recent progress on tiling proofs of q--series identities · proof. construct tilings in the...

Recent Progress on Tiling Proofs of

q–Series Identities

David P. Little

November 11, 2008

www.math.psu.edu/dlittle

Weighted Tilings

Definition

A tiling is a covering of an infinitely long board:

· · ·1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Weighted Tilings

Definition


· · ·1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

using different types of tiles:

Weighted Tilings

Definition


· · ·1 2 3 4 5 6 7 8 9 10 11 12 13 14 15


3 5

Weighted Tilings

Definition


· · ·1 2 3 4 5 6 7 8 9 10 11 12 13 14 15


3 5

The weight of a tiling T is given by

w(T ) =∏

t∈T

w(t)

where w(t) is the weight of the tile t. The weight of a white square willalways be 1. Each tiling will have a finite number of weighted tiles.

Lebesgue Identities

The weight of tile t:

w(t) =

qi if t is a in position i

zqi if t is a in position i

1 if t is a in position i

Theorem

∞∑

n=0

(−z; q)n(q; q)n

q(n+1

2 ) =

∞∏

n=1

(1 + qn)(1 + zq2n−1)

where (z; q)n = (1 − z)(1 − zq) · · · (1 − zqn−1).

Eight Identities of Rogers

∑

n≥0

znqn2

(q; q)n= (−zq2; q2)∞

∑

n≥0

znqn2

(q2; q2)n(−zq2; q2)n

∑

n≥0

znqn2+n

(q; q)n= (−zq2; q2)∞

∑

n≥0

znqn2+2n

(q2; q2)n(−zq2; q2)n

∑

n≥0

z2nq4n2+2n

(q4; q4)n= (zq2; q2)∞

∑

n≥0

znqn2+n

(q2; q2)n(zq2; q2)n

∑

n≥0

znq2n2+n

(q2; q2)n= (zq2; q2)∞

∑

n≥0

znq(3n2+n)/2

(q; q)n(zq2; q2)n

Five More Identities

∑

n≥0

znq2n2

(q2; q2)n= (zq; q2)∞

∑

n≥0

znq(3n2+n)/2

(q; q)n(zq; q2)n+1

∑

n≥0

znqn2

(q; q)n= (−zq; q)∞

∑

n≥0

(−1)nz2nq3n2

(q2; q2)n(−zq; q)2n

∑

n≥0

znqn2+n(1 − z2q2n+3)

(q; q)n= (−zq; q)∞

∑

n≥0

(−1)nz2nq3n2

(q2; q2)n(−zq; q)2n+1

An Example


w(t) =

−zqi if t is a in position i



An Example


w(t) =




Theorem (Cauchy)

∞∑

n=0

znqn2

(q; q)n(zq; q)n=

1

(zq; q)∞

An Example


w(t) =




Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1

Proof. Construct tilings in the following manner:

STEP 1: Place n in positions 1, 3, 5, . . . , 2n − 1.

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1


STEP 1: Place n in positions 1, 3, 5, . . . , 2n − 1.

· · ·

This accounts for a weight of

znq1+3+5+···+(2n−1) = znqn2

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1


STEP 2: Pick j ≥ 0 and insert immediately after the jth tile.

︷︸︸︷j ≥ 0 tiles

· · ·

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1




· · ·︸︷︷︸

k dominoes

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1




· · ·︸︷︷︸

k dominoes︸︷︷︸

n − k dominoes

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1




· · ·︸︷︷︸


n − k dominoes

position j + k + 1

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1




· · ·︸︷︷︸


n − k dominoes

position j + k + 1

Increases the weight of the tiling by −zqj+k+1qn−k = −zqn+j+1

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1




· · ·︸︷︷︸


n − k dominoes

position j + k + 1

Increases the weight of the tiling by −zqj+k+1qn−k = −zqn+j+1

∏

j≥0

(1 − zqn+j+1) =(zq; q)∞(zq; q)n

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1


STEP 3: Project the dominoes

· · ·· · · · · ·

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1



· · ·· · · · · ·

↓

· · ·· · · · · ·

Increases the weight of the tiling by a factor of q.

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1



· · ·· · · · · ·

↓

· · ·· · · · · ·

Increases the weight of the tiling by a factor of q. Therefore, the left–handside is the generating function for all weighted tilings.

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1



When projecting tiles, always work in a right to left, weakly increasingmanner. In other words, make sure that each domino is projected at leastas many times as the dominoes to its left are projected.

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1



When projecting tiles, always work in a right to left, weakly increasingmanner. In other words, make sure that each domino is projected at leastas many times as the dominoes to its left are projected.

This process accounts for a weight of

1

(1 − q)(1 − q2)(1 − q3) · · · (1 − qn)=

1

(q; q)n

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1

Proof. Cancel out all non-empty tilings:

STEP 4:Find first occurrence of: or

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1



and replace with: or (respectively)

Theorem (Cauchy)

(zq; q)∞

∞∑

n=0

znqn2

(q; q)n(zq; q)n= 1




The only remaining tiling is the empty tiling, which has weight 1.

q-analog of the binomial series

Weight tiles in the following manner:

w(t) =

aqi if t is a with i or to its left

bqi if t is a with i or to its left

1 if t is a

q-analog of the binomial series


w(t) =

aqi if t is a with i or to its left


1 if t is a

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn

Proof. PART I: Interpret infinite series

STEP 1: Place n black or gray squares in positions 1, 2, 3, . . . , n.

· · ·

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn


STEP 1: Place n black or gray squares in positions 1, 2, 3, . . . , n.

· · ·

A in position i accounts for a weight of a.

A in position i accounts for a weight of bqn−i.This process accounts for a weight of

n∏

i=1

(a + bqn−i) = (−b/a; q)nan

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn


STEP 2: Project the tiles.

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn



↓

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn



↓ ↓

This process increases the weight of the tiling by a factor of q.

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn



↓ ↓

This process increases the weight of the tiling by a factor of q. Therefore,the left-hand side is the generating function for all weighted tilings.

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn

Proof. PART II: Interpret infinite product

Each tiling can be broken up into segments:

· · · · · ·

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn



· · · · · ·︷︸︸︷

j ≥ 0 black squares

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn



· · · · · ·︷︸︸︷


The weight of the nth segment for n ≥ 0 is given by

(1 + bqn)

∞∑

j=0

ajqnj =1 + bqn

1 − aqn

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn



· · · · · ·︷︸︸︷


The weight of the nth segment for n ≥ 0 is given by

(1 + bqn)

∞∑

j=0

ajqnj =1 + bqn

1 − aqn

Multiplying over n ≥ 0 completes the construction.

A Few Observations

The generating function for all weighted tilings is given by

∞∑

n=0

(−b/a; q)nan

(q; q)n

A Few Observations


∞∑

n=0

(−b/a; q)nan

(q; q)n

Adding the parameter c in the following manner allows us to countnumber of white squares before the last weighted tile.

∞∑

n=0

(−b/a; q)nan

(cq; q)n

A Few Observations


∞∑

n=0

(−b/a; q)nan

(q; q)n

Adding the parameter c in the following manner allows us to countnumber of white squares before the last weighted tile.

∞∑

n=0

(−b/a; q)nan

(cq; q)n

Multiplication by a produces the generating function for tilings that startwith a black square.

∞∑

n=0

(−b/a; q)nan+1

(cq; q)n

Rogers-Fine Identity

Theorem

∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0

(−b/a; q)n(−b/c; q)nancnqn2

(1 + bq2n)

(a; q)n+1(cq; q)n

Rogers-Fine Identity

Theorem

∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(a; q)n+1(cq; q)n

Multiplying both sides by (1 − a) yields:

(1 − a)

∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n

In this form, the left hand side counts tilings that do not start with a blacksquare where the power of c keeps track of the number of white squaresbefore the last weighted tile.

Theorem (Rogers-Fine)

(1 − a)∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n

Proof. Interpret right–hand side

Front segments:

· · ·


(1 − a)∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n


Front segments:

· · ·︸︷︷︸



(1 − a)∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n


Front segments:

· · ·︸︷︷︸


Back segments:


(1 − a)∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n


Front segments:

· · ·︸︷︷︸


Back segments:

︸︷︷︸

k ≥ 0 white squares


(1 − a)∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n


Front segments:

· · ·︸︷︷︸


Back segments:

︸︷︷︸

k ≥ 0 white squares

· · ·

The center of a tiling marks the transition between front segments andback segments. The center can either be empty or a gray square.


(1 − a)

∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n


Tilings that start with or and have n front/back segments.


(1 − a)

∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n


Tilings that start with or and have n front/back segments.Generating function for n front segments:

(c + b)

(1 − aq)

(c + bq)

(1 − aq2)· · ·

(c + bqn−1)

(1 − aqn)=

(−b/c; q)ncn

(aq; q)n


(1 − a)

∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n


Tilings that start with or and have n front/back segments.Generating function for n back segments:

(aqn + bq2n−1)

(1 − cqn)· · ·

(aqn + bqn+1)

(1 − cq2)

(aqn + bqn)

(1 − cq)=

(−b/a; q)nanqn2

(cq; q)n


(1 − a)

∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n


The center can either be empty or a gray square.

· · · · · ·︸︷︷︸

n front segments︸︷︷︸

n back segments

If the center is a gray square, then it has weight bqn and increases theweight of the back segments by qn.


(1 − a)

∞∑

n=0

(−b/a; q)nan

(cq; q)n=

∞∑

n=0


(1 + bq2n)

(aq; q)n(cq; q)n


The center can either be empty or a gray square.

· · · · · ·︸︷︷︸

n front segments︸︷︷︸

n back segments

If the center is a gray square, then it has weight bqn and increases theweight of the back segments by qn.In other words, the factor (1 + bq2n) represents the choice of the center.

Recall:

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn

Recall:

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn

Theorem

∞∑

n=0

(−a/b; q)nbnq(n

2)

(q; q)n(a; q)n=

∞∏

n=0

1 + bqn

1 − aqn

Recall:

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn

Theorem

∞∑

n=0

(−c/a; q)nanq(n

2)

(q; q)n(c; q)n=

∞∏

n=0

1 + aqn

1 − cqn

Recall:

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn

Theorem

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

1 + aqn

1 − cqn

Recall:

Theorem (Cauchy)

∞∑

n=0

(−b/a; q)nan

(q; q)n=

∞∏

n=0

1 + bqn

1 − aqn

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)

∞∑

n=0

(a + c)(a + cq) · · · (a + cqn−1)q(n+1

2 )

(q; q)n(1 − cqn+1)(1 − cqn+2) · · ·

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)

∞∑

n=0

(a + c)(a + cq) · · · (a + cqn−1)q(n+1

2 )

(q; q)n(1 − cqn+1)(1 − cqn+2) · · ·


w(t) =

aqi if t is a in position i

cqi if t is a in position i

1 if t is a

−cqi if t is a in position i

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)


STEP 1: Place in positions 1, 2, 3, . . . , n.

· · ·

A in position i accounts for a weight of aqi.

This process accounts for a weight of

anq1+2+···+n = anq(n+1

2 )

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)


STEP 2: Place or in positions i > n.

· · ·

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)


STEP 2: Place or in positions i > n.

· · ·

A in position i accounts for a weight of 1.

A in position i accounts for a weight of −cqi.This process accounts for a weight of

∏

i>n

(1 − cqi) =(cq; q)∞(cq; q)n

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)


STEP 3: Project the black tiles.

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)



↓

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)



↓ ↓

This process increases the weight of the tiling by a factor of q.

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)


STEP 3a: Convert into .If a square is converted into a circle, project each of the tiles to its right.

· · ·

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)


STEP 3a: Convert into .If a square is converted into a circle, project each of the tiles to its right.

· · ·

The factor (1 + cqn−i/a) represents the choice of converting the ithsquare into a circle.

n∏

i=1

(1 + cqn−i/a) = (−c/a; q)n

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)


STEP 3b: Project the black tiles.

· · ·

Constructs all tilings where every circle is followed by a white tile.

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)


Cancel out any tilings with a or

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)



Find first occurrence of: or

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)





Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)





Remaining tilings cannot have any circles.

Theorem

(cq; q)∞

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(cq; q)n=

∞∏

n=1

(1 + aqn)





Remaining tilings cannot have any circles.

Therefore, each tiling can be constructed by simply deciding whether ornot to place a black square in each position n ≥ 1.

q-analog of Gauss’s Theorem


w(t) =



abqi if t is a with i or to its left

bcqi if t is a with i or to its left

1 if t is a


q-analog of Gauss’s Theorem


w(t) =



abqi if t is a with i or to its left


1 if t is a


Theorem (Heine)

(cq; q)∞

∞∑

n=0

(−c/a; q)n(−q/b; q)nanbn

(q; q)n(cq; q)n=

∞∏

n=1

(1 + bcqn−1)(1 + aqn)

(1 − abqn−1)

q-analog of Kummer’s Theorem


w(t) =





−bcqi if t is a in position i + 1

1 if t is a

q-analog of Kummer’s Theorem


w(t) =





−bcqi if t is a in position i + 1

1 if t is a

Theorem (Bailey)

(bc; q)∞

∞∑

n=0

(−c; q)n(−q/b; q)nbn

(q; q)n(bc; q)n=

∞∏

n=1

(1 + qn)(1 + cq2n−1)(1 + cb2q2n−2)

1 − bqn−1

A q-Series Symmetry Result

Theorem (Ramanujan)

(−bq; q)n

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(−bq; q)n= (−aq; q)n

∞∑

n=0

(−c/b; q)nbnq(n+1

2 )

(q; q)n(−aq; q)n

A q-Series Symmetry Result

Theorem (Ramanujan)

(−bq; q)n

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(−bq; q)n= (−aq; q)n

∞∑

n=0

(−c/b; q)nbnq(n+1

2 )

(q; q)n(−aq; q)n

Equivalently, the following function is symmetric in the variables a and b:

(−bq; q)n

∞∑

n=0

(−c/a; q)nanq(n+1

2 )

(q; q)n(−bq; q)n

Labeled Tilings


w(t) =

ajqi if t is a j in position i, 1 ≤ j ≤ k + 1

cjqi if t is a j in position i, 1 ≤ j ≤ k


Labeled Tilings


w(t) =

ajqi if t is a j in position i, 1 ≤ j ≤ k + 1

cjqi if t is a j in position i, 1 ≤ j ≤ k


Consider the following generating function:

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).

STEP 1: Place n1 ≥ 0 1-squares in positions 1, 2, . . . , n1, immediatelyfollowed by n2 ≥ 0 2–squares, and so on.

· · ·

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


· · ·1 1 1

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


· · ·1 1 1 2 2

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


· · ·1 1 1 2 2 3 3 3 3

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


· · ·1 1 1 2 2 3 3 3 3 4 4

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


· · ·1 1 1 2 2 3 3 3 3 4 4

A j in position i accounts for a weight of ajqi.


an1

1 · · · ank

k q(n1+n2+···+nk+1

2)

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).

STEP 2: Arbitrarily place k +1-squares in positions i > n1 +n2 + · · ·+nk.

· · ·1 1 1 2 2 3 3 3 3 4 4

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


· · ·1 1 1 2 2 3 3 3 3 4 4 5

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


· · ·1 1 1 2 2 3 3 3 3 4 4 5 5

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


· · ·1 1 1 2 2 3 3 3 3 4 4 5 5 5

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


· · ·1 1 1 2 2 3 3 3 3 4 4 5 5 5 5

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


· · ·1 1 1 2 2 3 3 3 3 4 4 5 5 5 5

A j in position i accounts for a weight of aqi.


∏

i>n1+n2+···+nk

(1 + ak+1qi) =

(−ak+1q; q)∞(−ak+1q; q)n1+n2+···+nk

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).

STEP 3: Projectiles, j and j.

j > j ≥ j j > j > j

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


j > j ≥ j j > j > j

↓

> j j ≥ j

↓

> j j > j

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).


j > j ≥ j j > j > j

↓

> j j ≥ j

↓

> j j > j

This process increases the weight of a tiling by a factor of q.

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).

STEP 3a: Decide whether or not to convert each j into j for eachj = k, . . . , 2, 1. If so, project every j-tile that appears to its right.This accounts for a weight of

k∏

i=1

nj∏

j=1

(1 + ciqj−1/ai) = (−c1/a1; q)n1

· · · (−ck/ak; q)nk

Note that every circle must be followed by a white square or a tile with alarger label.

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).

STEP 3b: Project the j-tiles for j = k, . . . , 2, 1. As usual, work in a rightto left manner and make sure to project each j-tile at least as many timesas the j-tiles to its left are projected.

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).

STEP 3b: Project the j-tiles for j = k, . . . , 2, 1. As usual, work in a rightto left manner and make sure to project each j-tile at least as many timesas the j-tiles to its left are projected. This accounts for a weight of

1

(q; q)n1· · · (q; q)nk

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).

STEP 3b: Project the j-tiles for j = k, . . . , 2, 1. As usual, work in a rightto left manner and make sure to project each j-tile at least as many timesas the j-tiles to its left are projected. This accounts for a weight of

1

(q; q)n1· · · (q; q)nk

Remark: We have constructed all labeled tilings that consist of weaklyincreasing sequences of weighted tiles separated by a white square wherethe label must strictly increase after a circle.

Labeled Tilings

Theorem

The generating function

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).

is symmetric in the variables (a1, a2, . . . , ak+1) and (c1, c2, . . . , ck).

Labeled Tilings

Theorem

The generating function

(−ak+1q)∞∑

n1,...,nk≥0

(−c1/a1)n1· · · (−ck/ak)nk

an1

1 · · · ank

k

(q)n1· · · (q)nk

(−ak+1q)n1+···+nk

q(n1+···+nk+1

2).

is symmetric in the variables (a1, a2, . . . , ak+1) and (c1, c2, . . . , ck).

Note that the following sequences of tiles have the same weight:

· · · · · ·1 1 2 2 2 3 3 3 3 4 4 5 5 5

· · · · · ·3 3 3 4 1 2 5 5 5 1 2 2 3 4

Tiling Statistics

Let Pn denote the set of Pell tilings, (i.e., tilings with , , ) ofa 1 × n board.

Definition

The number of or that are immediately followed by a intiling T is the number of descents in T , denoted des(T ).

Tiling Statistics


Definition


Find the generating function for descents

∑

T∈Pn

xdes(T )

Tiling Statistics


Definition


Find the generating function for descents

∑

T∈Pn

xdes(T )

Weight tilings in the following manner:

w(t) =

{

x if t is the last tile on the board

1 or − x otherwise

Tilings with no Descents

Let Fn(x) denote the G.F. for tilings of a 1×n board using and .



F0(x) = 1 ∅

F1(x) = x

F2(x) = (1 − x)x + x



F0(x) = 1 ∅

F1(x) = x

F2(x) = (1 − x)x + x

For n ≥ 3,

Fn(x) = (1 − x)Fn−1(x) + (1 − x)Fn−2(x)

F (x, t) =∞∑

n=0

Fn(x)tn

=1 + (2x − 1)t + (2x − 1)t2

1 − (1 − x)t + (1 − x)t2

F (x, t) =∞∑

n=0

Fn(x)tn

=1 + (2x − 1)t + (2x − 1)t2

1 − (1 − x)t + (1 − x)t2

Let Gn(x) denote the G.F. for tilings of a 1 × n board using , and

, where no or is followed by a .

F (x, t) =∞∑

n=0

Fn(x)tn

=1 + (2x − 1)t + (2x − 1)t2

1 − (1 − x)t + (1 − x)t2

Let Gn(x) denote the G.F. for tilings of a 1 × n board using , and

, where no or is followed by a .

G(x, t) =∞∑

n=0

Gn(x)tn

=1

1 − (1 − x)tF (x, t) −

(1 − x)t

1 − (1 − x)t+

xt

1 − (1 − x)t

=1 + 2(2x − 1)t + x(2x − 1)t2 + (x − 1)(2x − 1)t3

(1 − (1 − x)t)(1 − (1 − x)t + (1 − x)t2)

Symmetric Functions

Definition

Let XN = (x1, x2, . . . , xn). We say that f(XN ) is symmetric if

f(x1, x2, . . . , xN ) = f(xσ1, . . . , xσN

)

for all σ ∈ SN .

Elementary Symmetric Functions:

ek(XN ) =∑

1≤i1<···<ik≤N

xi1 · · · xik

eλ = eλ1eλ2

· · · eλl

Symmetric Functions

Definition

Let XN = (x1, x2, . . . , xn). We say that f(XN ) is symmetric if

f(x1, x2, . . . , xN ) = f(xσ1, . . . , xσN

)

for all σ ∈ SN .

Elementary Symmetric Functions:

ek(XN ) =∑

1≤i1<···<ik≤N

xi1 · · · xik

eλ = eλ1eλ2

· · · eλl

Homogeneous Symmetric Functions

hk(XN ) =∑

1≤i1≤···≤ik≤N

xi1 · · · xik

hλ = hλ1hλ2

· · ·hλl

Theorem

hn(X) =∑

λ⊢n

(−1)n−l(λ)Bλeλ(X)

where Bλ is the number of compositions that are rearrangements of the

parts of λ.

For example:

h3(X3) = x1x1x1 + x1x1x2 + x1x1x3 + x1x2x2 + x1x2x3

+ x1x3x3 + x2x2x2 + x2x2x3 + x2x3x3 + x3x3x3

= x1x2x3 − 2(x1x2 + x1x3 + x2x3)(x1 + x2 + x3)

+ (x1 + x2 + x3)3

= e3(X3) − 2e2(X3)e1(X3) + e1(X3)3

= e3(X3) − 2e2,1(X3) + e1,1,1(X3)

Theorem

hn(X) =∑

λ⊢n

(−1)n−l(λ)Bλeλ(X)

where Bλ is the number of compositions that are rearrangements of the

parts of λ.

Consider B4,2,2,1 = 12

A Ring Homomorphism

Define ζ(en) = (−1)n−1G(x, t)∣∣∣tn

.

ζ(hn) =∑

λ⊢n

(−1)n−l(λ)Bλζ(eλ(X))

=∑

λ⊢n

(−1)n−l(λ)Bλ

l(λ)∏

i=1

(−1)λi−1G(x, t)∣∣∣tλi

=∑

λ⊢n

Bλ

l(λ)∏

i=1

G(x, t)∣∣∣tλi

An Involution

Find first tile weighted by −x or consecutive bricks with no descent.

1 1 1 −x −x x

An Involution


1 1 1 −x −x x

l1 1 1 x −x x

An Involution


1 1 1 −x −x x

l1 1 1 x −x x

Fixed Points:

1 1 1 x 1 1 x 1 1 x 1 1 1 x

An Involution


1 1 1 −x −x x

l1 1 1 x −x x

Fixed Points:

1 1 1 x 1 1 x 1 1 x 1 1 1 x

Conclusion:

ζ(hn) =∑

T∈Pn

xdes(T )+1

Theorem

∞∑

n=0

hn(X)tn =1

1 +∑

n≥1(−1)nen(X)tn

∞∑

n=0

ζ(hn)tn =1

1 +∑∞

n=1(−1)nζ(en)tn

=1

1 +∑∞

n=1(−1)n(−1)n−1G(x, t)∣∣tn

tn

=1

1 −∑∞

n=1 G(x, t)∣∣tn

tn

=1

1 − (G(x, t) − 1)

=1

2 − G(x, t)

Recap

G(x, t) =1 + 2(2x − 1)t + x(2x − 1)t2 + (x − 1)(2x − 1)t3

(1 − (1 − x)t)(1 − (1 − x)t + (1 − x)t2)

ζ(hn) =∑

T∈Pn

xdes(T )+1

∞∑

n=0

ζ(hn)tn =1

2 − G(x, t)

Theorem

∑

T∈Pn

xdes(T )+1 =1

2 − G(x, t)

∣∣∣tn

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