4.11 Hankelians 4 137
in which the column difference is
1
3
(v
3
−u
3
).
Let the determinant of the elements in the firstnrows and the firstn
columns of the matrix be denoted byAn. Prove that
An=
Knn!
3
(2n)!
(u−v)
n(n+1)
2.Define a HankelianBnas follows:
Bn=
∣
∣
∣
∣
φm
(m+ 1)(m+2)
∣
∣
∣
∣
n
, 0 ≤m≤ 2 n− 2 ,
where
φm=
m
∑
r=0
(m+1−r)u
m−r
v
r
Prove that
Bn=
An+1
n!(u−v)
2 n
,
whereAnis defined in Exercise 1.
4.11 Hankelians 4
Throughout this section,Kn=Kn(0), the simple Hilbert determinant.
4.11.1 v-Numbers
The integersvnidefined by
vni=Vni(0) =
(−1)
n+i
(n+i−1)!
(i−1)!
2
(n−i)!
(4.11.1)
=(−1)
n+i
i
(
n− 1
i− 1
)(
n+i− 1
n− 1
)
, 1 ≤i≤n, (4.11.2)
are of particular interest and will be referred to asv-numbers.
A few values of thev-numbersvniare given in the following table:
i
n 123 45
1 1
2 − 26
3 3 − 24 30
4 − 460 − 180 140
5 5 − 120 630 −1120 630