Andrzej Nowicki Toru´n, 11.10.2017
1 Introduction
A Hankel matrix, named after Hermann Hankel (1839 − 1873), is a square matrix in which each ascending skew-diagonal from left to right is constant ([1]), e.g.:
a b c b c d c d e
,
a b c d
b c d e c d e f d e f g
.
If a = (an)n>1 is a sequence, and n is a positive integer, then the determinant
a1 a2 · · · an a2 a3 · · · an+1
... ... ... an an+1 · · · a2n−1
is called the n-th Hankel determinant of a. In [5], we described Hankel determinants of several sequences associated with arithmetic functions. In this article we present other Hankel determinants of some sequences. First we consider polynomial sequences, and next we consider sequences connected with several rational functions. In particular, we prove that
1 1·2
1
2·3 . . . n(n+1)1
1 2·3
1
3·4 . . . (n+1)(n+2)1
... ... ...
1 n(n+1)
1
(n+1)(n+2) . . . (2n−1)2n1
= Ψ2n−1Ψ2n Ψ2n ,
where Ψm is the product 1!2! · · · m!.
2 Polynomials and determinants
In this section R[x] and R[x, y] are the polynomial rings over an arbitrary commu- tative ring R with identity. The following theorem appears in many books.
Theorem 2.1. ([2] 210, [6] 356). Let f1(x), . . . , fn(x) ∈ R[x], s = max
deg f1(x), . . . , deg fn(x) ,
and let b1, . . . , bn ∈ R. Let A be the n × n matrix [fi(bj)], that is,
A =
f1(b1) f1(b2) · · · f1(bn) f2(b1) f2(b2) · · · f2(bn)
... ... ...
fn(b1) fn(b2) · · · fn(bn)
.
If n > s + 1, then det A = 0.
Proof. Put fi(x) = ai0+ ai1x1+ ai2x2+ · · · + aisxs for i = 1, . . . , n, and let
Aj =
f1(bj) f2(bj)
... fn(bj)
(for j = 1, . . . , n), Bk =
a1k a2k ... ank
(for k = 0, . . . , s).
The matrices A1, . . . , Anare the successive columns of A, that is, A = [A1, . . . , An]. For every j ∈ {1, . . . , n}, we have the matrix equality Aj = b0jB0+ b1jB1+ b2jB2· · · + bsjBs. Hence,
det A = det [A1, A2, . . . , An]
= det
s P
k1=0
bk11Bk1,
s
P
k2=0
bk22Bk2, . . . ,
s
P
kn=0
bknnBkn
=
s
P
k1=0 s
P
k2=0
· · ·
s
P
kn=0
bk11bk22· · · bnkn · det [Bk1, Bk2, . . . , Bkn] .
Since all indexes k1, . . . , kn belong to the set {0, 1, . . . , s} and n > s + 1, every matrix of the form [Bk1, Bk2, . . . , Bkn] has two identical columns, so the determinant of such matrix is equal to zero. Therefore, det A = 0.
Theorem 2.2. Let F (x, y) ∈ R[x, y] and s = min
degxF (x, y), degyF (x, y)
, and let a1, . . . , an, b1, . . . , bn be elements belonging to R. Let B be the n × n matrix [F (ai, bj)], that is,
B =
F (a1, b1) F (a1, b2) · · · F (a1, bn) F (a2, b1) F (a2, b2) · · · F (a2, bn)
... ... ...
F (an, b1) F (an, b2) · · · F (an, bn)
.
If n > s + 1, then det B = 0.
Proof. Assume that s = degyF (x, y) and put fi(y) := F (ai, y) for i = 1, . . . , n.
Then f1(y), . . . , fn(y) ∈ R[y], and s > max
deg f1(y), . . . , deg fn(y)
. In this case the matrix B is equal to the matrix [fi(bj)]. Now, by Theorem 2.1, we have det B = 0.
By a similar way we obtain the same in the case when s = degxF (x, y).
Theorem 2.3. Assume that F (x, y) ∈ R[x, y] and s = min
degxF (x, y), degyF (x, y) . Let α, β : R → R be functions, and let u1, . . . , un, v1, . . . , vn be elements belonging to R. Let C be the n × n matrix [F (α(ui), β(vj)], that is,
C =
F
α(u1), β(v1) F
α(u1), β(v2)
· · · F
α(u1), β(vn) F
α(u2), β(v1) F
α(u2), β(v2)
· · · F
α(u2), β(vn)
... ... ...
F
α(un), β(v1) F
α(un), β(v2)
· · · F
α(un), β(vn)
.
If n > s + 1, then det C = 0.
Proof. Let ai := α(u1), bi := β(vi) for i = 1, . . . , n. Then the matrix C is equal to the matrix [F (ai, bj)] and this assertion follows from Theorem 2.2.
2.1 Polynomial Hankel determinants
Theorem 2.4. Let f (x) be a polynomial in one variable, and let H(f )n be the deter- minant of the (n + 1) × (n + 1) Hankel matrix [f (i + j)](06i,j6n), that is,
Hn(f ) =
f1(0) f (1) · · · f (n) f (1) f (2) · · · f (n + 1)
... ... ...
f (n) f (n + 1) · · · f (n + n) .
If n > deg f , then Hn(f ) = 0.
Proof. Put F (x, y) = f (x + y), and ai = bi = i − 1 for all i = 1, 2, . . . , n + 1.
Then this assertion follows from Theorem 2.2.
Theorem 2.5. Let f (x) be a polynomial in one variable, and let Dnbe the determinant of the n × n Hankel matrix [f (i + j − 1)](16i,j6n), that is,
Dn=
f1(1) f (2) · · · f (n) f (2) f (3) · · · f (n + 1)
... ... ...
f (n) f (n + 1) · · · f (n + n) .
If n > 1 + deg f , then Dn = 0.
Proof. Put F (x, y) = f (x + y − 1), and ai = bi = i for all i = 1, 2, . . . , n. Then this assertion follows from Theorem 2.2.
2.2 Examples
The following examples are consequences of the above theorems.
2.6. Let an = a + (n − 1)b be an arithmetic progression, and let
Dn =
a1 a2 · · · an a2 a3 · · · an+1
... ... ... an an+1 · · · a2n−1
.
Then D1 = a, D2 = −b2, and Dn= 0 for n > 3.
2.7. Dn =
1 2 3 · · · n
2 3 4 · · · n + 1
... ... ... ... n n + 1 n + 2 · · · 2n − 1
;
D1 = 1, D2 = −1,
Dn= 0 for n > 3.
2.8. Dn =
12 22 · · · n2 22 33 · · · (n + 1)2
... ... ...
n2 (n + 1)2 · · · (2n − 1)2
;
D1 = 1, D2 = −7, D3 = −8
Dn= 0 for n > 4.
2.9. Dn =
13 23 · · · n3 23 33 · · · (n + 1)3
... ... ...
n3 (n + 1)3 · · · (2n − 1)3
;
D1 = 1, D2 = −37, D3 = −756, D4 = 1296 = 2434, Dn= 0 for n > 5.
2.10. Let An = det h
ti+j
i
16i,j6n and Bn = det h
ti + tj
i
16i,j6n, where each tm is the m-th triangular number, that is, tm = 12m(m + 1). Then
A1 = 3, A2 = −6, A3 = −1,
An= 0 for n > 4.
B1 = 2, B2 = −4,
Bn = 0 for n > 3.
2.11. Let An = deth Ti+ji
16i,j6n
and Bn = deth
Ti+ Tji
16i,j6n
, where each Tm is the m-th tetrahedral number, that is, Tm= 16m(m + 1)(m + 2). Then
A1 = 4, A2 = −20, A3 = −20, A4 = 1,
An= 0 for n > 5.
B1 = 2, B2 = −9,
Bn = 0 for n > 3.
Next examples are consequences of Theorems 2.2 and 2.3.
2.12. Dn =
a1+ b1 a1+ b2 · · · a1+ bn a2+ b1 a2+ b2 · · · a2+ bn
... ... ...
an+ b1 an+ b2 · · · an+ bn
;
D1 = a1+ b1,
D2 = (a2− a1)(b1− b2), Dn= 0 for n > 3.
Theorem 2.2 for F (x, y) = x + y
.
2.13. Dn =
a1− b1 a1− b2 · · · a1− bn a2− b1 a2− b2 · · · a2− bn
... ... ...
an− b1 an− b2 · · · an− bn
;
D1 = a1− b1,
D2 = (a2 − a1)(b2− b1), Dn= 0 for n > 3.
Theorem 2.2 for F (x, y) = x − y
. 2.14. Let
Dn =
(a1+ b1)2 (a1+ b2)2 · · · (a1+ bn)2 (a2+ b1)2 (a2+ b2)2 · · · (a2+ bn)2
... ... ...
(an+ b1)2 (an+ b2)2 · · · (an+ bn)2 .
Then, by Theorem 2.2 for F (x, y) = (x + y)2, we have:
D1 = (a1+ b1)2,
D2 = (a2− a1)(b1 − b2)(a1b1 + a1b2+ a2b1+ a2b2+ 2a1a2+ 2b1b2), D3 = −2(a2− a1)(a3− a1)(a3− a2)(b2− b1)(b3− b1)(b3− b2), Dn = 0 for n > 4.
2.15.
a2 (a + 1)2 (a + 2)2 (a + 3)2 b2 (b + 1)2 (b + 2)2 (b + 3)2 c2 (c + 1)2 (c + 2)2 (c + 3)2 d2 (d + 1)2 (d + 2)2 (d + 3)2
= 0.
2.16.([2] 293).
(a0+ b0)n (a0+ b1)n · · · (a0+ bn)n (a1+ b0)n (a1+ b1)n · · · (a1+ bn)n
... ... ...
(an+ b0)n (an+ b1)n · · · (an+ bn)n
= n1 n
2 · · · nn Y
i>k
(ai− ak)(bk− bi).
2.17. Dn =
a21+ b21 a21+ b22 · · · a21+ b2n a22+ b21 a22+ b22 · · · a22+ b2n
... ... ...
a2n+ b21 a2n+ b22 · · · a2n+ b2n
;
D1 = a21+ b21,
D2 = (a22− a21)(b21− b22), Dn= 0 for n > 3.
Theorem 2.3 for F (x, y) = x + y .
2.18. Dn =
1 + a1b1 1 + a1b2 · · · 1 + a1bn 1 + a2b1 1 + a2b2 · · · 1 + a2bn
... ... ...
1 + anb1 1 + anb2 · · · 1 + anbn
;
D1 = 1 + a1b1,
D2 = (a2− a1)(b2− b1), Dn = 0 for n > 3.
Theorem 2.3, F (x, y) = 1 + xy .
2.19. Dn =
a31+ b31 a31+ b32 · · · a31+ b3n a32+ b31 a32+ b32 · · · a32+ b3n
... ... ...
a3n+ b31 a3n+ b32 · · · a3n+ b3n
;
D1 = a31+ b31,
D2 = (a32− a31)(b31− b32), Dn= 0 for n > 3.
Theorem 2.3 for F (x, y) = x + y
.
3 Upper factorials and determinants
If a is an element of a commutative with unity ring R and k is a nonnegative integer, then we denote by a(k) the polynomial of R defined by
a(k) =
( 1, for k = 0,
a(a + 1)(a + 2) · · · (a + k − 1), for k > 1.
In particular, a(0) = 1, a(1) = a, a(2) = a2+ a, a(3) = a(a + 1)(a + 2) = a3+ 3a2+ 2a.
Note the following obvious identities:
Lemma 3.1.
(1) (a + 1)(k)− a(k) = k(a + 1)k−1, (2) 1
a(k) − 1
(a + 1)(k) = k
a(k+1) for k > 1.
For every p > 1, consider the determinant
Wp(x) =
(x + 1)(0) (x + 2)(0) · · · (x + p)(0) (x + 1)(1) (x + 2)(1) · · · (x + p)(1) (x + 1)(2) (x + 2)(2) · · · (x + p)(2)
... ... ...
(x + 1)(p−1) (x + 2)(p−1) · · · (x + p)(p−1)
.
It is the determinant of a (p × p) polynomial matrix.
Lemma 3.2. Wp(x) = (p − 1)! · Wp−1(x + 1) for all p > 1.
Proof. Starting from the second column, we subtract the previous column and, using Lemma 3.1, we obtain that Wp(x) is equal to
1 0 0 · · · 0
(x + 1)(1) (x + 2)(0) (x + 3)(0) · · · (x + p)(0) (x + 1)(2) 2(x + 2)(1) 2(x + 3)(1) · · · 2(x + p)(1) (x + 1)(3) 3(x + 2)(2) 3(x + 3)(2) · · · 3(x + p)(2)
... ... ... ...
(x + 1)(p−1) (p − 1)(x + 2)(p−2) (p − 1)(x + 3)(p−2) · · · (p − 1)(x + p)(p−2)
= (p − 1)!
(x + 2)(0) (x + 3)(0) · · · (x + p)(0) (x + 2)(1) (x + 3)(1) · · · (x + p)(1) (x + 2)(2) (x + 3)(2) · · · (x + p)(2)
... ... ...
(x + 2)(p−2) (x + 3)(p−2) · · · (x + p)(p−2)
= (p − 1)! · Wp−1(x + 1).
This completes the proof.
Hence, Wp(x) = (p − 1)! · Wp−1(x + 1) = (p − 1)!(p − 2)! · Wp−2(x + 2) and, in general, Wp(x) = 0!1!2! · · · (p − 1)! for every p > 1. We proved:
Proposition 3.3.
Wp(x) =
(x + 1)(0) (x + 2)(0) · · · (x + p)(0) (x + 1)(1) (x + 2)(1) · · · (x + p)(1) (x + 1)(2) (x + 2)(2) · · · (x + p)(2)
... ... ...
(x + 1)(p−1) (x + 2)(p−1) · · · (x + p)(p−1)
= 0!1!2! · · · (p − 1)! .
Let us denote by Ψn the product 0!1!2! · · · n!. Thus, Wp(x) = Ψp−1. Now we use the above proposition and we obtain:
Proposition 3.4.
1(1) 2(1) · · · n(1) 1(2) 2(2) · · · n(2) ... ... ... 1(n) 2(n) · · · n(n)
= 1!2! · · · n! = Ψn .
Proof. Denote this determinant by An. Observe that
An=
1 1 1 · · · 1
0 1(1) 2(1) · · · n(1) 0 1(2) 2(2) · · · n(2) ... ... ... 0 1(n) 2(n) · · · n(n)
=
0(0) 1(0) 2(0) · · · n(0) 0(1) 1(1) 2(1) · · · n(1) 0(2) 1(2) 2(2) · · · n(2) ... ... ... 0(n) 1(n) 2(n) · · · n(n)
= Wn+1(−1).
We know, by Proposition 3.3, that Wn+1(−1) = Ψn. Hence, An = Ψn= 1!2! · · · n!. Proposition 3.5. ([7], [1]).
(0 + 0)! (0 + 1)! · · · (0 + n)!
(1 + 0)! (1 + 1)! · · · (1 + n)!
(2 + 0)! (2 + 1)! · · · (2 + n)!
... ... ...
(n + 0)! (n + 1)! · · · (n + n)!
=
0!1!2! · · · n!2
= Ψ2n.
Proof. Denote this determinant by Bn. Observe that Bn= ΨnCn, where
Cn=
(0+0)!
0!
(0+1)!
1!
(0+2)!
2! · · · (0+n)!n!
(1+0)!
0!
(1+1)!
1!
(1+2)!
2! · · · (1+n)!n!
(2+0)!
0!
(2+1)!
1!
(2+2)!
2! · · · (2+n)!n!
... ... ...
(n+0)!
0!
(n+1)!
1!
(n+2)!
2! · · · (n+n)!n!
=
1(0) 2(0) · · · (n + 1)(0) 1(1) 2(1) · · · (n + 1)(1) 1(2) 2(2) · · · (n + 1)(2)
... ... ... 1(n) 2(n) · · · (n + 1)(n)
= Wn+1(0).
It follows from Proposition 3.3 that Cn= Ψn. Therefore, Bn = ΨnCn= Ψ2n. Using the same proofs we obtain
3.6.
(1 + 1)! (1 + 2)! · · · (1 + n)!
(2 + 1)! (2 + 2)! · · · (2 + n)!
(3 + 1)! (3 + 2)! · · · (3 + n)!
... ... ...
(n + 1)! (n + 2)! · · · (n + n)!
=
0!1!2! · · · (n−1)!2
n!(n+1)! = Ψn−1Ψn+1.
3.7.
1! 2! · · · n!
2! 3! · · · (n + 1)!
3! 4! · · · (n + 2)!
... ... ...
n! (n + 1)! · · · (2n − 1)!
=
0!1!2! · · · (n − 1)!2
n! = Ψn−1Ψn .
3.8.([7], [1]). Let D(n) denote the number of derangements of an n-element set, that is, the number of permutations without fixed points. Then
deth
D(i + j)i
06i,j6n
= deth
(i + j)!i
06i,j6n
=
n
Y
k=0
k!
!2
= Ψ2n.
Every element of the form a(k) is called a upper factorial of a. It is well known (see for example [4]) that
(a + b)(n)=
n
X
k=0
n k
a(k)b(n−k) for all n > 0 and for all elements a, b from a commutative ring,
4 Determinants H(n,k)
Let n, k be positive integers. We denote by H(n, k) the determinant of the n × n matrix h
1 (i+j−1)(k)
i
16i,j6n, that is,
H(n, k) =
1 1(k)
1
2(k) . . . n(k)1
1 2(k)
1
3(k) . . . (n+1)1 (k)
... ... ...
1 n(k)
1
(n+1)(k) . . . (2n−1)1 (k)
.
We present a method for calculations of H(n, k). First we explain this method for n = 4. Put A1 = H(4, k). Thus
A1 =
1 1(k)
1 2(k)
1 3(k)
1 4(k) 1
2(k) 1 3(k)
1 4(k)
1 5(k) 1
3(k) 1 4(k)
1 5(k)
1 6(k) 1
4(k) 1 5(k)
1 6(k)
1 7(k)
.
From the first row we subtract the second row. From the second row we subtract the third row, and from the third row we subtract the fourth row. Using the identity
1
a(k) − (a+1)1(k) = a(k+1)k (see Lemma 3.1), we see that A1 = k3A2, where
A2 =
1 1(k+1)
1 2(k+1)
1 3(k+1)
1 4(k+1) 1
2(k+1) 1 3(k+1)
1 4(k+1)
1 5(k+1) 1
3(k+1) 1 4(k+1)
1 5(k+1)
1 6(k+1) 1
4(k) 1 5(k)
1 6(k)
1 7(k)
.
Now from the first row we subtract the second row, from the second row we subtract the third row, and we have A2 = (k + 1)2A3 where
A3 =
1 1(k+2)
1 2(k+2)
1 3(k+2)
1 4(k+2) 1
2(k+2) 1 3(k+2)
1 4(k+2)
1 5(k+2) 1
3(k+1) 1 4(k+1)
1 5(k+1)
1 6(k+1) 1
4(k) 1 5(k)
1 6(k)
1 7(k)
.
Next, from the first row we subtract the second row, and we have A3 = (k + 2)A4, where
A4 =
1 1(k+3)
1 2(k+3)
1 3(k+3)
1 4(k+3) 1
2(k+2) 1 3(k+2)
1 4(k+2)
1 5(k+2) 1
3(k+1) 1 4(k+1)
1 5(k+1)
1 6(k+1) 1
4(k) 1 5(k)
1 6(k)
1 7(k)
=
0!
(k+3)!
1!
(k+4)!
2!
(k+5)!
3!
(k+6)!
1!
(k+3)!
2!
(k+4)!
3!
(k+5)!
4!
(k+6)!
2!
(k+3)!
3!
(k+4)!
4!
(k+5)!
5!
(k+6)!
3!
(k+3)!
4!
(k+4)!
5!
(k+5)!
6!
(k+6)!
.
Now we use Proposition 3.5, and we obtain
(k + 3)!(k + 4)!(k + 5)!(k + 6)!A4 =
0! 1! 2! 3!
1! 2! 3! 4!
2! 3! 4! 5!
3! 4! 5! 6!
=
1!2!3!2
.
So, A4 =
1!2!3!
2
(k+3)!(k+4)!(k+5)!(k+6)!. Thus we have
H(4, k) =
k3(k + 1)2(k + 2)1·
1!2!3!2
(k + 3)!(k + 4)!(k + 5)!(k + 6)!. Theorem 4.1. H(n, k) = ab, where
a =
n−1
Y
j=1
(k + j − 1)n−j
! n−1 Y
j=1
j!
!2
, b =
n
Y
j=1
(k + n − 2 + j)!.
Proof. We use the same method as in the case n = 4. Put A1 = H(n, k). Thus
A1 =
1 1(k)
1
2(k) . . . n1(k)
1 2(k)
1
3(k) . . . (n+1)1 (k)
... ... ...
1 n(k)
1
(n+1)(k) . . . (2n−1)1 (k)
.
From the first row we subtract the second row. From the second row we subtract the third row, and so on. Using the identity a(k)1 −(a+1)1 (k) = a(k+1)k (see Lemma 3.1), we see that A1 = kn−1A2, where
A2 =
1 1(k+1)
1
2(k+1) . . . n(k+1)1
1 2(k+1)
1
3(k+1) . . . (n+1)1(k+1)
... ... ...
1 (n−1)(k+1)
1
n(k+1) . . . (2n−2)1(k+1)
1 n(k)
1
(n+1)(k) . . . (2n−1)1 (k)
.
Now from the first row we subtract the second row, from the second row we subtract the third row, and so on. Then we have A2 = (k + 1)n−2A3 where
A3 =
1 1(k+2)
1
2(k+2) . . . n(k+2)1
1 2(k+2)
1
3(k+2) . . . (n+1)1(k+2)
... ... ...
1 (n−2)(k+2)
1
(n−1)(k+2) . . . (2n−3)1(k+2)
1 (n−1)(k+1)
1
n(k+1) . . . (2n−2)1(k+1)
1 n(k)
1
(n+1)(k) . . . (2n−1)1 (k)
.
We repeat this procedure several times, and we obtain the identity A1 = cB, where c =
n−1
Q
j=1
(k + j − 1)n−j, and
B =
1 1(k+n−1)
1
2(k++n−1) . . . n(k+n−1)1
1 2(k+n−2)
1
3(k+n−2) . . . (n+1)(k+n−2)1
... ... ...
1 (n−2)(k+2)
1
(n−1)(k+2) . . . (2n−3)1(k+2)
1 (n−1)(k+1)
1
n(k+1) . . . (2n−2)1(k+1)
1 n(k)
1
(n+1)(k) . . . (2n−1)1 (k)
=
0!
(k+n−1)!
1!
(k+n)! . . . (k+2n−2)!(n−1)!
1!
(k+n−1)!
2!
(k+n)! . . . (k+2n−2)!n!
... ... ...
(n−3)!
(k+n−1)!
(n−2)!
(k+n)! . . . (k+2n−2)!(2n−4)!
(n−2)!
(k+n−1)!
(n−1)!
(k+n)! . . . (k+2n−2)!(2n−3)!
(n−1)!
(k+n−1)!
n!
(k+n)! . . . (k+2n−2)!(2n−2)!
Now we use Proposition 3.5, and we obtain
bB =
0! 1! . . . (n − 1)!
1! 2! . . . n!
... ... ... (n − 1)! n! . . . (2n − 2)!
=
n−1
Y
j=1
j!
!2
,
where b =
n
Q
j=1
(k + n − 2 + j)!. Put u =
n−1
Q
j=1
j!
!2
. Then we have
H(n, k) = A1 = cB = uc b = a
b, where a = c · u =
n−1
Q
j=1
(k + j − 1)n−j
! n−1
Q
j=1
j!
!2
. This completes the proof.
4.1 The case k = 1
Proposition 4.2. ([3] 3.28.5, [6] 418, [2] 333).
1 1
1
2 . . . n1
1 2
1
3 . . . n+11 ... ... ...
1 n
1
n+1 . . . 2n−11
=
1!2! · · · (n − 1)!3
n!(n + 1)!(n + 2)! · · · (2n − 1)! = Ψ4n−1 Ψ2n−1 .
Denote this determinant by Dn. The first few values: D1 = 1, D2 = 121, D3 = 21601 , D4 = 60480001 , D5 = 2667168000001 . Every Dn is of the form c1
n, where the sequence (cn) is defined by
c1 = 1 and cn+1=2n n
2
(2n + 1)cn for n > 1.
Proof. Use Theorem 4.1 for k = 1. The assertion concerning cn is obvious. Now we present a second proof. The above proposition is a special case of the following, more general, theorem.
Theorem 4.3. ([6] 416, [2] 323).
1 a1+b1
1
a1+b2 . . . a 1
1+bn
1 a2+b1
1
a2+b2 . . . a 1
2+bn
... ... ...
1 an+b1
1
an+b2 . . . a 1
n+bn
= Q
16i<k6n
(ai − ak)(bi− bk)
n
Q
i=1 n
Q
k=1
(ai+ bk) .
Proof. Let A be this determinant. From every row, except first, we subtract the first row, and next from every column, except first, we subtract the first column. Then we obtain the equality A = abcdB, where
a = (a1− a2)(a1 − a3) · · · (a1− an), b = (b2− b1)(b3− b1) · · · (bn− b1), c = (a1+ b1)(a1+ b2) · · · (a1 + bn), d = (a2+ b1)(a3+ b1) · · · (an+ b1),
B =
1 a2+b2
1
a2+b3 . . . a 1
2+bn
1 a3+b2
1
a3+b3 . . . a 1
3+bn
... ... ...
1 an+b2
1
an+b3 . . . a 1
n+bn
.
Repeating this procedure several times, we are done.
Second Proof of Proposition 4.2. Use Theorem 4.3 for the sequences (a1, . . . , an) and (b1, . . . , bn) defined by ai = i − 1, bi = i, i = 1, . . . , n.
4.4.
1 1+1
1
1+2 . . . 1+n1
1 2+1
1
2+2 . . . 2+n1 ... ... ...
1 n+1
1
n+2 . . . n+n1
= n!
1!2! · · · (n − 1)!3
(n + 1)!(n + 2)! · · · (2n)!.
Denote this determinant by Bn. Then we have: B1 = 12, B2 = 721 , B3 = 432001 , B4 = 4233600001 , B5 = 672126336000001 .
Proof. Use Theorem 4.3 for ai = bi = i, i = 1, . . . , n.
4.2 The case k = 2
Proposition 4.5.
1 1·2
1
2·3 . . . n(n+1)1
1 2·3
1
3·4 . . . (n+1)(n+2)1
... ... ...
1 n(n+1)
1
(n+1)(n+2) . . . (2n−1)2n1
=
1!2! · · · n!
1!2! · · · (n − 1)!2
(n + 1)!(n + 2)! · · · (2n)! = Ψ2n−1Ψ2n Ψ2n .
Denote this determinant by Dn. We have: D1 = 12, D2 = 721, D3 = 432001 , D4 =
1
423360000, D5 = 672126336000001 . Every Dn is of the form c1
n, where the sequence (cn) is defined by
c1 = 2 and cn+1 =2n + 1 n
2
(2n + 2)cn for n > 1.
Proof. Use Theorem 4.1 for k = 2. The assertion concerning cn is obvious. Let tm be the m-th triangular number, that is, tm = m(m+1)2 . As a consequence of the above proposition we obtain:
Proposition 4.6.
1 t1
1
t2 . . . t1
n
1 t2
1
t3 . . . t1 .. n+1
. ... ...
1 tn
1
tn+1 . . . t 1
2n−1
= 2n·
1!2! · · · n!
1!2! · · · (n − 1)!2
(n + 1)!(n + 2)! · · · (2n)! = 2n·Ψ2n−1Ψ2n Ψ2n .
Denote this determinant by Dn. We have: D1 = 1, D2 = 181, D3 = 54001 , D4 = 264600001 , D5 = 21003948000001 .
4.3 The case k = 3
Proposition 4.7.
1 1·2·3
1
2·3·4 . . . n(n+1)(n+2)1 1
2·3·4
1
3·4·5 . . . (n+1)(n+2)(n+3)1
... ... ...
1 n(n+1)(n+2)
1
(n+1)(n+2)(n+3) . . . (2n−1)2n(2n+1)1
= Ψ2n−1Ψ2n+1 2nΨ2n+1 .
Denote this determinant by Dn. We have: D1 = 16, D2 = 9601 , D3 = 30240001 , D4 =
1
170698752000, D5 = 1656119291904000001 . Every Dn is of the form c1
n, where the sequence (cn) is defined by
c1 = 6 and cn+1 = 22n + 2 n
2
(2n + 3)cn for n > 1.
Proof. Use Theorem 4.1 for k = 3. The assertion concerning cn is obvious. Let Tmbe the m-th tetrahedral number, that is, tm = m(m+1)(m+2)
6 . As a consequence of the above proposition we obtain:
Proposition 4.8.
1 T1
1
T2 . . . T1
n
1 T2
1
T3 . . . T1 .. n+1
. ... ...
1 Tn
1
Tn+1 . . . T 1
2n−1
= 3n· Ψ2n−1Ψ2n+1 Ψ2n+1 .
Acknowledgments. The author would like to thank Prof. Stanis law Spodzieja and Mr. Miko laj Marciniak for many valuable scientific discussions and comments concerning the determinats H(n, 2).
References
[1] R. Ehrenborg, The Hankel determinant of exponential polynomials, The American Mathematical Monthly, 107(6) (2000), 557-560.
[2] D. K. Faddeev, I. S. Sominsky, Problems in Higher Algebra (Russian), Mir, Moscow, 1968.
[3] L. Je´smanowicz, J. Lo´s, Collection of Problems in Algebra, (in Polish), PWN, Warsaw, 1965.
[4] A. Nowicki, Binomial sequences, Toru´n, 2017.
[5] A. Nowicki, Hankel deterinants with arithemic functions, Toru´n, 2017.
[6] I. V. Proskuryakov, Collection of Exercises in Linear Algebra (Russian), Nauka, Moscow, 1974.
[7] C. Radoux, Det´erminant de Hankel construit sur des polynomes li´es aux nombres de d´erangements, European J. Comibin. 12 (1991) 327-329.
[8] V. A. Sadovnichy, A. A. Grigoryan, S. V. Konyagin, Tasks Student Mathematical Olympiads (Russian), Moscow, 1987.
Nicolaus Copernicus University, Faculty of Mathematics and Computer Science, 87-100 Toru´n, Poland, (e-mail: anow@mat.uni.torun.pl).
