4.11 Hankelians 4 137in which the column difference is1
3(v
3
−u
3
).Let the determinant of the elements in the firstnrows and the firstncolumns of the matrix be denoted byAn. Prove thatAn=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 ,whereφm=m
∑r=0(m+1−r)um−r
vrProve thatBn=An+1n!(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− 1i− 1)(
n+i− 1n− 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:in 123 451 12 − 263 3 − 24 304 − 460 − 180 1405 5 − 120 630 −1120 630