A hyperfactorial divisibility Darij Grinberg *long version*

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1 A hyperfactorial divisibility Darij Grinberg *long version* Let us define a function H : N N by n H n k! for every n N Our goal is to prove the following theorem: Theorem 0 MacMahon We have H b + c H c + a H a + b H a H b H c H a + b + c for every a N, every b N and every c N Remark: Here, we denote by N the set 0,, 2, } and not the set, 2, 3, }, as some authors do Before we come to the proof, first let us make some definitions: Notations [ ] j For any matrix A, we denote by A the entry in the j-th column and i the i-th row of A [This is usually denoted by A ij or by A i,j ] Let R be a ring Let u N and v N, and let a i,j be an element of R for every i, j, 2,, u}, 2,, v} Then, we denote by a i,j j v i u the u v matrix A R u v which satisfies [ ] j A a i i,j for every i, j, 2,, u}, 2,, v} Let R be a commutative ring with unity Let P R [X] be a polynomial Let j N Then, we denote by coeff j P the coefficient of the polynomial P before X j In particular, this implies coeff j P 0 for every j > deg P Thus, for every P R [X] and every d N satisfying deg P d, we have P X d coeff k P X k m If n and m are two integers, then the binomial coefficient n defined by m m m m n +, if n 0; n! n 0, if n < 0 m It is well-known that Z for all n Z and m Z n Q is

2 We recall a fact from linear algebra: Theorem Vandermonde erminant Let R be a commutative ring with unity Let m N Let a, a 2,, a m be m elements of R Then, a j j m i a i a j i m i,j,2,,m} 2 ; We are more interested in a corollary - and generalization - of this fact: Theorem 2 generalized Vandermonde erminant Let R be a commutative ring with unity Let m N For every j, 2,, m}, let P j R [X] be a polynomial such that deg P j j Let a, a 2,, a m be m elements of R Then, m P j a i j m i m coeff j P j a i a j j i,j,2,,m} 2 ; Both Theorems and 2 can be deduced from the following lemma: Lemma 3 Let R be a commutative ring with unity Let m N For every j, 2,, m}, let P j R [X] be a polynomial such that deg P j j Let a, a 2,, a m be m elements of R Then, P j a i j m i m m coeff j P j j a j i j m i m Proof of Lemma 3 For every j, 2,, m}, we have P j X m coeff k P j X k since deg P j j m, since j m Thus, for every i, 2,, m} and j, 2,, m}, we have m m m P j a i coeff k P j a k i a k i coeff k P j a k i coeff k P j k here we substituted k for k in the sum Hence, P j a i j m i m a j j m i coeff i m i P j j m i m 2 according to the definition of the product of two matrices Here is the proof of 2 in more ail: The definition of the product of two matrices yields a j i j m i m This proves 2 coeff i P j j m i m m a k i coeff k P j k 2 P ja i by j m i m P j a i j m i m

3 But the matrix coeff i P j j m i m is upper triangular since coeff i P j 0 for every i, 2,, m} and j, 2,, m} satisfying i > j 2 ; hence, coeff i P j j m i m m coeff j P j since the erminant of an upper j triangular matrix equals the product of its diagonal entries Now, 2 yields a P j a i j m j j m i m i coeff i m i P j j m i m a j j m i coeff i m i P j j m i m j a j i j m i m m coeff j P j j m coeff j P j j m a j coeff j P j i j m i m and thus, Lemma 3 is proven Proof of Theorem For every j, 2,, m}, define a polynomial P j R [X] by P j X j X a k Then, P j is a monic polynomial of degree j k since P j is a product of j monic polynomials of degree each 3 In other words, deg P j j and coeff j P j for every j, 2,, m} Obviously, deg P j j yields deg P j j for every j, 2,, m} Thus, Lemma 3 yields P j a i j m i m m coeff j P j j a j i, j m i m 3 But the matrix P j a i j m i m is lower triangular since P j a i 0 for every i, 2,, m} and j, 2,, m} satisfying i < j 4 ; hence, P j a i j m i m m P j a j since the erminant of a lower triangular matrix equals the product j 2 because i > j yields i > j, thus i > deg P j since deg P j j and therefore coeff i P j 0 3 because X a k is a monic polynomial of degree for every k, 2,, j }, and we have P j X j k X a k 4 because i < j yields i j since i and j are integers and thus P j a i 0 j k a i a k since the factor of the product j k since P j X j k X a k a i a k for k i equals a i a i 0 3

4 of its diagonal entries Thus, 3 rewrites as m m a j P j a j coeff j P j i j j Since m coeff j P j m, this simplifies to j j Thus, a j i j m i m m a j P j a j i j m P j a j j m j m i m j j m i m j k j,2,,m} k,2,,j } k,2,,m}; k<j a j a k j j since P j X X a k yields P j a j a j a k j,2,,m} i,j,2,,m} 2 ; j<i k,2,,m}; k<j a i a j k a j a k j,k,2,,m} 2 ; k<j k a j a k here we renamed j and k as i and j in the product a i a j i,j,2,,m} 2 ; Hence, Theorem is proven Proof of Theorem 2 Lemma 3 yields m a P j a i j m j j m i m coeff j P j i i m j m coeff j P j j Hence, Theorem 2 is proven A consequence of Theorem 2 is the following fact: i,j,2,,m} 2 ; a i a j by Theorem i,j,2,,m} 2 ; a i a j 4

5 Corollary 4 Let R be a commutative ring with unity Let m N Let a, a 2,, a m be m elements of R Then, j j m a i k a i a j k i m i,j,2,,m} 2 ; Proof of Corollary 4 For every j, 2,, m}, define a polynomial P j R [X] by P j X j X k Then, P j is a monic polynomial of degree j k since P j is a product of j monic polynomials of degree each 5 In other words, deg P j j and coeff j P j for every j, 2,, m} Obviously, deg P j j yields deg P j j for every j, 2,, m} Thus, Theorem 2 yields P j a i j m i m Since P j a i j P j X j k j k m coeff j P j j i,j,2,,m} 2 ; a i a j a i k for every i, 2,, m} and j, 2,, m} since X k, this rewrites as a i k k j m i m m coeff j P j j Since m coeff j P j m, this simplifies to j j j j m a i k k i m Hence, Corollary 4 is proven We shall need the following simple lemma: Lemma 5 Let m N Then, i,j,2,,m} 2 ; i,j,2,,m} 2 ; i j H m i,j,2,,m} 2 ; a i a j a i a j 5 because X k is a monic polynomial of degree for every k, 2,, j }, and we have P j X j k X k 5

6 Proof of Lemma 5 We have i,j,2,,m} 2 ; i j i,j 0,,,m } 2 ; i+>j+ i,j 0,,,m } 2 ; since i+>j+ is equivalent to i + j + i j here we substituted i + and j + for i and j in the product i j i,j 0,,,m } 2 ; i 0,,,m } j 0,,,m }; i 0,,,m } i 0,,,m } i 0,,,m } j 0,,,m }; j N; j m and since j 0,,,m } is equivalent to j N and j m i j j N; j m and j N; since the assertion j m and is equivalent to because if, then j m since i 0,,,m } yields i m j N; i j N; j0 j<i i j i j i 0,,,m } j0 i i j here we substituted i j for j in the second product i 0,,,m } } } m i0 H m i! m i0 i! m i j i 0,,,m } j i! k! here we renamed i as k in the product Hence, Lemma 5 is proven Now let us prove Theorem 0: Let a N, let b N and let c N 6

7 We have H a + b + c H b + c H c + a b+c c+a a+b+c H a + b k! a+b Ha+b k! a+b+c ka+b c a + b + i! i k! H a + b a+b+c ka+b here we substituted a + b + i for k in the product, 4 b k! k! Hb b+c kb k! H b b+c kb k! H b k! c b + i! here we substituted b + i for k in the product, 5 a k! k! Ha Next, we show a lemma: c+a ka k! H a c+a ka k! H a i c a + i! here we substituted a + i for k in the product 6 Lemma 6 For every i N and j N satisfying i and j, we have a j + b + i a + b + i! a + i! b + j! a + i k Proof of Lemma 6 Let i N and j N be such that i and j One of the following two cases must hold: Case : We have 0 Case 2: We have < 0 In Case, we have a + i! a+i k k a+i j k a+i j! k j! a + i k k a+i ka+i j+ k k! i a+i ka+i j+ here we substituted a + i k for k in the product, k 7

8 so that Now, a + b + i j a + i!! a + i k 7 k a + b + i!! a + b + i! a + b + i!! b + j! since a + b + i b + j a + b + i! a + i! a + i! b + j!! j a + b + i! a + i! b + j! a + i k by 7 Hence, Lemma 6 holds in Case In Case 2, we have j a + i k k so that 0, a+i k a + i k a+i ka+i k a + i k } } a+i a+i0 since < 0 yields a + i < j a + b + i 0 since < 0 a + b + i! a + i! b + j! 0 j ka+i+ j a+i k k j a + b + i! a + i! b + j! a + i k k a + i k Hence, Lemma 6 holds in Case 2 Hence, in both cases, Lemma 6 holds Thus, Lemma 6 always holds, and this completes the proof of Lemma 6 Another trivial lemma: Lemma 7 Let R be a commutative ring with unity Let u N and v N, and let a i,j be an element of R for every i, j, 2,, u}, 2,, v} a Let α, α 2,, α u be u elements of R Then, αi, if j i; j u i u a i,j j v i u α ia i,j j v i u 8

9 b Let β, β 2,, β v be v elements of R Then, a i,j j v i u βi, if j i; j v i v a i,j β j j v i u c Let α, α 2,, α u be u elements of R Let β, β 2,, β v be v elements of R Then, αi, if j i; j u a i,j j v βi, if j i; i u i u j v i v α i a i,j β j j v i u Thus, d Let α, α 2,, α u be u elements of R Let β, β 2,, β v be v elements of R If u v, then u v α i a i,j β j j v i u α i β i a i,j j v i u i Proof of Lemma 7 a For every i, 2,, u} and j, 2,, v}, we have u l l,2,,u} l,2,,u}; li l,2,,u}; li α i a i,j αi, if l i; 0, if l i αi, if l i; 0, if l i α i, since li α i a l,j + l,2,,u}; l i a l,j i l,2,,u} a l,j + 0 a l,j l,2,,u}; l i αi, if l i; 0, if l i l,2,,u}; li αi, if l i; 0, if l i 0, since l i a l,j a l,j α i a l,j l i} α i a l,j } } 0 since i, 2,, u} yields l, 2,, u} l i} i} αi, if j i; j u a i,j j v i u i u u l αi, if l i; 0, if l i a l,j j v i u α i a i,j j v i u, and thus, Lemma 7 a is proven 9

10 b For every i, 2,, u} and j, 2,, v}, we have v βl, if j l; a i,l a 0, if j l i,l Thus, l l,2,,v} l,2,,v}; lj l,2,,v}; lj a i,j β j a i,l a i,l β l + a i,j j v βi, if j i; i u βl, if j l; 0, if j l β l, since lj yields jl l,2,,v}; l j l,2,,v} + l,2,,v}; l j a i,l 0 a i,l l,2,,v}; lj βl, if j l; 0, if j l βl, if j l; 0, if j l 0, since l j yields j l a i,l β l a i,l β l l j} } } 0 since j, 2,, v} yields l, 2,, v} l j} j} j v i v and thus, Lemma 7 b is proven c We have j u αi, if j i; v a i,l l a i,j j v i u } i u } α i a i,j j v i u by Lemma 7 a α i a i,j j v i u βi, if j i; βl, if j l; 0, if j l βi, if j i; j v i v j v i u j v α i a i,j β j j v i u i v a i,j β j j v i u, by Lemma 7 b applied to α i a i,j instead of a i,j Thus, Lemma 7 c is proven j u αi, if j i; αi, if j i; d The matrix is diagonal since i u 0 for every i, 2,, u} and j, 2,, u} satisfying j i Since the erminant of a diagonal matrix equals the product of its diagonal entries, this yields j u αi, if j i; u αi, if i i; u α 0, if i i i i u i i α i, since ii Similarly, βi, if j i; j v i v v β i i 0

11 Lemma 7 c yields α i a i,j β j j v i u αi, if j i; j u i u a i,j j v i u βi, if j i; j v i v Thus, if u v, then α i a i,j β j j v i u j u j v αi, if j i; a i,j j v i u βi, if j i; i u i v j u αi, if j i; j v a i,j j v βi, if j i; i u i u i v u α i i u v α i β i i i a i,j j v i u v β i i Thus, Lemma 7 d is proven Now, let us prove Theorem 0: Proof of Theorem 0 Let a N, b N and c N We have a j c + b + i i c j j c a + b + i! a + i! b + j! a + i k by Lemma 6 k i c j j c a + b + i! a + i k a + i! b + j! c i a + b + i! a + i! k c i b + i! j i c by Lemma 7 d, applied to R Q, u c, v c, a i,j a + i k k j a + b + i! α i and β i Since a + i! b + i! j j c a + i k a + i a + j k i c i j i,j,2,,c} 2 ; k j c i c a + i k, by Corollary 4, applied to R Z, m c and a i a + i for every i, 2,, c} i j H c by Lemma 5, applied to m c, i,j,2,,c} 2 ;

12 this becomes a + b + i since Now, j c i c c i c a + b + i! H c a + i! b + i! i 8 H a H b H c H a + b + c H b + c H c + a H a + b c H a H b H c H a + b a + b + i! i c c H b b + i! H a a + i! H a + b i by 4, 5 and 6 c a + b + i! i c c H c b + i! a + i! i i c a + b + i! i c c H c a + i! b + i! i i a + b + i! c c i a + i! i b + i! c c a + b + i! H c a + i! b + i! i i a j c + b + i by 8 9 Z i c i j c a + b + i Z c c In other words, i c H b + c H c + a H a + b H a H b H c H a + b + c Thus, Theorem 0 is proven Remarks Theorem 0 was briefly mentioned with a combinatorial interpretation, but without proof on the first page of [] It also follows from the formula H a H b H c H a + b + c 2 in [3] since c a + b + i! i! H b + c H c + a H a + b i a + i! b + i!, or, equivalently, the formula 27 in [4] It is also generalized in [2], Section 429 where one has to consider the limit x 2 We can prove more: 2

13 Theorem 8 For every a N, every b N and every c N, we have H a H b H c H a + b + c H b + c H c + a H a + b a + b + i j c i c We recall a useful fact to help us in the proof: a + b j c i c Theorem 9, the Vandermonde convolution identity Let x Z and y Z Let q Z Then, x + y x y q k q k k Z The sum on the right hand side is an infinite sum, but only finitely many of its addends are nonzero Proof of Theorem 8 For every i, 2,, c} and every j, 2,, c}, we have a + b + i a + b i k k k Z by Theorem 9, applied to x a + b, y i and q a + b i a j + l a j + l l Z l Z here we substituted a j + l for k in the sum a + b i a j + l i l i l Z; 0 i l i is true l Z; 0 i l i l Z; l i since 0 i l i is equivalent to l i l Z; l i l a + b a j + l a + b a j + l i + i l i l + l Z; 0 i l i is false l Z; 0 i l i is false a + b 0 a j + l } } 0 a + b a j + l l Z; l i i l i i l 0, since i 0 and 0 i l i is false a + b i a j + l i l } } i a + b i l a j + l i i a + b c i a + b i l a j + l i l a j + l l i here we replaced the sign by an c sign, since all addends for l > i l l i are zero as 0 for l > i, since i l < 0 for l > i and since c i i l 3

14 Thus, j c a + b + i i c c l j c i a + b i l a j + l i c j c i i j i c a + b a j + i a + b j c i a + b i j i c j c i c j c i c 0 j c i i Now, the matrix is lower triangular since 0 for every i j i c i j i, 2,, m} and j, 2,, m} satisfying i < j 6 Since the erminant of an lower triangular matrix equals the product of its diagonal entries, this yields i j c m j m i j i c j j j } } j j Now, a + b + i j c i c Combined with 9, this yields 0 i j c a + b i j i c by j c i c by 0 i j c a + b i j i c H a H b H c H a + b + c H b + c H c + a H a + b a j c + b + i i c i 6 because i < j yields i j < 0 and thus i j a + b 4 0 j c i c a + b j c i c j c i c

15 Thus, Theorem 8 is proven 3 We notice a particularly known consequence of Corollary 4: Corollary 0 Let m N Let a, a 2,, a m be m integers Then, ai j m H m a i a j 2 j i m i,j,2,,m} 2 ; In particular, H m i,j,2,,m} 2 ; a i a j Proof of Corollary 0 Corollary 4 applied to R Z yields j j m a i k a i a j 3 k i m i,j,2,,m} 2 ; Now, for every i, 2,, m} and j, 2,, m}, we have ai j j j! a i k j! j! j a i k k a i k j a i k here we substituted k for k in the product j a i k a i k j j! k k j! 4 5

16 Therefore, ai j m j a i k j i m j! k m m j j m a i k i! i i k i m a i a j m i,j,2,,m} 2 ; i! so that i by 3 by Lemma 7 d, applied to R Q, u m, v m, a i,j j a i k, α i and β i k i! m a i a j, i! i i,j,2,,m} 2 ; Thus, i,j,2,,m} 2 ; a i a j since ai j Z ai j ai j j m i m j m i m m i! i m k! Hm j m i m here we substituted k for i in the product ai j m H m j H m j m i m i,j,2,,m} 2 ; i m a i a j Z Thus, Corollary 0 is proven Corollary Let m N Let a, a 2,, a m be m integers Then, j m ai H m a i a j j i m i,j,2,,m} 2 ; 6

17 Proof of Corollary The equality 2 applied to a i + instead of a i yields ai + j j m i m H m i,j,2,,m} 2 ; i,j,2,,m} 2 ; a i + a j + a i a j a i a j Since a i + a i for every i, 2,, m}, this rewrites as j m ai H m a i a j j This proves Corollary i m i,j,2,,m} 2 ; References [] Theresia Eisenkölbl, Rhombus tilings of a hexagon with three fixed border tiles, J Combin Theory Ser A , ; arxiv:math/97226v2 [mathco] [2] Percy Alexander MacMahon, Combinatory Analysis, vol 2, Cambridge University Press, 96; reprinted by Chelsea, New York, [3] Christian Krattenthaler, Advanced Determinant Calculus: A Complement, Linear Algebra Appl , 68-66; arxiv:math/ v2 [mathco] [4] Christian Krattenthaler, Advanced Determinant Calculus, S\ eminaire Lotharingien Combin The Andrews Festschrift, paper B42q, 67 pp; arxiv:math/ v3 [mathco] 7 See also: Percy Alexander MacMahon, Combinatory Analysis, vol, Cambridge University Press, 95, Percy Alexander MacMahon, An introduction to Combinatory analysis, Cambridge University Press, 920, introductiontoco00macmrich 7