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)!
(m + n)!
,
m! n!
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)!
(m + n)!
,
m! n!
n! (6n)!
,
(2n)!2 (3n)!
(2m)! (2n)!
,
m! n! (m + n)!
n! (30n)!
(6n)! (10n)! (15n)!
Integer quotients of factorials
There are many formulas for integer quotients of factorials:
(2n)!
,
n! (n + 1)!
(m + n)!
,
m! n!
n! (6n)!
,
(2n)!2 (3n)!
(2m)! (2n)!
,
m! n! (m + n)!
n! (30n)!
(6n)! (10n)! (15n)!
(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)!
(m + n)!
,
m! n!
n! (6n)!
,
(2n)!2 (3n)!
(2m)! (2n)!
,
m! n! (m + n)!
n! (30n)!
(6n)! (10n)! (15n)!
(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
I It is straightforward to check that a factorial quotient is
integral (Landau’s theorem).
Some known facts
I It is straightforward to check that a factorial quotient is
integral (Landau’s theorem).
I In all “interesting” cases where
Q
(a m + bi n)!
Qi i
j (cj m + dj n)!
P
P
P
P
is integral, i ai = j cj and i bi = j dj (and similarly for
more than two parameters).
The one-parameter case
Q
Q
I Suppose that un = ri=1 (ai n)!/ sj=1 (bj n)! is integral. Then
P
∞
n
n=0 un x is algebraic if and only if s = r + 1.
(Rodriguez-Villegas, based on work of Beukers and Heckman).
The one-parameter case
Q
Q
I Suppose that un = ri=1 (ai n)!/ sj=1 (bj n)! is integral. Then
P
∞
n
n=0 un x is algebraic if and only if s = r + 1.
(Rodriguez-Villegas, based on work of Beukers and Heckman).
I These algebraic cases are all known. There are three
two-parameter formulas and 52 sporadic cases. (Bober, based
on work of Beukers and Heckman.)
As a corollary to the classification, we have
The one-parameter case
Q
Q
I Suppose that un = ri=1 (ai n)!/ sj=1 (bj n)! is integral. Then
P
∞
n
n=0 un x is algebraic if and only if s = r + 1.
(Rodriguez-Villegas, based on work of Beukers and Heckman).
I These algebraic cases are all known. There are three
two-parameter formulas and 52 sporadic cases. (Bober, based
on work of Beukers and Heckman.)
As a corollary to the classification, we have
Qr
Qs
I
If
(a
n)!/
i
i=1
j=1 (bj n)! is integral and nontrivial (and
P
P
i ai =
j bj ) then s ≥ r + 1.
The one-parameter case
Q
Q
I Suppose that un = ri=1 (ai n)!/ sj=1 (bj n)! is integral. Then
P
∞
n
n=0 un x is algebraic if and only if s = r + 1.
(Rodriguez-Villegas, based on work of Beukers and Heckman).
I These algebraic cases are all known. There are three
two-parameter formulas and 52 sporadic cases. (Bober, based
on work of Beukers and Heckman.)
As a corollary to the classification, we have
Qr
Qs
I
If
(a
n)!/
i
i=1
j=1 (bj n)! is integral and nontrivial (and
P
P
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
1
m! (2m + 2n)!
n 2n −m − 2
= (−1) 2
(2m)! n! (m + n)!
n
1
(2m)! (2n)!
n 2m+2n m − 2
= (−1) 2
m! n! (m + n)!
m+n
They are clearly integers, because they are coefficients of odd
powers of
∞ X
1
2n n
√
=
z .
n
1 − 4z
n=0
They are clearly integers, because they are coefficients of odd
powers of
∞ X
1
2n n
√
=
z .
n
1 − 4z
n=0
More precisely, for m a nonnegative integer,
1
(1 − 4z)−m− 2 =
∞
X
m! (2m + 2n)! n
z
(2m)! n! (m + n)!
n=0
1
m− 2
(1 − 4z)
=
m−1
X
(−1)n
n=0
(m − n)! (2m)! n
z
(2m − 2n)! n! m!
+ (−1)m
∞
X
(2m)! (2n)! m+n
z
m! n! (m + n)!
n=0
The formula
1
−m− 2
(1 − 4z)
=
X
∞ k =0
∞
2k k 2m+1 X m! (2m + 2n)! n
=
z
z
k
(2m)! n! (m + n)!
n=0
gives a combinatorial interpretation to the number
m! (2m + 2n)!
.
(2m)! n! (m + n)!
The formula
1
−m− 2
(1 − 4z)
=
X
∞ k =0
∞
2k k 2m+1 X m! (2m + 2n)! n
=
z
z
k
(2m)! n! (m + n)!
n=0
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
−m− 2
(1 − 4z)
=
X
∞ k =0
∞
2k k 2m+1 X m! (2m + 2n)! n
=
z
z
k
(2m)! n! (m + n)!
n=0
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
(1 − 4z)m− 2 =
m−1
X
(−1)n
n=0
(m − n)! (2m)! n
z
(2m − 2n)! n! m!
+ (−1)m
∞
X
(2m)! (2n)! m+n
z
m! n! (m + n)!
n=0
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
f g α = 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
of 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)
(−1)n
.
(α)−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)
(−1)n
.
(α)−n =
(1 − α)n
If α is a nonnegative integer then
(α + 1)n =
and
22n (α + 12 )n =
(α + n)!
α!
α! (2α + 2n)!
.
(2α)! (α + n)!
A simple example is
1
(1 − x − y )2 −
α+ 1
2
4xy
=
X (2α + 1)m+n (α + 1)m+n
m,n
m! n! (α + 1)m (α + 1)n
where (α)n = α(α + 1) · · · (α + n − 1).
x my n,
A simple example is
1
(1 − x − y )2 −
α+ 1
2
4xy
=
X (2α + 1)m+n (α + 1)m+n
m,n
m! n! (α + 1)m (α + 1)n
x my n,
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
=
X l! (2l + m + n)! (l + m + n)!
x my n
(2l)!
m!
n!
(l
+
m)!
(l
+
n)!
m,n
A simple example is
1
(1 − x − y )2 −
α+ 1
2
4xy
=
X (2α + 1)m+n (α + 1)m+n
m,n
m! n! (α + 1)m (α + 1)n
x my n,
where (α)n = α(α + 1) · · · (α + n − 1). Taking α to be a
nonnegative integer l gives
1
(1 − x − y )2 − 4xy
Thus
l+ 1
2
=
X l! (2l + m + n)! (l + m + n)!
x my n
(2l)!
m!
n!
(l
+
m)!
(l
+
n)!
m,n
l! (2l + m + n)! (l + m + n)!
is integral, since
(2l)! m! n! (l + m)! (l + n)!
X m + n2
1
=
x my n
1
m
m,n
(1 − x − y )2 − 4xy 2
We also have a combinatorial interpretation (though not an
especially interesting one) of the coefficients. Since their
generating function is
X i + j 2 i j 2l+1
xy
,
i
i,j
and
i+j 2
i
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
X i + j 2 i j 2l+1
xy
,
i
i,j
and
i+j 2
i
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
(1 − 4z)−l− 2
We can also look at the integral powers of
−1
(1 − x − y )2 − 4xy
. We have (for l > 0)
1
l
(1 − x − y )2 − 4xy
X (2l + 2m + 2n)! (2l + m + n − 1)! (l + m)! (l + n)!
=
2l
x my n
l!
m!
n!
(2l
+
2m)!
(2l
+
2n)!
(l
+
m
+
n)!
m,n
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
l+1
(1 − x − y )2 − 4xy
X (2l + 2m + 2n)! (2l + m + n)! (l + m)! (l + n)!
=
x my n
l!
m!
n!
(2l
+
2m)!
(2l
+
2n)!
(l
+
m
+
n)!
m,n
Here are some similar two-variable formulas. They can all be
obtained from the previous formula by the method of 2 F1
transformations which I’ll explain later.
1
X
(2α)n (α + 12 )m+n
(1 + x)α− 2
α =
x my n
1
1
(1 − y )2 − 4xy
m!
n!
(α
+
)
(α
+
)
2 n−m
2 m
m,n
1
X
(2n)! (α)n−m (1 − α)n
(1 − 4xy )α− 2
=
x my n
m! n! (2n − m)! (1 − α)n−m
(1 − (1 + x)2 y )α
m≤2n
1
∞
X
(1 − 4xy )α− 2
(1 − α)m+n (α)m−n m n
=
x y
α
(1 − x − y )
m! n! (1 − α)m−n
m,n=0
Note that in these cases there is no simple formula for the
coefficients of simple powers such as
1 − 4xy α
.
1−x −y
The last formula is especially interesting, because it shows that
the coefficients of
1
(1 − 4xy )l+ 2
(1 − x − y )l+1
are nonnegative when l is a nonnegative integer, and thus so
are are coefficients of
p
1
∞
X
1 − 4xy
(1 − 4xy )l+ 2 l
z =
l+1
1 − x − y − z + 4xyz
(1 − x − y )
l=0
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
(1 − 4xy )l+ 2
(1 − x − y )l+1
are nonnegative when l is a nonnegative integer, and thus so
are are coefficients of
p
1
∞
X
1 − 4xy
(1 − 4xy )l+ 2 l
z =
l+1
1 − x − y − z + 4xyz
(1 − x − y )
l=0
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
=
X (2α + 1)l+m (α + 1)l+m
l,m
l! m! (α + 1)l (α + 1)m
x l y m,
can be generalized to
(1 + z)α
α+ 1
2
(1 − x − y )2 − 4xy (1 + z)
X (2α + 1)l+m (α + 1)l+m−n
=
x l y mzn.
l! m! n! (α + 1)l−n (α + 1)m−n
l,m,n
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
√
∞
X
1 − 1 − 4x
n
c(x) =
Cn x =
.
2x
n=0
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
√
∞
X
1 − 1 − 4x
n
c(x) =
Cn x =
.
2x
n=0
Here’s a simple example:
1
2
1
p
1+ √
1 − 4x 1 − 4y
=
!
√
!α+1
2
1 − 4x +
p
1 − 4y
X (1 + 1 α)m (1 + 1 α)n ( 1 + 1 α)m+n 22m+2n
2
2
2
2
x my n
m!
n!
(1
+
α)
m+n
m,n
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
p
=
.
1
−
xc(x)
− yc(y )
1 − 4x + 1 − 4y
Here we also have the nice formula
√
2
!α
p
1 − 4y
X ( 1 + 1 α)m ( 1 + 1 α)n ( 1 α)m+n 22m+2n
2
2
2
2
2
=
x my n
m!
n!
(1
+
α)
m+n
m,n
1 − 4x +
In the case α = 1, this reduces to
X ( 1 )m+n
1
2
=
22m+2n x m y n
1 − xc(x) − yc(y )
(2)m+n
m,n≥0
X
Cm+n x m y n
=
m,n≥0
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
x m y n 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
x m y n 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
!α
2
1+
p
(1 − 4x)(1 − 4y )
X (α + 1)2m (α)2n
=
x my n,
m! n! (α + 1)m+n
m,n≥0
which has the more symmetric variant
1
2
1
1
√
+p
1 − 4x
1 − 4y
!
2
1+
!α
p
(1 − 4x)(1 − 4y )
X (α)2m (α)2n
=
x my n
m! n! (α)m+n
m,n≥0
Here it is not obvious from the generating function that the
coefficients are positive, and in fact, the coefficients of
1
2
p
=
1 − xc(x) − yc(y ) + 2xyc(xy )
1 + (1 − 4x)(1 − 4y )
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
1
2
p
=
1 − xc(x) − yc(y ) + 2xyc(xy )
1 + (1 − 4x)(1 − 4y )
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
∞
X
(α)k +l (α + 1)k +l+m (α + 1)2m k l m
u vz
k ! l! m! (α + 1)m+k (α + 1)m+l
m,k ,l=0
1
=√
1 − 4z
2
p
√
1 − u − v + 1 − 4z (1 − u − v )2 − 4uv
If α is a nonnegative integer a, the coefficient of u k v l z m 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 +l
k +l +1
.
k +1
k
k
!α
How can we prove these formulas? How can we discover
them?
How can we prove these formulas? How can we discover
them?
P
Suppose the identity is f g α = m,n · · · Several methods:
How can we prove these formulas? How can we discover
them?
P
Suppose the identity is f g α = m,n · · · Several methods:
I Show that both sides satisfy the same differential equation.
How can we prove these formulas? How can we discover
them?
P
Suppose the identity is f g α = m,n · · · Several methods:
I Show that both sides satisfy the same differential equation.
This always works, but it isn’t very interesting.
I Expand the left side and evaluate coefficients using known
identities such Chu-Vandermonde or Pfaff-Saalschütz.
I 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
=
X (2α + 1)m+n (α + 1)m+n
m,n
m! n! (α + 1)m (α + 1)n
x my n,
If we expand the left side by the binomial theorem and extract
the coefficient of x m y n , we get a sum that we can evaluate by
the Chu-Vandermonde theorem.
I Start with the right side, write the sum with respect to one of
the variables as a 2 F1 , 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.
2 F1
transformations
The most important hypergeometric series (sometimes called
the hypergeometric series) is the 2 F1 , defined by
2 F1
a, b
;x
c
=
∞
X
(a)n (b)n
n=0
n! (c)n
x n.
There are two important transformations that convert one 2 F1 to
another.
Pfaff’s transformation:
a, b
a, c − b
x
−a
; x = (1 − x) 2 F1
;
2 F1
c
c
x −1
Euler’s transformation:
a, b
c − a, c − b
c−a−b
; x = (1 − x)
;x
2 F1
2 F1
c
c
We can sometimes write a multiple hypergeometric series as a
sum of 2 F1 s to which we can apply these transformations.
We can sometimes write a multiple hypergeometric series as a
sum of 2 F1 s to which we can apply these transformations. For
example,
1
α+ 1
=
X (2α + 1)m+n (α + 1)m+n
m! n! (α + 1)m (α + 1)n
x my n,
2
m,n
(1 − x − y )2 − 4xy
X (2α + 1)n X (2α + n + 1)m (α + n + 1)m
yn
xm
=
n!
m!
(α
+
1)
m
m
n
X (2α + 1)n
2α + n + 1, α + n + 1
n
=
y 2 F1
;x
n!
α+1
n
We can sometimes write a multiple hypergeometric series as a
sum of 2 F1 s to which we can apply these transformations. For
example,
1
α+ 1
=
X (2α + 1)m+n (α + 1)m+n
m! n! (α + 1)m (α + 1)n
x my n,
2
m,n
(1 − x − y )2 − 4xy
X (2α + 1)n X (2α + n + 1)m (α + n + 1)m
yn
xm
=
n!
m!
(α
+
1)
m
m
n
X (2α + 1)n
2α + n + 1, α + n + 1
n
=
y 2 F1
;x
n!
α+1
n
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.
I By a change of variables, get a formula with rational
parameters that can be proved by equating coefficients, using
known identities.
2
2
For
√ example if we set x = X /(1 +pX ) , y = Y /(1 + Y ) then
1 − 4x = (1 − X )/(1 + X ) and 1 − 4y = (1 − Y )/(1 + Y ).
Thus the formula
!α
1
2
√
p
1 − 4x 1 + (1 − 4x)(1 − 4y )
X (α + 1)2m (α)2n
=
x my n,
m! n! (α + 1)m+n
m,n≥0
becomes
m n
X (α + 1)2m (α)2n X
Y
m! n! (α + 1)m+n (1 + X )2
(1 + Y )2
m,n≥0
=
1 (1 + X )α+1 (1 + Y )α
1−X
(1 + XY )α
We rewrite this as
X (α + 1)2m (α)2n
X m (1 + X )−2m−α−1 Y n (1 + Y )−2n−α
m! n! (α + 1)m+n
m,n≥0
=
1
,
(1 − X )(1 + XY )α
which we can prove by expanding the left side and applying
Chu-Vandermonde twice.
I We can derive identities using the Catalan and Narayana
generating functions from well poised summation formulas
using the formula
∞
X
m=k
(−m)k
(α + 1)2m m
(−x)k
·
x =√
c(x)α+2k
(α + m + 1)k m! (α + 1)m
1 − 4x
I We can derive identities using the Catalan and Narayana
generating functions from well poised summation formulas
using the formula
∞
X
m=k
(−m)k
(α + 1)2m m
(−x)k
·
x =√
c(x)α+2k
(α + m + 1)k m! (α + 1)m
1 − 4x
(and a similar formula for the Narayana generating function).
I We can derive identities using the Catalan and Narayana
generating functions from well poised summation formulas
using the formula
∞
X
m=k
(−m)k
(α + 1)2m m
(−x)k
·
x =√
c(x)α+2k
(α + m + 1)k m! (α + 1)m
1 − 4x
(and a similar formula for the Narayana generating function).
For example, a form of Dixon’s theorem is
X (α)k
k
(−m)k
(−n)k
k ! (α + m + 1)k (α + n + 1)k
=
(α + 1)m (α + 1)n ( 12 α + 1)m+n
( 12 α + 1)m ( 12 α + 1)n (α + 1)m+n
This gives the identity
X ( 1 + 1 α)m ( 1 + 1 α)n (1 + 1 α)m+n 22m+2n
2
2
2
2
2
x my n
m!
n!
(1
+
a)
m+n
m,n
α
c(x)c(y )
1
p
=√
1 − 4x 1 − 4y 1 − xyc(x)2 c(y )2
1
p
=√
1 − 4x 1 − 4y
2
p
√
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) =
then
∞
X
an x n
n=0
an = CT f (x)/x n = res f (x)/x n+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) =
then
∞
X
an x n
n=0
an = CT f (x)/x n = res f (x)/x n+1 ,
where CT denotes the constant term.
The residue change of variables theorem (equivalent to
Lagrange inversion) says that if g(x) = g1 x + g2 x 2 + · · · , where
g1 6= 0, then
res f (x) = res f (g(x))g 0 (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
1
(2m)! (2n)!
= CT(−1)m (1 − 4x)m− 2 /x m+n .
m! n! (m + n)!
Applying the change of variables formula with
g(x) = x/(1 + x)2 , we get that
(2m)! (2n)!
= CT(−1)m (1 − x)2m (1 + x)2n /x m+n
m! n! (m + 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 −
l+ 1
4xy 2
=
X l! (2l + m + n)! (l + m + n)!
x my n,
(2l)!
m!
n!
(l
+
m)!
(l
+
n)!
m,n
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)!
(1 + x)2l+m+n (1 + y )2l+m+n
= CT
.
(2l)! m! n! (l + m)! (l + n)!
(1 − xy )2l x m y 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 −
l+ 1
4xy 2
=
X l! (2l + m + n)! (l + m + n)!
x my n,
(2l)!
m!
n!
(l
+
m)!
(l
+
n)!
m,n
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)!
(1 + x)2l+m+n (1 + y )2l+m+n
= CT
.
(2l)! m! n! (l + m)! (l + n)!
(1 − xy )2l x m y n
If we equate coefficients, we get Dixon’s theorem.
q-analogues: a conjecture
It’s easy to show, using Landau’s theorem, that if
Q
(a m + bi n)!
Qi i
j (cj m + dj n)!
is integral then
Q
(a m + bi n)!q
Qi i
j (cj m + dj n)!q
is a polynomial in q.
q-analogues: a conjecture
It’s easy to show, using Landau’s theorem, that if
Q
(a m + bi n)!
Qi i
j (cj m + dj n)!
is integral then
Q
(a m + bi n)!q
Qi i
j (cj m + dj n)!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