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