A hyperfactorial divisibility Darij Grinberg *long version* Let us define a function H : N → N by H (n) = n−1 Y for every n ∈ N. k! k=0 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, 1, 2, ...} (and not the set {1, 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 Aij or by Ai,j .] • Let R be a ring. Let u ∈ N and v ∈ N, and let ai,j be an element of R for every (i, j) ∈ {1, 2, ..., u} × {1, 2, ..., v} . Then, we denote by (ai,j )1≤j≤v 1≤i≤u the u×v u × v matrix A ∈ R which satisfies j A = ai,j for every (i, j) ∈ {1, 2, ..., u} × {1, 2, ..., v} . i • 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 X coeff k (P ) · X k . k=0 m • If n and m are two integers, then the binomial coefficient ∈ Q is n defined by m (m − 1) · · · (m − n + 1) m , if n ≥ 0; . = n! 0, n if n < 0 m It is well-known that ∈ Z for all n ∈ Z and m ∈ Z. n 1 We recall a fact from linear algebra: Theorem 1 (Vandermonde determinant). Let R be a commutative ring with unity. Let m ∈ N. Let a1 , a2 , ..., am be m elements of R. Then, Y 1≤j≤m det aj−1 = (ai − aj ) . i 1≤i≤m (i,j)∈{1,2,...,m}2 ; i>j We are more interested in a corollary - and generalization - of this fact: Theorem 2 (generalized Vandermonde determinant). Let R be a commutative ring with unity. Let m ∈ N. For every j ∈ {1, 2, ..., m}, let Pj ∈ R [X] be a polynomial such that deg (Pj ) ≤ j − 1. Let a1 , a2 , ..., am be m elements of R. Then, ! m Y Y det (Pj (ai ))1≤j≤m = coeff (P ) · (ai − aj ) . j−1 j 1≤i≤m (i,j)∈{1,2,...,m}2 ; i>j j=1 Both Theorems 1 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 ∈ {1, 2, ..., m}, let Pj ∈ R [X] be a polynomial such that deg (Pj ) ≤ j − 1. Let a1 , a2 , ..., am be m elements of R. Then, ! m Y j−1 1≤j≤m det (Pj (ai ))1≤j≤m = coeff (P ) · det a . j−1 j i 1≤i≤m 1≤i≤m j=1 Proof of Lemma 3. For every j ∈ {1, 2, ..., m}, we have Pj (X) = m−1 P coeff k (Pj )· k=0 X k (since deg (Pj ) ≤ j −1 ≤ m−1, since j ≤ m). Thus, for every i ∈ {1, 2, ..., m} and j ∈ {1, 2, ..., m}, we have Pj (ai ) = m−1 X coeff k (Pj ) · aki = k=0 m−1 X aki · coeff k (Pj ) = k=0 m X ak−1 · coeff k−1 (Pj ) (1) i k=1 (here we substituted k − 1 for k in the sum) . Hence, j−1 (Pj (ai ))1≤j≤m 1≤i≤m = ai 1≤j≤m 1≤i≤m · (coeff i−1 (Pj ))1≤j≤m 1≤i≤m (2) (according to the definition of the product of two matrices)1 . 1 Here is the proof of (2) in more detail: The definition of the product of two matrices yields 1≤j≤m aj−1 i 1≤j≤m 1≤i≤m 1≤j≤m · (coeff i−1 (Pj ))1≤i≤m m X = ak−1 · coeff (P ) k−1 j i k=1 | {z } =Pj (ai ) (by (1)) This proves (2). 2 1≤i≤m 1≤j≤m = (Pj (ai ))1≤i≤m . But the matrix (coeff i−1 (Pj ))1≤j≤m 1≤i≤m is upper triangular (since coeff i−1 (Pj ) = 0 2 for every i ∈ {1, 2, ..., m} and j ∈ {1, 2, ..., m} satisfying i > j ); hence, Q m det (coeff i−1 (Pj ))1≤j≤m = coeff j−1 (Pj ) (since the determinant of an upper 1≤i≤m j=1 triangular matrix equals the product of its diagonal entries). Now, (2) yields 1≤j≤m j−1 1≤j≤m det (Pj (ai ))1≤j≤m = det a · (coeff (P )) i−1 j 1≤i≤m i 1≤i≤m 1≤i≤m 1≤j≤m j−1 1≤j≤m = det ai 1≤i≤m · det (coeff i−1 (Pj ))1≤i≤m {z } | m Q = coeff j−1 (Pj ) j=1 = det aj−1 i m 1≤j≤m Y · coeff j−1 (Pj ) 1≤i≤m j=1 = m Y ! coeff j−1 (Pj ) · det 1≤j≤m aj−1 i 1≤i≤m , j=1 and thus, Lemma 3 is proven. Proof of Theorem 1. For every j ∈ {1, 2, ..., m}, define a polynomial Pj ∈ j−1 Q (X − ak ). Then, Pj is a monic polynomial of degree j − 1 R [X] by Pj (X) = k=1 (since Pj is a product of j − 1 monic polynomials of degree 1 each3 ). In other words, deg (Pj ) = j−1 and coeff j−1 (Pj ) = 1 for every j ∈ {1, 2, ..., m}. Obviously, deg (Pj ) = j − 1 yields deg (Pj ) ≤ j − 1 for every j ∈ {1, 2, ..., m}. Thus, Lemma 3 yields ! m Y j−1 1≤j≤m det (Pj (ai ))1≤j≤m = coeff (P ) · det a . (3) j−1 j i 1≤i≤m 1≤i≤m j=1 1≤j≤m But the matrix (Pj (ai ))1≤i≤m is lower triangular (since Pj (ai ) = 0 for every 1≤j≤m 4 i ∈ {1, 2, ..., m} and j ∈ {1, 2, ..., m} satisfying i < j ); hence, det (Pj (ai ))1≤i≤m = m Q Pj (aj ) (since the determinant of a lower triangular matrix equals the product j=1 2 because i > j yields i−1 > j −1, thus i−1 > deg (Pj ) (since deg (Pj ) ≤ j −1) and therefore coeff i−1 (Pj ) = 0 3 because X − ak is a monic polynomial of degree 1 for every k ∈ {1, 2, ..., j − 1}, and we j−1 Q have Pj (X) = (X − ak ) 4 k=1 because i < j yields i ≤ j − 1 (since i and j are integers) and thus Pj (ai ) = j−1 Y (ai − ak ) since Pj (X) = j−1 Y ! (X − ak ) k=1 k=1 =0 (since the factor of the product j−1 Q (ai − ak ) for k = i equals ai − ai = 0) k=1 3 of its diagonal entries). Thus, (3) rewrites as ! m m Y Y 1≤j≤m Pj (aj ) = coeff j−1 (Pj ) · det aj−1 . i 1≤i≤m j=1 Since j=1 m Q m Q coeff j−1 (Pj ) = 1 = 1, this simplifies to {z } j=1 j=1 | =1 m Y Pj (aj ) = det 1≤j≤m aj−1 i 1≤i≤m . j=1 Thus, det aj−1 i 1≤j≤m 1≤i≤m = m Y j−1 Y m Y Pj (aj ) = j=1 j=1 = |{z} Q Q = j∈{1,2,...,m} since Pj (X) = (aj − ak ) k=1 |{z} Q = k∈{1,2,...,m}; k<j k∈{1,2,...,j−1} j−1 Y (X − ak ) yields Pj (aj ) = k=1 = Y Y Y ! (aj − ak ) k=1 Y (aj − ak ) = (aj − ak ) 2 j∈{1,2,...,m} k∈{1,2,...,m}; k<j = j−1 Y (j,k)∈{1,2,...,m} ; k<j (ai − aj ) 2 (i,j)∈{1,2,...,m} ; j<i = (here we renamed j and k as i and j in the product) Y (ai − aj ) . (i,j)∈{1,2,...,m}2 ; i>j Hence, Theorem 1 is proven. Proof of Theorem 2. Lemma 3 yields det (Pj (ai ))1≤j≤m = 1≤i≤m m Y ! coeff j−1 (Pj ) j=1 · det | 1≤j≤m aj−1 i 1≤i≤m {z } Q · Y = (ai −aj ) (i,j)∈{1,2,...,m}2 ; i>j by Theorem 1 = m Y ! coeff j−1 (Pj ) (ai − aj ) . 2 j=1 (i,j)∈{1,2,...,m} ; i>j Hence, Theorem 2 is proven. A consequence of Theorem 2 is the following fact: 4 Corollary 4. Let R be a commutative ring with unity. Let m ∈ N. Let a1 , a2 , ..., am be m elements of R. Then, !1≤j≤m j−1 Y Y = det (ai − k) (ai − aj ) . k=1 (i,j)∈{1,2,...,m}2 ; i>j 1≤i≤m Proof of Corollary 4. For every j ∈ {1, 2, ..., m}, define a polynomial Pj ∈ j−1 Q R [X] by Pj (X) = (X − k). Then, Pj is a monic polynomial of degree j − 1 k=1 (since Pj is a product of j − 1 monic polynomials of degree 1 each5 ). In other words, deg (Pj ) = j−1 and coeff j−1 (Pj ) = 1 for every j ∈ {1, 2, ..., m}. Obviously, deg (Pj ) = j − 1 yields deg (Pj ) ≤ j − 1 for every j ∈ {1, 2, ..., m}. Thus, Theorem 2 yields ! m Y Y (ai − aj ) . det (Pj (ai ))1≤j≤m = · coeff (P ) j−1 j 1≤i≤m (i,j)∈{1,2,...,m}2 ; i>j j=1 Since Pj (ai ) = j−1 Q (ai − k) for every i ∈ {1, 2, ..., m} and j ∈ {1, 2, ..., m} (since k=1 Pj (X) = j−1 Q (X − k)), this rewrites as k=1 j−1 Y det !1≤j≤m (ai − k) k=1 Since = m Y ! coeff j−1 (Pj ) (ai − aj ) . 2 j=1 1≤i≤m Y · (i,j)∈{1,2,...,m} ; i>j m Q m Q coeff j−1 (Pj ) = 1 = 1, this simplifies to {z } j=1 j=1 | =1 det j−1 Y k=1 !1≤j≤m (ai − k) Y = (ai − aj ) . 2 1≤i≤m (i,j)∈{1,2,...,m} ; i>j Hence, Corollary 4 is proven. We shall need the following simple lemma: Lemma 5. Let m ∈ N. Then, Y (i − j) = H (m) . 2 (i,j)∈{1,2,...,m} ; i>j 5 because X − k is a monic polynomial of degree 1 for every k ∈ {1, 2, ..., j − 1}, and we have j−1 Q Pj (X) = (X − k) k=1 5 Proof of Lemma 5. We have Y Y (i − j) = 2 (i + 1) − (j + 1) | {z } 2 (i,j)∈{1,2,...,m} ; i>j (i,j)∈{0,1,...,m−1} ; i+1>j+1 | {z Q =i−j } = (i,j)∈{0,1,...,m−1}2 ; i>j (since i+1>j+1 is equivalent to i>j) (here we substituted i + 1 and j + 1 for i and j in the product) Y (i − j) = (i,j)∈{0,1,...,m−1}2 ; i>j = |Q {z } Q i∈{0,1,...,m−1} j∈{0,1,...,m−1}; i>j Y Y i∈{0,1,...,m−1} j∈{0,1,...,m−1}; i>j = | {z Q = (i − j) } j∈N; j≤m−1 and i>j (since j∈{0,1,...,m−1} is equivalent to (j∈N and j≤m−1)) Y Y i∈{0,1,...,m−1} j∈N; j≤m−1 and i>j = | (i − j) {z } Q = j∈N; i>j (since the assertion (j≤m−1 and i>j) is equivalent to (i>j) (because if i>j, then j≤m−1 (since i∈{0,1,...,m−1} yields i≤m−1))) Y Y i∈{0,1,...,m−1} j∈N; i>j = Y (i − j) = i−1 Y i∈{0,1,...,m−1} j=0 (i − j) = Y = |{z} = j∈N; j<i j i∈{0,1,...,m−1} j=1 =i! |{z} Q i Y i−1 Q j=0 (here we substituted i − j for j in the second product) Y = i! = i∈{0,1,...,m−1} | {z = m−1 Q m−1 Y i=0 i! = m−1 Y k! (here we renamed i as k in the product) k=0 } i=0 = H (m) . Hence, Lemma 5 is proven. Now let us prove Theorem 0: Let a ∈ N, let b ∈ N and let c ∈ N. 6 We have a+b+c−1 Y H (a + b + c) = k=0 a+b−1 a+b+c−1 Y Y Y a+b+c−1 k! = k! = H (a + b) · k! · k! k=0 k=a+b k=a+b | {z } =H(a+b) = H (a + b) · c Y (a + b + i − 1)! i=1 (here we substituted a + b + i − 1 for k in the product) , (4) H (b + c) = b+c−1 Y k=0 b−1 b+c−1 b+c−1 c Y Y Y Y k! = H (b) · k! = H (b) · (b + i − 1)! k! = k! · k=0 k=b i=1 k=b | {z } =H(b) (here we substituted b + i − 1 for k in the product) , H (c + a) = c+a−1 Y k=0 (5) Y a−1 c+a−1 c Y Y Y c+a−1 · k! = H (a) · k! = H (a) · (a + i − 1)! k! = k! k=0 k=a i=1 k=a | {z } =H(a) (here we substituted a + i − 1 for k in the product) . (6) Next, we show a lemma: Lemma 6. For every i ∈ N and j ∈ N satisfying i ≥ 1 and j ≥ 1, we have j−1 Y a+b+i−1 (a + b + i − 1)! · (a + i − k) . = (a + i − 1)! · (b + j − 1)! k=1 a+i−j Proof of Lemma 6. Let i ∈ N and j ∈ N be such that i ≥ 1 and j ≥ 1. One of the following two cases must hold: Case 1: We have a + i − j ≥ 0. Case 2: We have a + i − j < 0. In Case 1, we have (a + i − 1)! = a+i−1 Y k=1 a+i−j a+i−1 a+i−1 Y Y Y k= k · k = (a + i − j)! · k k=1 k=a+i−j+1 k=a+i−j+1 | {z } =(a+i−j)! = (a + i − j)! · j−1 Y (a + i − k) k=1 (here we substituted a + i − k for k in the product) , 7 so that j−1 (a + i − 1)! Y = (a + i − k) . (a + i − j)! k=1 (7) Now, a+b+i−1 (a + b + i − 1)! (a + b + i − 1)! = = a+i−j (a + i − j)! · ((a + b + i − 1) − (a + i − j))! (a + i − j)! · (b + j − 1)! (since (a + b + i − 1) − (a + i − j) = b + j − 1) (a + i − 1)! (a + b + i − 1)! · = (a + i − 1)! · (b + j − 1)! (a + i − j)! j−1 Y (a + b + i − 1)! = · (a + i − k) (by (7)) . (a + i − 1)! · (b + j − 1)! k=1 Hence, Lemma 6 holds in Case 1. In Case 2, we have j−1 Y (a + i − k) = k=1 a+i−1 Y k=1 ! a+i j−1 Y Y (a + i − k) (a + i − k) · (a + i − k) · k=a+i+1 k=a+i {z } | =a+i−(a+i)=0 (since a + i − j < 0 yields a + i < j) = 0, so that a+b+i−1 =0 a+i−j = (since a + i − j < 0) (a + b + i − 1)! · (a + i − 1)! · (b + j − 1)! 0 |{z} = j−1 Q (a+i−k) k=1 j−1 Y (a + b + i − 1)! · (a + i − k) . = (a + i − 1)! · (b + j − 1)! k=1 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 ai,j be an element of R for every (i, j) ∈ {1, 2, ..., u} × {1, 2, ..., v} . (a) Let α1 , α2 , ..., αu be u elements of R. Then, 1≤j≤u αi , if j = i; 1≤j≤v · (ai,j )1≤j≤v 1≤i≤u = (αi ai,j )1≤i≤u . 0, if j 6= i 1≤i≤u 8 (b) Let β1 , β2 , ..., βv be v elements of R. Then, (ai,j )1≤j≤v 1≤i≤u · 1≤j≤v βi , if j = i; 0, if j 6= i = (ai,j βj )1≤j≤v 1≤i≤u . 1≤i≤v (c) Let α1 , α2 , ..., αu be u elements of R. Let β1 , β2 , ..., βv be v elements of R. Then, αi , if j = i; 0, if j 6= i 1≤j≤u βi , if j = i; 0, if j 6= i ·(ai,j )1≤j≤v 1≤i≤u · 1≤i≤u 1≤j≤v = (αi ai,j βj )1≤j≤v 1≤i≤u . 1≤i≤v (d) Let α1 , α2 , ..., αu be u elements of R. Let β1 , β2 , ..., βv be v elements of R. If u = v, then ! ! u v Y Y 1≤j≤v det (αi ai,j βj )1≤i≤u = αi · βi · det (ai,j )1≤j≤v 1≤i≤u . i=1 i=1 Proof of Lemma 7. (a) For every i ∈ {1, 2, ..., u} and j ∈ {1, 2, ..., v}, we have u X `=1 |{z} P = αi , if ` = i; · a`,j = 0, if ` 6= i X `∈{1,2,...,u} αi , if ` = i; · a`,j 0, if ` 6= i `∈{1,2,...,u} X αi , if ` = i; αi , if ` = i; = ·a`,j + ·a`,j 0, if ` 6= i 0, if ` 6= i `∈{1,2,...,u}; | `∈{1,2,...,u}; | {z } {z } X `=i = `6=i =αi , since `=i X X αi a`,j + `∈{1,2,...,u}; `=i 0 · a`,j = αi a`,j = `∈{1,2,...,u}; `=i `∈{1,2,...,u}; `6=i | =0, since `6=i X {z X αi a`,j `∈{i} } =0 (since i ∈ {1, 2, ..., u} yields {` ∈ {1, 2, ..., u} | ` = i} = {i}) = αi ai,j . Thus, αi , if j = i; 0, if j 6= i 1≤j≤u ·(ai,j )1≤j≤v 1≤i≤u = 1≤i≤u u X αi , if ` = i; · a`,j 0, if ` 6= i `=1 and thus, Lemma 7 (a) is proven. 9 !1≤j≤v = (αi ai,j )1≤j≤v 1≤i≤u , 1≤i≤u (b) For every i ∈ {1, 2, ..., u} and j ∈ {1, 2, ..., v}, we have v X X β` , if j = `; β` , if j = `; ai,` · = ai,` · 0, if j 6= ` 0, if j 6= ` `=1 `∈{1,2,...,v} |{z} P = `∈{1,2,...,v} X β` , if j = `; β` , if j = `; + ai,` · = ai,` · 0, if j 6= ` 0, if j 6= ` `∈{1,2,...,v}; | {z } `∈{1,2,...,v}; {z } | X `=j X = `6=j =β` , since `=j yields j=` X ai,` β` + `∈{1,2,...,v}; `=j | X ai,` · 0 = `∈{1,2,...,v}; `6=j =0, since `6=j yields j6=` ai,` β` = `∈{1,2,...,v}; `=j {z X ai,` β` `∈{j} } =0 (since j ∈ {1, 2, ..., v} yields {` ∈ {1, 2, ..., v} | ` = j} = {j}) = ai,j βj . Thus, 1≤j≤v βi , if j = i; 1≤j≤v (ai,j )1≤i≤u · = 0, if j 6= i 1≤i≤v v X ai,` · `=1 β` , if j = `; 0, if j 6= ` !1≤j≤v = (ai,j βj )1≤j≤v 1≤i≤u , 1≤i≤u and thus, Lemma 7 (b) is proven. (c) We have 1≤j≤v 1≤j≤u βi , if j = i; αi , if j = i; 1≤j≤v · (ai,j )1≤i≤u · 0, if j 6= i 0, if j 6= i 1≤i≤v 1≤i≤u {z } | =(αi ai,j )1≤j≤v 1≤i≤u by Lemma 7 (a) = (αi ai,j )1≤j≤v 1≤i≤u · βi , if j = i; 0, if j 6= i 1≤j≤v = (αi ai,j βj )1≤j≤v 1≤i≤u 1≤i≤v by Lemma 7 (b) (applied to αi ai,j instead of ai,j ). Thus, Lemma 7 (c) is proven. 1≤j≤u αi , if j = i; αi , if j = i; (d) The matrix is diagonal (since = 0, if j 6= i 0, if j 6= i 1≤i≤u 0 for every i ∈ {1, 2, ..., u} and j ∈ {1, 2, ..., u} satisfying j 6= i). Since the determinant of a diagonal matrix equals the product of its diagonal entries, this yields 1≤j≤u ! Y u u Y αi , if i = i; αi , if j = i; det = = αi . 0, if i 6= i 0, if j 6= i 1≤i≤u i=1 | i=1 {z } =αi , since i=i Similarly, det βi , if j = i; 0, if j 6= i 10 1≤j≤v ! = 1≤i≤v v Y i=1 βi . Lemma 7 (c) yields 1≤j≤u 1≤j≤v αi , if j = i; βi , if j = i; 1≤j≤v 1≤j≤v (αi ai,j βj )1≤i≤u = · (ai,j )1≤i≤u · . 0, if j 6= i 0, if j 6= i 1≤i≤u 1≤i≤v Thus, if u = v, then det (αi ai,j βj )1≤j≤v 1≤i≤u 1≤j≤u 1≤j≤v ! αi , if j = i; βi , if j = i; = det · (ai,j )1≤j≤v 1≤i≤u · 0, if j 6= i 0, if j 6= i 1≤i≤u 1≤i≤v ! 1≤j≤u 1≤j≤v ! αi , if j = i; βi , if j = i; = det · det (ai,j )1≤j≤v 1≤i≤u · det 0, if j 6= i 0, if j 6= i 1≤i≤u 1≤i≤v | {z } {z } | = u Q = αi i=1 u Y = ! αi · i=1 v Y βi i=1 ! βi v Q · det (ai,j )1≤j≤v 1≤i≤u . i=1 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 1≤j≤c ! a+b+i−1 det a+i−j 1≤i≤c !1≤j≤c j−1 Y (a + b + i − 1)! · (a + i − k) (by Lemma 6) = det (a + i − 1)! · (b + j − 1)! k=1 1≤i≤c !1≤j≤c j−1 Y (a + b + i − 1)! 1 = det · (a + i − k) · (a + i − 1)! (b + j − 1)! k=1 1≤i≤c ! ! !1≤j≤c j−1 c c Y Y Y (a + b + i − 1)! 1 = · · det (a + i − k) (a + i − 1)! (b + i − 1)! i=1 i=1 k=1 1≤i≤c (by Lemma 7 (d), applied to R = Q, u = c, v = c, ai,j = j−1 Q (a + i − k), k=1 (a + b + i − 1)! 1 and βi = ). Since (a + i − 1)! (b + i − 1)! !1≤j≤c j−1 Y Y = (a + i) − (a + j) det (a + i − k) | {z } 2 αi = k=1 = 1≤i≤c (i,j)∈{1,2,...,c} ; i>j =i−j (by Corollary 4, applied to R = Z, m = c and ai = a + i for every i ∈ {1, 2, ..., c}) Y (i − j) = H (c) (by Lemma 5, applied to m = c) , (i,j)∈{1,2,...,c}2 ; i>j 11 this becomes 1≤j≤c ! a+b+i−1 det = a+i−j 1≤i≤c c Y (a + b + i − 1)! i=1 ! (a + i − 1)! · c Y i=1 ! 1 ·H (c) . (b + i − 1)! (8) Now, H (a) H (b) H (c) H (a + b + c) H (b + c) H (c + a) H (a + b) c Q H (a) H (b) H (c) H (a + b) · (a + b + i − 1)! i=1 = c c Q Q H (b) · (b + i − 1)! · H (a) · (a + i − 1)! · H (a + b) i=1 i=1 (by (4), (5) and (6)) c Q (a + b + i − 1)! i=1 c · H (c) = c Q Q (b + i − 1)! · (a + i − 1)! i=1 c Q i=1 (a + b + i − 1)! i=1 c Q · Q c 1 ·H (c) (a + i − 1)! (b + i − 1)! | i=1 {z } |i=1 {z } 1 c c (a + b + i − 1)! Q Q = = i=1 (b + i − 1)! i=1 (a + i − 1)! ! ! c c Y Y (a + b + i − 1)! 1 = · · H (c) (a + i − 1)! (b + i − 1)! i=1 i=1 1≤j≤c ! a+b+i−1 (by (8)) = det a+i−j 1≤i≤c = (9) ∈Z 1≤j≤c a+b+i−1 (since ∈ Zc×c ). In other words, a+i−j 1≤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. 1. Theorem 0 was briefly mentioned (with a combinatorial interpretation, but without proof) on the first page of [1]. It also follows from the formula c (a + b + i − 1)! (i − 1)! Q H (a) H (b) H (c) H (a + b + c) (2.1) in [3] (since = ), H (b + c) H (c + a) H (a + b) i=1 (a + i − 1)! (b + i − 1)! or, equivalently, the formula (2.17) in [4]. It is also generalized in [2], Section 429 (where one has to consider the limit x → 1). 2. We can prove more: 12 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) 1≤j≤c ! 1≤j≤c ! a+b+i−1 a+b = det = det . a+i−j a+i−j 1≤i≤c 1≤i≤c We recall a useful fact to help us in the proof: Theorem 9, the Vandermonde convolution identity. Let x ∈ Z and y ∈ Z. Let q ∈ Z. Then, X 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 ∈ {1, 2, ..., c} and every j ∈ {1, 2, ..., c}, we have X a+b+i−1 a+b i−1 = a+i−j k a+i−j−k k∈Z (by Theorem 9, applied to x = a + b, y = i − 1 and q = a + i − j) X a + b i−1 = a − j + ` a + i − j − (a − j + `) `∈Z (here we substituted a − j + ` for k in the sum) X a + b i − 1 = a−j+` i−` `∈Z X X a+b i−1 + = a−j+` i−` `∈Z; `∈Z; = P {z P a+b a−j+` (0≤i−`≤i−1 is false) (0≤i−`≤i−1 is true) | =0, since i−1≥0 and (0≤i−`≤i−1 is false) } = `∈Z; `∈Z; 0≤i−`≤i−1 1≤`≤i (since 0≤i−`≤i−1 is equivalent to 1≤`≤i) X a + b i − 1 = + a−j+` i−` `∈Z; 1≤`≤i i−1 i−` | {z } X a + b i − 1 a+b ·0= a−j+` i−` a−j+` `∈Z; `∈Z; | {z } 1≤`≤i (0≤i−`≤i−1 is false) a+b | {z } |{z} i − 1 = i =0 P i−` a−j+` = X `=1 = i X `=1 X c i−1 a+b i−1 a+b = i−` a−j+` i−` a−j+` `=1 i c P P here we replaced the sign by an sign, since all addends for ` > i `=1 `=1 i−1 are zero (as = 0 for ` > i, since i − ` < 0 for ` > i) and since c ≥ i i−` 13 . Thus, 1≤j≤c a+b+i−1 = a+i−j 1≤i≤c !1≤j≤c c X i−1 a+b i−` a−j+` `=1 1≤i≤c 1≤j≤c 1≤j≤c i−1 a+b = · i−j j+i 1≤i≤c | a −{z } a+b = a+i−j 1≤i≤c 1≤j≤c 1≤j≤c i−1 a+b = · . (10) i−j a+i−j 1≤i≤c 1≤i≤c 1≤j≤c i−1 i−1 Now, the matrix is lower triangular (since = 0 for every i−j i−j 1≤i≤c i ∈ {1, 2, ..., m} and j ∈ {1, 2, ..., m} satisfying i < j 6 ). Since the determinant of an lower triangular matrix equals the product of its diagonal entries, this yields 1≤j≤c ! Y m m Y j−1 i−1 det = = 1 = 1. (11) j−j i−j 1≤i≤c j=1 | {z } j=1 j−1 = =1 0 Now, det 1≤j≤c ! 1≤j≤c 1≤j≤c ! a+b+i−1 i−1 a+b = det · a+i−j i−j a+i−j 1≤i≤c 1≤i≤c 1≤i≤c (by (10)) 1≤j≤c ! 1≤j≤c ! a+b i−1 = det · det a+i−j i−j 1≤i≤c 1≤i≤c | {z } =1 by (11) = det a+b a+i−j 1≤j≤c ! . 1≤i≤c Combined with (9), this yields H (a) H (b) H (c) H (a + b + c) H (b + c) H (c + a) H (a + b) 1≤j≤c ! 1≤j≤c ! a+b+i−1 a+b = det = det . a+i−j a+i−j 1≤i≤c 1≤i≤c 6 because i < j yields i − j < 0 and thus i−1 i−j 14 =0 Thus, Theorem 8 is proven. 3. We notice a particularly known consequence of Corollary 4: Corollary 10. Let m ∈ N. Let a1 , a2 , ..., am be m integers. Then, 1≤j≤m ! Y ai − 1 det · H (m) = (ai − aj ) . (12) j−1 1≤i≤m 2 (i,j)∈{1,2,...,m} ; i>j In particular, H (m) | Y (ai − aj ) . 2 (i,j)∈{1,2,...,m} ; i>j Proof of Corollary 10. Corollary 4 (applied to R = Z) yields !1≤j≤m j−1 Y Y = det (ai − k) (ai − aj ) . k=1 (13) 2 1≤i≤m (i,j)∈{1,2,...,m} ; i>j Now, for every i ∈ {1, 2, ..., m} and j ∈ {1, 2, ..., m}, we have (j−1)−1 Q ai − 1 = j−1 k=0 ((ai − 1) − k) (j−1)−1 Y 1 = ((ai − 1) − k) (j − 1)! k=0 (j − 1)! j−1 Y 1 (ai − 1) − (k − 1) = {z } (j − 1)! k=1 | =ai −k (here we substituted k − 1 for k in the product) ! j−1 j−1 Y Y 1 1 (ai − k) = 1 · (ai − k) · . = (j − 1)! k=1 (j − 1)! k=1 15 (14) Therefore, !1≤j≤m ! 1≤j≤m ! j−1 Y ai − 1 1 det = det 1 · (ai − k) · j−1 (j − 1)! 1≤i≤m k=1 1≤i≤m ! !1≤j≤m j−1 m m Y Y Y 1 = 1 · · det (ai − k) (i − 1)! i=1 k=1 1≤i≤m } | | i=1 {z |{z} {z } Q =1 1 = (ai −aj ) = m (i,j)∈{1,2,...,m}2 ; Q i>j (i − 1)! (by (13)) i=1 by Lemma 7 (d), applied to R = Q, u = m, v = m, j−1 Q 1 ai,j = (ai − k) , αi = 1 and βi = (i − 1)! k=1 Y 1 = Q · (ai − aj ) , m 2 (i − 1)! (i,j)∈{1,2,...,m} ; i>j i=1 so that Y 1≤j≤m ! Y m ai − 1 · (i − 1)! j−1 1≤i≤m i=1 1≤j≤m ! m−1 Y ai − 1 k! · j−1 1≤i≤m k=0 | {z } (ai − aj ) = det 2 (i,j)∈{1,2,...,m} ; i>j = det =H(m) (here we substituted k for i − 1 in the product) 1≤j≤m ! ai − 1 · H (m) . = det j−1 1≤i≤m Thus, H (m) | Y (ai − aj ) (i,j)∈{1,2,...,m}2 ; i>j 1≤j≤m a − 1 i (since det ∈ Z). Thus, Corollary 10 is proven. j−1 | {z } ∈Z 1≤i≤m Corollary 11. Let m ∈ N. Let a1 , a2 , ..., am be m integers. Then, 1≤j≤m ! Y ai det · H (m) = (ai − aj ) . j − 1 1≤i≤m 2 (i,j)∈{1,2,...,m} ; i>j 16 Proof of Corollary 11. The equality (12) (applied to ai +1 instead of ai ) yields det 1≤j≤m ! (ai + 1) − 1 · H (m) = j−1 1≤i≤m Y 2 (i,j)∈{1,2,...,m} ; i>j Y = ((ai + 1) − (aj + 1)) | {z } =ai −aj (ai − aj ) . 2 (i,j)∈{1,2,...,m} ; i>j Since (ai + 1) − 1 = ai for every i ∈ {1, 2, . . . , m}, this rewrites as 1≤j≤m ! Y ai det · H (m) = (ai − aj ) . j − 1 1≤i≤m 2 (i,j)∈{1,2,...,m} ; i>j This proves Corollary 11. References [1] Theresia Eisenkölbl, Rhombus tilings of a hexagon with three fixed border tiles, J. Combin. Theory Ser. A 88 (1999), 368-378; arXiv:math/9712261v2 [math.CO]. http://arxiv.org/abs/math/9712261v2 [2] Percy Alexander MacMahon, Combinatory Analysis, vol. 2, Cambridge University Press, 1916; reprinted by Chelsea, New York, 1960. http://www.archive.org/details/combinatoryanaly02macmuoft 7 [3] Christian Krattenthaler, Advanced Determinant Calculus: A Complement, Linear Algebra Appl. 411 (2005), 68-166; arXiv:math/0503507v2 [math.CO]. http://arxiv.org/abs/math.CO/0503507v2 [4] Christian Krattenthaler, Advanced Determinant Calculus, S\’eminaire Lotharingien Combin. 42 (1999) (The Andrews Festschrift), paper B42q, 67 pp; arXiv:math/9902004v3 [math.CO]. http://arxiv.org/abs/math/9902004v3 7 See also: • Percy Alexander MacMahon, Combinatory Analysis, vol. 1, Cambridge University Press, 1915, http://www.archive.org/details/combinatoryanal01macmuoft; • Percy Alexander MacMahon, An introduction to Combinatory analysis, Cambridge University Press, 1920, http://www.archive.org/details/ introductiontoco00macmrich. 17