Hankel determinants, continued fractions,
orthgonal polynomials,
and hypergeometric series
Ira M. Gessel
with Jiang Zeng and Guoce Xin
LaBRI
June 8, 2007
Continued fractions and Hankel determinants
There is a close relationship between continued fractions,
Hankel determinants, and orthogonal polynomials.
Continued fractions and Hankel determinants
There is a close relationship between continued fractions,
Hankel determinants, and orthogonal polynomials.
Why are we interested in these things?
Continued fractions and Hankel determinants
There is a close relationship between continued fractions,
Hankel determinants, and orthogonal polynomials.
Why are we interested in these things?
Continued fractions count certain “weighted Motzkin paths”
which encode objects of enumerative interest such as partitions
and permutations. (Flajolet)
Continued fractions and Hankel determinants
There is a close relationship between continued fractions,
Hankel determinants, and orthogonal polynomials.
Why are we interested in these things?
Continued fractions count certain “weighted Motzkin paths”
which encode objects of enumerative interest such as partitions
and permutations. (Flajolet)
Hankel determinants arise in some enumeration problems, for
example, counting certain kinds of tilings or alternating sign
matrices.
Orthogonal polynomials
A sequence of polynomials pn (x) n≥0 , where pn (x) has degree
n, is orthogonal if there is a linear functional L on polynomials
such that L pm (x)pn (x) = 0 for m 6= n, but L pm (x)2 6= 0.
We will assume that L(1) = 1.
Orthogonal polynomials
A sequence of polynomials pn (x) n≥0 , where pn (x) has degree
n, is orthogonal if there is a linear functional L on polynomials
such that L pm (x)pn (x) = 0 for m 6= n, but L pm (x)2 6= 0.
We will assume that L(1) = 1.
The moments of the sequence pn (x) are µn = L(x n ).
Orthogonal polynomials
A sequence of polynomials pn (x) n≥0 , where pn (x) has degree
n, is orthogonal if there is a linear functional L on polynomials
such that L pm (x)pn (x) = 0 for m 6= n, but L pm (x)2 6= 0.
We will assume that L(1) = 1.
The moments of the sequence pn (x) are µn = L(x n ). So
µ0 = 1.
Orthogonal polynomials
A sequence of polynomials pn (x) n≥0 , where pn (x) has degree
n, is orthogonal if there is a linear functional L on polynomials
such that L pm (x)pn (x) = 0 for m 6= n, but L pm (x)2 6= 0.
We will assume that L(1) = 1.
The moments of the sequence pn (x) are µn = L(x n ). So
µ0 = 1.
Theorem. The sequence pn (x) of monic polynomials
is
orthogonal if and only if there exist numbers an n≥0 and
bn n≥1 , with bn 6= 0 for all n ≥ 1 such that
xpn (x) = pn+1 (x) + an pn (x) + bn pn−1 (x),
with p0 (x) = 1 and p1 (x) = x − a0 .
n≥1
The connection
We then have the continued fraction
∞
X
k =0
1
µk x k =
1 − a0 x −
b1 x 2
1 − a1 x −
b2 x 2
1 − a2 x − · · ·
and the Hankel determinant det(µi+j )0≤i,j<n is equal to
2 b .
b1n−1 b2n−2 · · · bn−2
1
If we’re interested in Hankel determinants or continued
fractions, what good are the orthogonal polynomials?
If we’re interested in Hankel determinants or continued
fractions, what good are the orthogonal polynomials?
There is a systematic treatment of all the “classical” orthogonal
polynomials, that have explicit expressions as hypergeometric
series, called the Askey scheme. In a comprehensive paper (or
web site) by Koekoek and Swarttouw, you can look up all of
these orthogonal polynomials and find out everything you might
want to know about them. . .
If we’re interested in Hankel determinants or continued
fractions, what good are the orthogonal polynomials?
There is a systematic treatment of all the “classical” orthogonal
polynomials, that have explicit expressions as hypergeometric
series, called the Askey scheme. In a comprehensive paper (or
web site) by Koekoek and Swarttouw, you can look up all of
these orthogonal polynomials and find out everything you might
want to know about them. . . except the moments.
If we’re interested in Hankel determinants or continued
fractions, what good are the orthogonal polynomials?
There is a systematic treatment of all the “classical” orthogonal
polynomials, that have explicit expressions as hypergeometric
series, called the Askey scheme. In a comprehensive paper (or
web site) by Koekoek and Swarttouw, you can look up all of
these orthogonal polynomials and find out everything you might
want to know about them. . . except the moments.
What are the moments of the classical orthogonal polynomials?
If we’re interested in Hankel determinants or continued
fractions, what good are the orthogonal polynomials?
There is a systematic treatment of all the “classical” orthogonal
polynomials, that have explicit expressions as hypergeometric
series, called the Askey scheme. In a comprehensive paper (or
web site) by Koekoek and Swarttouw, you can look up all of
these orthogonal polynomials and find out everything you might
want to know about them. . . except the moments.
What are the moments of the classical orthogonal polynomials?
What are their generating functions?
If we’re interested in Hankel determinants or continued
fractions, what good are the orthogonal polynomials?
There is a systematic treatment of all the “classical” orthogonal
polynomials, that have explicit expressions as hypergeometric
series, called the Askey scheme. In a comprehensive paper (or
web site) by Koekoek and Swarttouw, you can look up all of
these orthogonal polynomials and find out everything you might
want to know about them. . . except the moments.
What are the moments of the classical orthogonal polynomials?
What are their generating functions?
Why do some, but not all of them, have nice exponential
generating functions?
If we’re interested in Hankel determinants or continued
fractions, what good are the orthogonal polynomials?
There is a systematic treatment of all the “classical” orthogonal
polynomials, that have explicit expressions as hypergeometric
series, called the Askey scheme. In a comprehensive paper (or
web site) by Koekoek and Swarttouw, you can look up all of
these orthogonal polynomials and find out everything you might
want to know about them. . . except the moments.
What are the moments of the classical orthogonal polynomials?
What are their generating functions?
Why do some, but not all of them, have nice exponential
generating functions?
What are the orthogonal polynomials whose moments are the
Genocchi numbers?
How do we find the moments of a sequence of orthogonal
polynomials? The key is to find numbers α0 , α1 , α2 , .. . such
that we can evaluate L (x + α0 )(x + α1 ) · · · (x + αk ) for each
k . (Usually this isn’t too hard to do.)
How do we find the moments of a sequence of orthogonal
polynomials? The key is to find numbers α0 , α1 , α2 , .. . such
that we can evaluate L (x + α0 )(x + α1 ) · · · (x + αk ) for each
k . (Usually this isn’t too hard to do.)
We then apply the following Lemma:
∞
X
(x +α0 )(x +α1 ) · · · (x +αn−1 )
n=0
tn
1
=
.
(1 + α0 t) · · · (1 + αn t)
1 − xt
Proof. We have the indefinite sum
m
X
n=0
(x + α0 ) · · · (x + αn−1 )
tn
(1 + α0 t) · · · (1 + αn t)
1
t m+1
=
1 − (x + α0 ) · · · (x + αn )
,
1 − xt
(1 + α0 t) · · · (1 + αn t)
which is easily proved by induction. The lemma follows by
taking m → ∞.
Now suppose that
L (x + α0 )(x + α1 ) · · · (x + αn−1 ) = Mn .
We apply L to
∞
X
(x +α0 )(x +α1 ) · · · (x +αn−1 )
n=0
tn
1
=
.
(1 + α0 t) · · · (1 + αn t)
1 − xt
to get
∞
X
n=0
tn
Mn
=L
(1 + α0 t) · · · (1 + αn t)
=
∞
X
k =0
1
1 − xt
L(x k )t k =
∞
X
k =0
µk t k .
Let’s look at an example. First, the standard notation:
p Fq
(α)n = α(α + 1) · · · (α + n − 1)
X
∞
(a1 )n · · · (ap )n n
a1 , . . . , ap z =
z
b , . . . , bp
n! (b )n · · · (bq )n
1
n=0
1
The Hahn polynomials are defined by
−n, n + α + β + 1, −x Qn (x; α, β, N) = 3 F2
1
α + 1, −N
They are orthogonal with respect to the linear functional L given
by
N X
α+x
β+N −x
L p(x) =
p(x),
x
N −x
x=0
By applying Vandermonde’s theorem we find that
(α + β + 2)N (α + 1)k (α + β + 2 + N)k
L (x + α + 1)k =
.
N!
(α + β + 2)k
Since we want L(1) to be 1, we normalize this by dividing L by
(α + β + 2)N /N!. (This is independent of k .) Then with our new
normalization,
(α + 1)k (α + β + 2 + N)k
L (x + α + 1)k =
.
(α + β + 2)k
Since the right side is a rational function of α, β, and N, we
don’t need N to be a nonnegative integer and we can therefore
make a change of variables, α = A − 1, β = C − A − 1,
N = B − C. Then
(A)n (B)n
L (x +A)n = L (x +A)(x +A+n) · · · (x +A+n−1) =
.
(C)n
Now we can apply our lemma to get the ordinary generating
function for the moments µn of the Hahn polynomials:
∞
X
n=0
µn t n =
∞
X
(A)n (B)n
n=0
(C)n
tn
.
j=0 (1 + (A + j)t)
Qn
This can be written as a rather strange-looking hypergeometric
series
∞
1 X
(A)n (Bn )
1
A, B, 1
1 .
=
3 F2
−1
−1
1 + At
1 + At
C, 1 + A + t (C)n (1 + A + t )n
n=0
This can be written as a rather strange-looking hypergeometric
series
∞
1 X
(A)n (Bn )
1
A, B, 1
1 .
=
3 F2
−1
−1
1 + At
1 + At
C, 1 + A + t (C)n (1 + A + t )n
n=0
How do we get an exponential generating function for the
moments?
This can be written as a rather strange-looking hypergeometric
series
∞
1 X
(A)n (Bn )
1
A, B, 1
1 .
=
3 F2
−1
−1
1 + At
1 + At
C, 1 + A + t (C)n (1 + A + t )n
n=0
How do we get an exponential generating function for the
moments?
We define a linear operator ε on formal power series by
ε
X
∞
n=0
un t
n
=
∞
X
n=0
un
tn
.
n!
This can be written as a rather strange-looking hypergeometric
series
∞
1 X
(A)n (Bn )
1
A, B, 1
1 .
=
3 F2
−1
−1
1 + At
1 + At
C, 1 + A + t (C)n (1 + A + t )n
n=0
How do we get an exponential generating function for the
moments?
We define a linear operator ε on formal power series by
ε
X
∞
un t
n
n=0
Then we apply ε to
P∞
n=0 µn t
=
∞
X
n=0
n.
un
tn
.
n!
Lemma.
(1 − e−t )n
tn
= e−At
.
ε
(1 + At)(1 + (A + 1)t) · · · (1 + (A + n)t)
n!
Proof #1. Expand the left side by partial fractions and the right
side by the binomial theorem.
Proof #2. Without explicitly computing the partial fraction on the
left, we can see that it will be a linear combination of e−At ,
e−(A+1)t , . . . , e−(A+n)t and the first nonzero term is t n /n!. The
right side is the only possibility.
Lemma.
(1 − e−t )n
tn
= e−At
.
ε
(1 + At)(1 + (A + 1)t) · · · (1 + (A + n)t)
n!
Proof #1. Expand the left side by partial fractions and the right
side by the binomial theorem.
Proof #2. Without explicitly computing the partial fraction on the
left, we can see that it will be a linear combination of e−At ,
e−(A+1)t , . . . , e−(A+n)t and the first nonzero term is t n /n!. The
right side is the only possibility.
Now we apply the lemma to
∞
X
n=0
n
µn t =
∞
X
(A)n (B)n
n=0
(C)n
tn
.
j=0 (1 + (A + j)t)
Qn
We have
∞
X
n=0
∞
µn
X (A)n (B)n
tn
(1 − e−t )n
=
e−At
n!
(C)n
n!
n=0
A, B −At
−t
= e 2 F1
1−e
.
C Let’s look at a few examples. First let’s see how this generalizes
the Meixner polynomials, which are the most general Sheffer
orthogonal polynomials. To get (one form of) the Meixner
polynomials, we take B = uC and then take the limit as C → ∞,
getting as the exponential generating function for the moments
(1 − u)et + u
−A
We have
∞
X
∞
X (A)n (B)n
tn
(1 − e−t )n
µn =
e−At
n!
(C)n
n!
n=0
n=0
A, B −t
−At
1−e
= e 2 F1
.
C As another example, if we take A = B = 1, C = 2, we get the
Bernoulli numbers:
∞
X
n=0
so µn = Bn .
µn
tn
t
= t
,
n!
e −1
We have
∞
X
∞
µn
n=0
X (A)n (B)n
(1 − e−t )n
tn
=
e−At
n!
(C)n
n!
n=0
A, B −t
−At
1−e
.
= e 2 F1
C Similarly, if we take A = 1, B = 2, C = 3, we get the Bernoulli
numbers shifted by 1:
∞
X
n=0
so µn = −2Bn+1 .
µn
tn
t
d
= −2
,
t
n!
dt e − 1
We can use the same approach on the Wilson polynomials,
which are the most general “classical” orthogonal polynomials
(with q = 1). They are defined by
−n, n + a + c + d − 1, a + ix, a − ix 2
Wn (x ) = 4 F3
1 ,
a + b, a + c, a + d
(in this form they are not monic). The corresponding linear
functional is
L p(x) =
Z ∞
Γ(a + ix)Γ(b + ix)Γ(c + ix)Γ(d + ix) 2
1
p(x 2 ) dx
2π 0 Γ(2ix)
It follows easily from known facts that
L (x + a2 )(x + (a + 1)2 ) · · · (x + (a + n − 1)2 )
(a + b)n (a + c)n (a + d)n
=
(a + b + c + d)n
so from our first lemma,
∞
X
n=0
n
µn t =
∞
X
(a + b)n (a + c)n (a + d)n t n
Qn
2t
(a
+
b
+
c
+
d)
1
+
(a
+
j)
n
j=0
n=0
It follows easily from known facts that
L (x + a2 )(x + (a + 1)2 ) · · · (x + (a + n − 1)2 )
(a + b)n (a + c)n (a + d)n
=
(a + b + c + d)n
so from our first lemma,
∞
X
n=0
n
µn t =
∞
X
(a + b)n (a + c)n (a + d)n t n
Qn
2t
(a
+
b
+
c
+
d)
1
+
(a
+
j)
n
j=0
n=0
What about an exponential generating function?
It follows easily from known facts that
L (x + a2 )(x + (a + 1)2 ) · · · (x + (a + n − 1)2 )
(a + b)n (a + c)n (a + d)n
=
(a + b + c + d)n
so from our first lemma,
∞
X
n=0
n
µn t =
∞
X
(a + b)n (a + c)n (a + d)n t n
Qn
2t
(a
+
b
+
c
+
d)
1
+
(a
+
j)
n
j=0
n=0
What about an exponential generating function?
If we replace t with −t 2 , then the denominator factors into linear
factors:
n
Y
j=0
1 − (a + j)2 t 2 =
1 − (a + n)t · · · 1 − (a + 1)t 1 − at
× 1 + at 1 + (a + 1 t · · · 1 + (a + n)t .
We want to apply our lemma
tm
(1 − e−t )m
ε
= e−At
.
(1 + At)(1 + (A + 1)t) · · · (1 + (A + m)t)
m!
to
t 2n
.
1 − (a + n)t · · · 1 − (a + 1)t 1 − at
× 1 + at 1 + (a + 1 t · · · 1 + (a + n)t
We want to apply our lemma
tm
(1 − e−t )m
ε
= e−At
.
(1 + At)(1 + (A + 1)t) · · · (1 + (A + m)t)
m!
to
t 2n
.
1 − (a + n)t · · · 1 − (a + 1)t 1 − at
× 1 + at 1 + (a + 1 t · · · 1 + (a + n)t
We can do this if a = 0 or if a = 1/2. (If a = 1/2, we must
multiply by t.) Our conclusion is
For a = 0,
∞
X
n=0
t 2n
µn
= 3 F2
(2n)!
2 t
b, c, d
sin
2
b + c + d, 12 and for a = 1/2,
∞
X
n=0
t
t 2n+1
= 2 sin 3 F2
µn
(2n + 1)!
2
!
b + 12 , c + 12 , d + 21 2 t
sin
2
b + c + d + 12 , 32 If we take the limit as d → ∞ in the Wilson polynomials
−n, n + a + c + d − 1, a + ix, a − ix 2
Wn (x ) = 4 F3
1 ,
a + b, a + c, a + d
we get the continuous dual Hahn polynomials
−n, a + ix, a − ix 2
1 .
pn (x ) = 3 F2
a + b, a + c The generating function for the moments of the continuous dual
Hahn polynomials is
∞
X
∞
X
(a + b)n (a + c)n t n
Qn
µn (a, b, c)t =
2
j=0 1 + (a + j) t
n=0
n=0
n
Then it is clear that the moments µn (a, b, c) are polynomials in
a, b, and c, and they have nonnegative coefficients, as is clear
from the corresponding continued fraction.
In fact µn (1, 1, 1) = G2n+4 and µn (0, 1, 1) = P
G2n+2 , where the
n
Genocchi numbers are given by x tan 12 x = ∞
n=2 Gn x /n!, and
the polynomials µn (a, b, c) are polynomials studied by Dumont
and Foata as refinements of the Genocchi numbers.
In fact µn (1, 1, 1) = G2n+4 and µn (0, 1, 1) = P
G2n+2 , where the
n
Genocchi numbers are given by x tan 12 x = ∞
n=2 Gn x /n!, and
the polynomials µn (a, b, c) are polynomials studied by Dumont
and Foata as refinements of the Genocchi numbers.
For a = 0 we have the generating function
∞
X
n=0
t 2n
= 2 F1
µn (0, b, c)
(2n)!
b, c 2 t
,
1 sin
2
2
In fact µn (1, 1, 1) = G2n+4 and µn (0, 1, 1) = P
G2n+2 , where the
n
Genocchi numbers are given by x tan 12 x = ∞
n=2 Gn x /n!, and
the polynomials µn (a, b, c) are polynomials studied by Dumont
and Foata as refinements of the Genocchi numbers.
For a = 0 we have the generating function
∞
X
n=0
t 2n
= 2 F1
µn (0, b, c)
(2n)!
b, c 2 t
,
1 sin
2
2
and it’s not hard to check that indeed,
X
∞
1, 1 2 t
d2
xn
x
=
=
Gn+2
x tan
2 F1
1 sin
2
2
n!
2
dx
2
n=0
using
2 F1
p
1, 1 2
1 − y 2 + y sin−1 y
.
=
1 y
(1 − y 2 )3/2
2
Another application of the continuous dual Hahn
polynomials
We have
∞
X
n=0
t 2n
µn ( 12 , − 12 γ, 21 γ)
= 2 F1
(2n)!
=
1
2 (1
!
− γ), 12 (1 + γ) 2 t
sin
3
2
2
sin 21 γt
γ sin 12 t
In the case γ = 1/3, the corresponding Hankel determinant
counts alternating sign matrices (Colomo and Pronko).
Hankel Determinants and Ternary Tree Numbers
Let
1
3n
an =
,
2n + 1 n
so that an is the number of trees with n vertices. Michael
Somos conjectured that det(ai+j )0≤i,j≤n−1 is the number of
cyclically symmetric transpose complement plane partitions
whose Ferrers diagrams fit in an n × n × n box. This number is
known to be
n
Y
(3i + 1) (6i)! (2i)!
.
(4i + 1)! (4i)!
i=1
Somos had similar conjectures for the Hankel determinants
det(ai+j )0≤i,j≤n−1 and det(a(i+j+1)/2 )0≤i,j≤n−1 , relating them to
alternating sign matrices invariant under vertical reflection and
alternating sign matrices invariant under both vertical and
horizontal reflection.
Let
g(x) =
∞
X
n=0
∞
1
3n n X
x =
an x n .
2n + 1 n
n=0
Then g(x) satisfies
g(x) = 1 + xg(x)3 .
If we compute the continued fraction for g(x), we find
empirically that it has a simple formula, which implies the
determinant evaluation. I found a (complicated) proof of this
continued fraction. Then Guoce Xin found a much nicer proof,
using Gauss’s continued fraction.
Let
g(x) =
∞
X
n=0
∞
1
3n n X
x =
an x n .
2n + 1 n
n=0
Then g(x) satisfies
g(x) = 1 + xg(x)3 .
If we compute the continued fraction for g(x), we find
empirically that it has a simple formula, which implies the
determinant evaluation. I found a (complicated) proof of this
continued fraction. Then Guoce Xin found a much nicer proof,
using Gauss’s continued fraction.
However, the same method had already been used by Ulrich
Tamm to evaluate these determinants, before Somos stated his
conjectures.
(But there is no known combinatorial connection between these
Hankel determinants and plane partitions or alternating sign
matrices.)
The key fact is that the continued fraction for g(x) is a special
case of Gauss’s continued fraction:
2 F1
a, b a, b + 1 x = S(x; λ1 , λ2 , . . .),
x
2 F1
c c+1 (1)
where
(a + n − 1)(c − b + n − 1)
,
(c + 2n − 2)(c + 2n − 1)
(b + n)(c − a + n)
=
,
(c + 2n − 1)(c + 2n)
λ2n−1 =
n = 1, 2, . . . ,
λ2n
n = 1, 2, . . . ,
and S(x; λ1 , λ2 , λ3 , . . .) denotes the continued fraction
S(x; λ1 , λ2 , λ3 , . . .) =
1
λ1 x
1−
λ2 x
1−
λ3 x
1−
..
.
(2)
In fact we have
g = 2 F1
2 4 3
3, 3; 2
27
4 x
.
2 F1
2 1 1
3, 3; 2
27
4 x
.
and this isn’t too hard to prove; if we set f = g − 1 by Lagrange
inversion one can show that the numerator is (1 + f )2 /(1 − 2f )
and the denominator is (1 + f )/(1 − 2f ). Xin and I tried to find
all related cases where an instance of Gauss’s continued
fraction is a polynomial in f (and thus the coefficients have an
explicit formula).
In fact we have
g = 2 F1
2 4 3
3, 3; 2
27
4 x
.
2 F1
2 1 1
3, 3; 2
27
4 x
.
and this isn’t too hard to prove; if we set f = g − 1 by Lagrange
inversion one can show that the numerator is (1 + f )2 /(1 − 2f )
and the denominator is (1 + f )/(1 − 2f ). Xin and I tried to find
all related cases where an instance of Gauss’s continued
fraction is a polynomial in f (and thus the coefficients have an
explicit formula).
We found exactly 10, but we couldn’t prove that there aren’t any
more.
Five of them are
1 + f = 2 F1
2 4 3
3, 3; 2
(1 + f )2 = 2 F1
4 5 5
3, 3; 2
(1 + f )(1 + 12 f ) = 2 F1
5 7 7
3, 3; 2
(1 + f )(1 − 12 f ) = 2 F1
5 7 5
3, 3; 2
(1 + f )(1 + 25 f ) = 2 F1
2 4 5
3, 3; 2
Why do they factor?
27
4 x
.
27
4 x
.
27
4 x
.
27
4 x
.
27
4 x
.
2 F1
2 F1
2 F1
2 F1
2 F1
2 1 1
3, 3; 2
4 2 3
3, 3; 2
5 4 5
3, 3; 2
5 4 3
3, 3; 2
2 1 3
3, 3; 2
27
4 x
27
4 x
27
4 x
27
4 x
27
4 x
Five of them are
1 + f = 2 F1
2 4 3
3, 3; 2
(1 + f )2 = 2 F1
4 5 5
3, 3; 2
(1 + f )(1 + 12 f ) = 2 F1
5 7 7
3, 3; 2
(1 + f )(1 − 12 f ) = 2 F1
5 7 5
3, 3; 2
(1 + f )(1 + 25 f ) = 2 F1
2 4 5
3, 3; 2
27
4 x
.
27
4 x
.
27
4 x
.
27
4 x
.
27
4 x
.
Why do they factor? We don’t know.
2 F1
2 F1
2 F1
2 F1
2 F1
2 1 1
3, 3; 2
4 2 3
3, 3; 2
5 4 5
3, 3; 2
5 4 3
3, 3; 2
2 1 3
3, 3; 2
27
4 x
27
4 x
27
4 x
27
4 x
27
4 x
Five of them are
1 + f = 2 F1
2 4 3
3, 3; 2
(1 + f )2 = 2 F1
4 5 5
3, 3; 2
(1 + f )(1 + 12 f ) = 2 F1
5 7 7
3, 3; 2
(1 + f )(1 − 12 f ) = 2 F1
5 7 5
3, 3; 2
(1 + f )(1 + 25 f ) = 2 F1
2 4 5
3, 3; 2
27
4 x
.
27
4 x
.
27
4 x
.
27
4 x
.
27
4 x
.
2 F1
2 F1
2 F1
2 F1
2 F1
2 1 1
3, 3; 2
4 2 3
3, 3; 2
5 4 5
3, 3; 2
5 4 3
3, 3; 2
2 1 3
3, 3; 2
27
4 x
27
4 x
27
4 x
27
4 x
27
4 x
Why do they factor? We don’t know.
The other five are the same as these, but with different constant
terms.