Integer quotients of factorials and algebraic generating functions - - PowerPoint PPT Presentation

Integer quotients of factorials and algebraic generating functions

Ira M. Gessel

Department of Mathematics Brandeis University

MIT Combinatorics Seminar September 30, 2011

Integer quotients of factorials

There are many formulas for integer quotients of factorials: (2n)! n! (n + 1)!,

Integer quotients of factorials

There are many formulas for integer quotients of factorials: (2n)! n! (n + 1)!, n! (6n)! (2n)!2 (3n)!,

Integer quotients of factorials

There are many formulas for integer quotients of factorials: (2n)! n! (n + 1)!, n! (6n)! (2n)!2 (3n)!, n! (30n)! (6n)! (10n)! (15n)!

Integer quotients of factorials

There are many formulas for integer quotients of factorials: (2n)! n! (n + 1)!, n! (6n)! (2n)!2 (3n)!, n! (30n)! (6n)! (10n)! (15n)! (m + n)! m! n! ,

Integer quotients of factorials

There are many formulas for integer quotients of factorials: (2n)! n! (n + 1)!, n! (6n)! (2n)!2 (3n)!, n! (30n)! (6n)! (10n)! (15n)! (m + n)! m! n! , (2m)! (2n)! m! n! (m + n)!,

Integer quotients of factorials

There are many formulas for integer quotients of factorials: (2n)! n! (n + 1)!, n! (6n)! (2n)!2 (3n)!, n! (30n)! (6n)! (10n)! (15n)! (m + n)! m! n! , (2m)! (2n)! m! n! (m + n)!, (4m)! (4n)! m! n! (2m + n)! (m + 2n)!

Integer quotients of factorials

There are many formulas for integer quotients of factorials: (2n)! n! (n + 1)!, n! (6n)! (2n)!2 (3n)!, n! (30n)! (6n)! (10n)! (15n)! (m + n)! m! n! , (2m)! (2n)! m! n! (m + n)!, (4m)! (4n)! m! n! (2m + n)! (m + 2n)! (k + 2l)! (k + 2m)! (k + 2n)! (k + l + m + n)! k! l! m! n! (k + l + m)! (k + l + n)! (k + m + n)!

Some known facts

◮ It is straightforward to check that a factorial quotient is integral (Landau’s theorem).

Some known facts

◮ It is straightforward to check that a factorial quotient is integral (Landau’s theorem). ◮ In all “interesting” cases where

• i(aim + bin)!
• j(cjm + djn)!

is integral,

i ai = j cj and i bi = j dj (and similarly for

more than two parameters).

The one-parameter case

◮ Suppose that un = r

i=1 (ain)!/s j=1(bjn)! is integral. Then

∞

n=0 unxn is algebraic if and only if s = r + 1.

(Rodriguez-Villegas, based on work of Beukers and Heckman).

The one-parameter case

◮ Suppose that un = r

i=1 (ain)!/s j=1(bjn)! is integral. Then

∞

n=0 unxn is algebraic if and only if s = r + 1.

(Rodriguez-Villegas, based on work of Beukers and Heckman). ◮ These algebraic cases are all known. There are three two-parameter formulas and 52 sporadic cases. (Bober, based

• n work of Beukers and Heckman.)

As a corollary to the classification, we have

The one-parameter case

◮ Suppose that un = r

i=1 (ain)!/s j=1(bjn)! is integral. Then

∞

n=0 unxn is algebraic if and only if s = r + 1.

(Rodriguez-Villegas, based on work of Beukers and Heckman). ◮ These algebraic cases are all known. There are three two-parameter formulas and 52 sporadic cases. (Bober, based

• n work of Beukers and Heckman.)

As a corollary to the classification, we have ◮ If r

i=1 (ain)!/s j=1(bjn)! is integral and nontrivial (and

• i ai =

j bj) then s ≥ r + 1.

The one-parameter case

◮ Suppose that un = r

i=1 (ain)!/s j=1(bjn)! is integral. Then

∞

n=0 unxn is algebraic if and only if s = r + 1.

(Rodriguez-Villegas, based on work of Beukers and Heckman). ◮ These algebraic cases are all known. There are three two-parameter formulas and 52 sporadic cases. (Bober, based

• n work of Beukers and Heckman.)

As a corollary to the classification, we have ◮ If r

i=1 (ain)!/s j=1(bjn)! is integral and nontrivial (and

• i ai =

j bj) then s ≥ r + 1. Is there a simple proof?

The two-parameter formulas

The two-parameter formulas are (m + n)! m! n! , m! (2m + 2n)! (2m)! n! (m + n)!, (2m)! (2n)! m! n! (m + n)!

The two-parameter formulas

The two-parameter formulas are (m + n)! m! n! , m! (2m + 2n)! (2m)! n! (m + n)!, (2m)! (2n)! m! n! (m + n)! They are all essentially binomial coefficients, since m! (2m + 2n)! (2m)! n! (m + n)! = (−1)n22n −m − 1

2

n

• (2m)! (2n)!

m! n! (m + n)! = (−1)n22m+2n m − 1

2

m + n

They are clearly integers, because they are coefficients of odd powers of 1 √ 1 − 4z =

∞

• n=0

2n n

• zn.

They are clearly integers, because they are coefficients of odd powers of 1 √ 1 − 4z =

∞

• n=0

2n n

• zn.

More precisely, for m a nonnegative integer, (1 − 4z)−m− 1

2 = ∞

• n=0

m! (2m + 2n)! (2m)! n! (m + n)!zn (1 − 4z)m− 1

2 = m−1

• n=0

(−1)n (m − n)! (2m)! (2m − 2n)! n! m!zn + (−1)m

∞

• n=0

(2m)! (2n)! m! n! (m + n)!zm+n

The formula (1 − 4z)−m− 1

2 =

∞

• k=0

2k k

• zk

2m+1 =

∞

• n=0

m! (2m + 2n)! (2m)! n! (m + n)!zn gives a combinatorial interpretation to the number m! (2m + 2n)! (2m)! n! (m + n)!.

The formula (1 − 4z)−m− 1

2 =

∞

• k=0

2k k

• zk

2m+1 =

∞

• n=0

m! (2m + 2n)! (2m)! n! (m + n)!zn gives a combinatorial interpretation to the number m! (2m + 2n)! (2m)! n! (m + n)!. It is the number of (2m + 1)-tuples of paths, each with as many up steps as down steps, with a total of 2n steps.

The formula (1 − 4z)−m− 1

2 =

∞

• k=0

2k k

• zk

2m+1 =

∞

• n=0

m! (2m + 2n)! (2m)! n! (m + n)!zn gives a combinatorial interpretation to the number m! (2m + 2n)! (2m)! n! (m + n)!. It is the number of (2m + 1)-tuples of paths, each with as many up steps as down steps, with a total of 2n steps. Note that the special case n = 2m gives m! (6m!) (2m)!2 (3m)!.

However, the formula (1 − 4z)m− 1

2 = m−1

• n=0

(−1)n (m − n)! (2m)! (2m − 2n)! n! m!zn + (−1)m

∞

• n=0

(2m)! (2n)! m! n! (m + n)!zm+n does not give a combinatorial interpretation, because of the sign changes.

What this talk is about

What this talk is about

There are some multivariable formulas of the form fgα = multiple hypergeometric series where f and g are algebraic functions (either rational or quadratic) with integer coefficients. Then taking α to be an integer (or sometimes a half-integer) gives an integral quotient

• f factorials.

What is a multiple hypergeometric series?

It is a power series whose terms are quotients of rising factorials: (α)n = α(α + 1) · · · (α + n − 1) (α)−n = (−1)n (1 − α)n .

What is a multiple hypergeometric series?

It is a power series whose terms are quotients of rising factorials: (α)n = α(α + 1) · · · (α + n − 1) (α)−n = (−1)n (1 − α)n . If α is a nonnegative integer then (α + 1)n = (α + n)! α! and 22n(α + 1

2)n = α! (2α + 2n)!

(2α)! (α + n)!.

A simple example is 1

• (1 − x − y)2 − 4xy

α+ 1

2

=

• m,n

(2α + 1)m+n(α + 1)m+n m! n! (α + 1)m(α + 1)n xmyn, where (α)n = α(α + 1) · · · (α + n − 1).

A simple example is 1

• (1 − x − y)2 − 4xy

α+ 1

2

=

• m,n

(2α + 1)m+n(α + 1)m+n m! n! (α + 1)m(α + 1)n xmyn, where (α)n = α(α + 1) · · · (α + n − 1). Taking α to be a nonnegative integer l gives 1

• (1 − x − y)2 − 4xy

l+ 1

2

=

• m,n

l! (2l + m + n)! (l + m + n)! (2l)! m! n! (l + m)! (l + n)! xmyn

A simple example is 1

• (1 − x − y)2 − 4xy

α+ 1

2

=

• m,n

(2α + 1)m+n(α + 1)m+n m! n! (α + 1)m(α + 1)n xmyn, where (α)n = α(α + 1) · · · (α + n − 1). Taking α to be a nonnegative integer l gives 1

• (1 − x − y)2 − 4xy

l+ 1

2

=

• m,n

l! (2l + m + n)! (l + m + n)! (2l)! m! n! (l + m)! (l + n)! xmyn Thus l! (2l + m + n)! (l + m + n)! (2l)! m! n! (l + m)! (l + n)! is integral, since 1

• (1 − x − y)2 − 4xy

1

2

=

• m,n

m + n m 2 xmyn

We also have a combinatorial interpretation (though not an especially interesting one) of the coefficients. Since their generating function is

• i,j

i + j i 2 xiyj 2l+1 , and i+j

i

2 counts pairs of paths ending at the same point. So (2l + m + n)! (l + m + n)! l! (2l)! m! n! (l + m)! (l + n)! counts (2l + 1)-tuples of such pairs of paths.

We also have a combinatorial interpretation (though not an especially interesting one) of the coefficients. Since their generating function is

• i,j

i + j i 2 xiyj 2l+1 , and i+j

i

2 counts pairs of paths ending at the same point. So (2l + m + n)! (l + m + n)! l! (2l)! m! n! (l + m)! (l + n)! counts (2l + 1)-tuples of such pairs of paths. Note also that setting y = x in (1 − x − y)2 − 4xy gives 1 − 4x, so these numbers are refinements of the coefficents of (1 − 4z)−l− 1

2

We can also look at the integral powers of

• (1 − x − y)2 − 4xy

−1. We have (for l > 0) 1

• (1 − x − y)2 − 4xy

l =

• m,n

2l (2l + 2m + 2n)! (2l + m + n − 1)! (l + m)! (l + n)! l! m! n! (2l + 2m)! (2l + 2n)! (l + m + n)! xmyn

We might hope that (2l + 2m + 2n)! (2l + m + n)! (l + m)! (l + n)! l! m! n! (2l + 2m)! (2l + 2n)! (l + m + n)! would be integral,

We might hope that (2l + 2m + 2n)! (2l + m + n)! (l + m)! (l + n)! l! m! n! (2l + 2m)! (2l + 2n)! (l + m + n)! would be integral, and in fact it is: 1 − x − y

• (1 − x − y)2 − 4xy

l+1 =

• m,n

(2l + 2m + 2n)! (2l + m + n)! (l + m)! (l + n)! l! m! n! (2l + 2m)! (2l + 2n)! (l + m + n)! xmyn

Here are some similar two-variable formulas. They can all be

• btained from the previous formula by the method of 2F1

transformations which I’ll explain later.

(1 + x)α− 1

2

• (1 − y)2 − 4xy

α =

• m,n

(2α)n(α + 1

2)m+n

m! n! (α + 1

2)n−m(α + 1 2)m

xmyn (1 − 4xy)α− 1

2

(1 − (1 + x)2y)α =

• m≤2n

(2n)! (α)n−m(1 − α)n m! n! (2n − m)! (1 − α)n−m xmyn (1 − 4xy)α− 1

2

(1 − x − y)α =

∞

• m,n=0

(1 − α)m+n(α)m−n m! n! (1 − α)m−n xmyn Note that in these cases there is no simple formula for the coefficients of simple powers such as 1 − 4xy 1 − x − y α .

The last formula is especially interesting, because it shows that the coefficients of (1 − 4xy)l+ 1

2

(1 − x − y)l+1 are nonnegative when l is a nonnegative integer, and thus so are are coefficients of

∞

• l=0

(1 − 4xy)l+ 1

2

(1 − x − y)l+1 zl =

• 1 − 4xy

1 − x − y − z + 4xyz and therefore so are the coefficients of 1 1 − x − y − z + 4xyz .

The last formula is especially interesting, because it shows that the coefficients of (1 − 4xy)l+ 1

2

(1 − x − y)l+1 are nonnegative when l is a nonnegative integer, and thus so are are coefficients of

∞

• l=0

(1 − 4xy)l+ 1

2

(1 − x − y)l+1 zl =

• 1 − 4xy

1 − x − y − z + 4xyz and therefore so are the coefficients of 1 1 − x − y − z + 4xyz . Unfortunately, I haven’t found any other examples as nice as this.

There are 3-variable generalizations of these formulas; for example, 1

• (1 − x − y)2 − 4xy

α+ 1

2

=

• l,m

(2α + 1)l+m(α + 1)l+m l! m! (α + 1)l(α + 1)m xlym, can be generalized to (1 + z)α

• (1 − x − y)2 − 4xy(1 + z)

α+ 1

2

=

• l,m,n

(2α + 1)l+m(α + 1)l+m−n l! m! n! (α + 1)l−n(α + 1)m−n xlymzn.

This gives the integral factorial quotient k! (2k + l + m)! (k + l + m − n)! (2k)! l! m! n! (k + l − n)! (k + m − n)!

The other identities that I want to talk about involve expressions with square roots, and can often be expressed in terms of the Catalan generating function c(x) =

∞

• n=0

Cnxn = 1 − √ 1 − 4x 2x .

The other identities that I want to talk about involve expressions with square roots, and can often be expressed in terms of the Catalan generating function c(x) =

∞

• n=0

Cnxn = 1 − √ 1 − 4x 2x . Here’s a simple example: 1 2

• 1 +

1 √ 1 − 4x

• 1 − 4y

2 √ 1 − 4x +

• 1 − 4y

α+1 =

• m,n

(1 + 1

2α)m(1 + 1 2α)n( 1 2 + 1 2α)m+n22m+2n

m! n! (1 + α)m+n xmyn

If we set α = 2l we get the integral factorial quotient (l + m)! (l + n)! (2l + 2m + 2n)! l! m! n! (l + m + n)! (2l + m + n)! These coefficients have a reasonably straightforward combinatorial interpretation in terms of lattice paths, since 2 √ 1 − 4x +

• 1 − 4y

= 1 1 − xc(x) − yc(y). Here we also have the nice formula

• 2

√ 1 − 4x +

• 1 − 4y

α =

• m,n

( 1

2 + 1 2α)m( 1 2 + 1 2α)n( 1 2α)m+n22m+2n

m! n! (1 + α)m+n xmyn

In the case α = 1, this reduces to 1 1 − xc(x) − yc(y) =

• m,n≥0

( 1

2)m+n

(2)m+n 22m+2nxmyn =

• m,n≥0

Cm+nxmyn which is essentially the Chung-Feller theorem, since we are counting paths with up and down steps ending on the horizontal axis, where up steps above the horizontal axis are weighted x and up steps below are weighted y:

x c(x) y c(y)

So for α a nonnegative integer, we are counting α-tuples of such paths, and in terms of factorials we have different formulas for α even and α odd.

So for α a nonnegative integer, we are counting α-tuples of such paths, and in terms of factorials we have different formulas for α even and α odd. If α = 2l + 1 then the number of α-tuples of paths weighted by xmyn is (2l + 1) (l + m)! (l + n)! (2l + 2m + 2n)! l! m! n! (l + m + n)! (2l + m + n + 1)!

So for α a nonnegative integer, we are counting α-tuples of such paths, and in terms of factorials we have different formulas for α even and α odd. If α = 2l + 1 then the number of α-tuples of paths weighted by xmyn is (2l + 1) (l + m)! (l + n)! (2l + 2m + 2n)! l! m! n! (l + m + n)! (2l + m + n + 1)! and if α = 2l then the number of such α-tuples is l l! (2m + 2m)! (2l + 2n)! (l + m + n − 1)! (2l)! m! n! (l + m)! (l + n)! (2l + m + n)! .

A similar-looking but in some ways very different identity is 1 √ 1 − 4x

• 2

1 +

• (1 − 4x)(1 − 4y)

α =

• m,n≥0

(α + 1)2m (α)2n m! n! (α + 1)m+n xmyn, which has the more symmetric variant 1 2

• 1

√ 1 − 4x + 1

• 1 − 4y

2 1 +

• (1 − 4x)(1 − 4y)

α =

• m,n≥0

(α)2m (α)2n m! n! (α)m+n xmyn

Here it is not obvious from the generating function that the coefficients are positive, and in fact, the coefficients of 2 1 +

• (1 − 4x)(1 − 4y)

= 1 1 − xc(x) − yc(y) + 2xyc(xy) are not positive. The second formula (with α = l + 1) gives the integral factorial quotient (l + 2m)! (l + 2n)! l! m! n! (l + m + n)!, which for l = 0 reduces to (2m)! (2n)!/m! n! (m + n)!.

Here it is not obvious from the generating function that the coefficients are positive, and in fact, the coefficients of 2 1 +

• (1 − 4x)(1 − 4y)

= 1 1 − xc(x) − yc(y) + 2xyc(xy) are not positive. The second formula (with α = l + 1) gives the integral factorial quotient (l + 2m)! (l + 2n)! l! m! n! (l + m + n)!, which for l = 0 reduces to (2m)! (2n)!/m! n! (m + n)!. If l = 0 and m = 1 then this is twice a Catalan number.

A similar formula related to Narayana numbers is

∞

• m,k,l=0

(α)k+l(α + 1)k+l+m(α + 1)2m k! l! m! (α + 1)m+k(α + 1)m+l ukvlzm = 1 √ 1 − 4z

• 2

1 − u − v + √ 1 − 4z

• (1 − u − v)2 − 4uv

α If α is a nonnegative integer a, the coefficient of ukvlzm is (a + k + l − 1)! (a + k + l + m)! (a + 2m)! (a − 1)! k! l! m! (a + k + m)! (a + l + m)! ; for a = 1, m = 0 this reduces to the Narayana number 1 k + 1 k + l k k + l + 1 k

• .

How can we prove these formulas? How can we discover them?

How can we prove these formulas? How can we discover them? Suppose the identity is fgα =

m,n · · · Several methods:

How can we prove these formulas? How can we discover them? Suppose the identity is fgα =

m,n · · · Several methods:

◮ Show that both sides satisfy the same differential equation.

How can we prove these formulas? How can we discover them? Suppose the identity is fgα =

m,n · · · Several methods:

◮ Show that both sides satisfy the same differential equation. This always works, but it isn’t very interesting.

◮ Expand the left side and evaluate coefficients using known identities such Chu-Vandermonde or Pfaff-Saalschütz.

◮ Expand the left side and evaluate coefficients using known identities such Chu-Vandermonde or Pfaff-Saalschütz. This works for our first formula, 1

• (1 − x − y)2 − 4xy

α+ 1

2

=

• m,n

(2α + 1)m+n(α + 1)m+n m! n! (α + 1)m(α + 1)n xmyn, If we expand the left side by the binomial theorem and extract the coefficient of xmyn, we get a sum that we can evaluate by the Chu-Vandermonde theorem.

◮ Start with the right side, write the sum with respect to one of the variables as a 2F1, and apply Pfaff’s or Euler’s transformation to change the identity to one that is already

• known. (Equivalently, apply known multiple hypergeometric

series transformations.) We can also use this method to discover identities, by starting with a known identity and applying a transformation.

2F1 transformations

The most important hypergeometric series (sometimes called the hypergeometric series) is the 2F1, defined by

2F1

a, b c ; x

• =

∞

• n=0

(a)n(b)n n! (c)n xn. There are two important transformations that convert one 2F1 to another. Pfaff’s transformation:

2F1

a, b c ; x

• = (1 − x)−a

2F1

a, c − b c ; x x − 1

• Euler’s transformation:

2F1

a, b c ; x

• = (1 − x)c−a−b

2F1

c − a, c − b c ; x

We can sometimes write a multiple hypergeometric series as a sum of 2F1s to which we can apply these transformations.

We can sometimes write a multiple hypergeometric series as a sum of 2F1s to which we can apply these transformations. For example, 1

• (1 − x − y)2 − 4xy

α+ 1

2

=

• m,n

(2α + 1)m+n(α + 1)m+n m! n! (α + 1)m(α + 1)n xmyn, =

• n

(2α + 1)n n! yn

m

(2α + n + 1)m(α + n + 1)m m! (α + 1)m xm =

• n

(2α + 1)n n! yn

2F1

2α + n + 1, α + n + 1 α + 1 ; x

We can sometimes write a multiple hypergeometric series as a sum of 2F1s to which we can apply these transformations. For example, 1

• (1 − x − y)2 − 4xy

α+ 1

2

=

• m,n

(2α + 1)m+n(α + 1)m+n m! n! (α + 1)m(α + 1)n xmyn, =

• n

(2α + 1)n n! yn

m

(2α + n + 1)m(α + n + 1)m m! (α + 1)m xm =

• n

(2α + 1)n n! yn

2F1

2α + n + 1, α + n + 1 α + 1 ; x

• In some cases, we can use these transformations to add

parameters to an identity. For example, we might replace x with x + z, thus getting a trivial generalization, and then transform this into a nontrivial generalization.

◮ By a change of variables, get a formula with rational parameters that can be proved by equating coefficients, using known identities.

For example if we set x = X/(1 + X)2, y = Y/(1 + Y)2 then √ 1 − 4x = (1 − X)/(1 + X) and

• 1 − 4y = (1 − Y)/(1 + Y).

Thus the formula 1 √ 1 − 4x

• 2

1 +

• (1 − 4x)(1 − 4y)

α =

• m,n≥0

(α + 1)2m (α)2n m! n! (α + 1)m+n xmyn, becomes

• m,n≥0

(α + 1)2m (α)2n m! n! (α + 1)m+n

• X

(1 + X)2 m Y (1 + Y)2 n = 1 1 − X (1 + X)α+1(1 + Y)α (1 + XY)α

We rewrite this as

• m,n≥0

(α + 1)2m (α)2n m! n! (α + 1)m+n X m(1 + X)−2m−α−1Y n(1 + Y)−2n−α = 1 (1 − X)(1 + XY)α , which we can prove by expanding the left side and applying Chu-Vandermonde twice.

◮ We can derive identities using the Catalan and Narayana generating functions from well poised summation formulas using the formula

∞

• m=k

(−m)k (α + m + 1)k · (α + 1)2m m! (α + 1)m xm = (−x)k √ 1 − 4x c(x)α+2k

◮ We can derive identities using the Catalan and Narayana generating functions from well poised summation formulas using the formula

∞

• m=k

(−m)k (α + m + 1)k · (α + 1)2m m! (α + 1)m xm = (−x)k √ 1 − 4x c(x)α+2k (and a similar formula for the Narayana generating function).

◮ We can derive identities using the Catalan and Narayana generating functions from well poised summation formulas using the formula

∞

• m=k

(−m)k (α + m + 1)k · (α + 1)2m m! (α + 1)m xm = (−x)k √ 1 − 4x c(x)α+2k (and a similar formula for the Narayana generating function). For example, a form of Dixon’s theorem is

• k

(α)k k! (−m)k (α + m + 1)k (−n)k (α + n + 1)k = (α + 1)m(α + 1)n( 1

2α + 1)m+n

( 1

2α + 1)m( 1 2α + 1)n(α + 1)m+n

This gives the identity

• m,n

( 1

2 + 1 2α)m( 1 2 + 1 2α)n(1 + 1 2α)m+n22m+2n

m! n! (1 + a)m+n xmyn = 1 √ 1 − 4x

• 1 − 4y
• c(x)c(y)

1 − xyc(x)2c(y)2 α = 1 √ 1 − 4x

• 1 − 4y
• 2

√ 1 − 4x +

• 1 − 4y

α

Constant term formulas

In some cases we can replace the algebraic generating functions with constant terms of rational functions.

Constant term formulas

In some cases we can replace the algebraic generating functions with constant terms of rational functions. To do this we apply the change of variables formula for residues:

Constant term formulas

In some cases we can replace the algebraic generating functions with constant terms of rational functions. To do this we apply the change of variables formula for residues: The residue of a formal Laurent series in x is the coefficent of x−1. So if we have a power series f(x) =

∞

• n=0

anxn then an = CT f(x)/xn = res f(x)/xn+1, where CT denotes the constant term.

Constant term formulas

In some cases we can replace the algebraic generating functions with constant terms of rational functions. To do this we apply the change of variables formula for residues: The residue of a formal Laurent series in x is the coefficent of x−1. So if we have a power series f(x) =

∞

• n=0

anxn then an = CT f(x)/xn = res f(x)/xn+1, where CT denotes the constant term. The residue change of variables theorem (equivalent to Lagrange inversion) says that if g(x) = g1x + g2x2 + · · · , where g1 = 0, then res f(x) = res f(g(x))g′(x).

We can use the change of variables formula to get rid of square roots.

We can use the change of variables formula to get rid of square roots. As a simple example take the formula (2m)! (2n)! m! n! (m + n)! = CT(−1)m(1 − 4x)m− 1

2 /xm+n.

Applying the change of variables formula with g(x) = x/(1 + x)2, we get that (2m)! (2n)! m! n! (m + n)! = CT(−1)m(1 − x)2m(1 + x)2n/xm+n

There is also a multivariable analogue of the change of variables formula. If we apply it to the formula 1

• (1 − x − y)2 − 4xy

l+ 1

2

=

• m,n

l! (2l + m + n)! (l + m + n)! (2l)! m! n! (l + m)! (l + n)! xmyn, changing x to x/(1 + x)/(1 + y) and y to y/(1 + x)/(1 + y), we get l! (2l + m + n)! (l + m + n)! (2l)! m! n! (l + m)! (l + n)! = CT (1 + x)2l+m+n(1 + y)2l+m+n (1 − xy)2l xmyn .

There is also a multivariable analogue of the change of variables formula. If we apply it to the formula 1

• (1 − x − y)2 − 4xy

l+ 1

2

=

• m,n

l! (2l + m + n)! (l + m + n)! (2l)! m! n! (l + m)! (l + n)! xmyn, changing x to x/(1 + x)/(1 + y) and y to y/(1 + x)/(1 + y), we get l! (2l + m + n)! (l + m + n)! (2l)! m! n! (l + m)! (l + n)! = CT (1 + x)2l+m+n(1 + y)2l+m+n (1 − xy)2l xmyn . If we equate coefficients, we get Dixon’s theorem.

q-analogues: a conjecture

It’s easy to show, using Landau’s theorem, that if

• i(aim + bin)!
• j(cjm + djn)!

is integral then

• i(aim + bin)!q
• j(cjm + djn)!q

is a polynomial in q.

q-analogues: a conjecture

It’s easy to show, using Landau’s theorem, that if

• i(aim + bin)!
• j(cjm + djn)!

is integral then

• i(aim + bin)!q
• j(cjm + djn)!q

is a polynomial in q.

• S. Ole Warnaar and Wadim Zudilin (2010) conjectured that

these polynomials always have nonnegative coefficients, and they proved that this is the case for m!q (2m + 2n)!q (2m)!q n!q (m + n)!q and (2m)!q (2n)!q m!q n!q (m + n)!q .