A simple approach to some Hankel determinants

In order to prove this we show first that for all , m n ∈

The proof is obvious by induction. For 0 n = and arbitrary m it is trivially true. If it is true for 1 nand arbitrary , m then we get ( ) ( ) ,0)

Then (3) implies .

This Lemma is well-known and intimately connected with the approach to Hankel determinants via orthogonal polynomials (see [6]). It is especially useful if for a given sequence ( ) 0 ( ,0) n a n ≥ explicit expressions for ( ), ( ) s n t n and ( , ) a n k exist. To find such expressions it is often convenient to compute the first values of the orthogonal polynomials ( , ) p n x (cf. e.g. [4] (1.10)) and their Favard resolution [4] (1.11) and try to guess ( ) s n and ( ) t n and the explicit form of ( , ) a n k . Other approaches to this lemma can be found in [1] and [8].

s n = identity (1) reduces to

In this case (2 , 2 1)

then it is easily verified that (1) holds with

Therefore it is convenient to consider first sequences of the form ( )

Let us first look at the situation for the famous example of the Catalan numbers

The table ( )

A n k can be found in [5], A053121. The first terms are 1 0 1 1 0 1 0 2 0 1 2 0 3 0 1 0 5 0 4 0 1 5 0 9 0 5 0 1 0 14 0 14 0 6 0 1

It is well known that ( ) 1

≡ and 0 else. This can also be written in the form

To prove this fact it suffices to verify (5). This is almost trivial:

from the recurrences of the binomial coefficients.

For the sequence ( ,0) .

But we cannot immediately deduce that ( ) We shall later use the trivial fact that ( 1) is equivalent with

.

I will not show here how to guess the following results (this the reader can do for himself using the above hint), but will only verify the relevant identities.

We shall use the usual qnotations, e.g. ,,,) . ;

Condition ( 5) is satisfied if we set

(1 (2 , 2 , , , )

.

We have to show that the identities

and

hold.

The left-hand side of ( 15) is

(1 )(1

c n k q a q b q k q b q q b q q q q q q q q q q q q q q a q q a q q q q -+ +

q b q q q a q a q a q q q q q a q q b q q q q q a q q a q q a q q q q q q a q

q q a q q a q q a q

we see that (15) is true.

In the same way the left-hand side of (16) reduces to

; ; ; ; ; ; ; ; ; ;

(1

n n c n k q a q b q c n k q a q b q k k n q q b q a c n k q a q b q k q a q a q b q q b q q q q q q q q q q q q q q a q q a q q q

q b q q q b q a q a q a q q q q q a q q b q q q q b q a q b q a q b q a q q q q q q a q

This implies ,,) . ;

Then the Hankel determinants ( )

( , , , , ) det ( , , , )

d n m a b q c i j m a b q

are given by 0,,,) ; ;

) ,,,) ( ,0, , , ) . ;

, , ) (2 1, , , ). ,,,) ;

By (19) we get

; ( ,0, , , ) 1 . ( ,0, , , ) 1 ;

; ( ,0, , , ). ;

d n qa qb q a a a q qa q b q q dn abq q a q -+ + +

Iterating this argument we get

)

( ,0, , , ). ;

b q d n m a b q d n q a q b q a q q b q q dn abq q a q

a q b q q q → → → and let 1 q → we get Condition ( 5) is satisfied if we set

and

, ( 1) , ). ,, ) . are given by ( ) ( )

)

m n j i

–

These Corollaries can of course also be proved directly.

Now let us consider again the Catalan numbers. Instead of n C we study Here we get 1 ( ) 4 T n = and find again that

. 4 2

In order to compute the higher Hankel determinants consider the factor ( ) ( )

This reduces in our case to

Therefore we get for the Catalan numbers themselves the well-known result (cf. [7], Theorem 33 and the historical comments given there) ( )

.

Then we get

, 0 0 det ; ; ;

.

As special case we note ( )

,

then we get (

. ( 1)

1 ( 1)

Therefore we have ( ) q q c n c n q q q q q + + = =

Here we get

;

q q q q q q d n q q q q q q q q q q q q q q q q q

This result is due to Carlitz [3]. For some generalizations see Krattenthaler [6], especially Theorem 26.

[ ]

Then (20) gives a qanalogue of the determinant of the Hilbert matrix [ ] ( ) ( )

(1 )

n n n j j j q j n q c n c n q q q q n q q n

Here we get

n n n n j n j j d n q q q q q –

) ,,,) ( ,0, , , ). ; n j m m n n j j n q q d n m q q q q d n q q q q q

The right-hand side can be simplified:

j m mn m n n j j j i j n m n j j m n j j q q j i j n j i n n j j q q n j n j j q q q n j n j j q q q + —-

n j n j n j n j q q q j q q q q q q j j n j n j q q q j j q q –

We note also that

( , , , )

n n n q T n q q q q q + = + + (41)

This implies that the Hankel determinants of the central binomial coefficients are given by

As a generalization of (9) we show, that with ( , )

A n k defined by (5) the following identity holds:

[ ]

For the left-hand side of ( 46

q b q n b q q k q a q q a q n b q q q a k q a q

Now replace a qa → and factor through ( ; ) . ;

q a q k q a q qa q