/x Ω Ωlog t /36. log fx log fx Therefore, it must be that m O 3 log t /. But in this range, the integrand we are concerned with is at least + o. Therefore, it must be that m O 3 which does not hold. Therefore, for sufficiently large t, there are at most solutions in region. 8 Region Three In this section we will show that there are at most 3 solutions in region 3 for sufficiently large t. This proof depends on the fact that if n and m are integers, then so are n + m and n m and applications of Lemma and results from section 3. Suppose that Γa + x + /Γa x + / t, where x Oa 3/. Then we have by our strong form of Stirling s formula that log t a + x + a + x log a + log π + Oa a + loga/ a + logπ + a + x + x log + a + a x + log + a x + log a + loga/ a + logπ + x a + Oa. a x x a + Oa 7

8 Therefore, we have that x a log t a aa + log + a + O. π Let a 0 be the positive real value so that Γa 0 / + t. So then we have that t a0 a 0 log a 0 a 0 + log + a 0 O. π It is also true that a 0 Olog t. If we pick an a so that t a a log aa + log + a + O < log t 6/. π Then there must be a unique complex x with x log t 6/ so that Γa + x + /Γa x + / t. Since the derivative of t a a log aa + log + a π is t a log a + log + a π and since its second derivative is Olog a, this should hold as long as a a 0 < Olog t 9/0. Furthermore, for a in this range, x attains all values with x log t 3/9. This allows use to define an analytic function ha defined on a a 0 < Olog t 9/0 so that Γ a + ha + Γ a ha + t. Furthermore, ha attains all values of absolute value at most log t 3/9 when a a 0 Olog t 5/9. Additionally, we have that t a ha a log aa + log + a + O. π Since the O is uniform in the region stated, its third derivative when a a 0 Olog t 3/9 notice that 4/9 < 3/9 < 9/0 can be written using Cauchy s Integral formula as Oz a 4 dz, C where C is a contour that traverses a circle centered at a of radius Ωlog t 9/0 once in the counter-clockwise direction. This is Olog t 7/0. Therefore, when ha < log t 3/9 we have that h a 3 t a a 3 a log aa + log + a + Olog t 7/0 π a + Olog t 7/0. 8

9 Since it is clear by Stirling s formula that a 0 Θ log t sufficiently large t, when ha log t 3/9 then 0 < h a < O we have that for. log t Now suppose for sake of contradiction that n i, m i are distinct region 3 solutions for i 4. Then n i + m i Z and hn i + m i n i m i Z. Furthermore, hn i + m i log t 3/9. Then since in the range between the log t, Lemma implies that the n i + m i we have that 0 < h a < O difference between the largest and smallest of the n i + m i is at least /6 log t Ω. Since this is larger than O, we have from results in section 3, that for sufficiently large t, this is impossible. Hence there are at most three solutions in region 3 for sufficiently large t. 9 Conclusions Hence we have proved our result that lim sup { n, m N 0 : n!m! t } 6. Notice that all of our statements about there being at most 6 solutions for sufficiently large t can be made effective, although this was not done in this paper. I do not believe that the effective bound that is achieved would be small enough to allow for a reasonable proof that there are at most 6 solutions for any t, at least without further insight. References [] D. Kane, New bounds on the number of representations of t as a binomial coefficient, Integers [] Hans Rademacher, Topics in Analytic Number Theory Springer-Verlag, Berlin-Heidelberg-New York