Two Binomial Coefficient Analogues · 2013-03-05 · Variation 2: q-binomial coe cients Variation 3: bonomial coe cients Coda: an open question and bibliography. Let N ... De ne the

Two Binomial Coefficient Analogues

Bruce SaganDepartment of Mathematics

Michigan State UniversityEast Lansing, MI 48824-1027

[email protected]/˜sagan

March 4, 2013

The theme

Variation 1: binomial coefficients

Variation 2: q-binomial coefficients

Variation 3: fibonomial coefficients

Coda: an open question and bibliography

Outline

The theme





Suppose we are given a sequence

a0, a1, a2, . . .

defined so that we only know that the an are rational numbers.

However, it appears as if the an are actually nonegative integers.How would we prove this? There are two standard techniques.

1. Find a recursion for the an and use induction.

2. Find a combinatorial interpretation for the an. In other words,find sets S0,S1, S2, . . . such that, for all n,

an = #Sn

where # denotes cardinality.


a0, a1, a2, . . .

defined so that we only know that the an are rational numbers.However, it appears as if the an are actually nonegative integers.

How would we prove this? There are two standard techniques.



an = #Sn



a0, a1, a2, . . .

defined so that we only know that the an are rational numbers.However, it appears as if the an are actually nonegative integers.How would we prove this?

There are two standard techniques.



an = #Sn



a0, a1, a2, . . .

defined so that we only know that the an are rational numbers.However, it appears as if the an are actually nonegative integers.How would we prove this? There are two standard techniques.



an = #Sn



a0, a1, a2, . . .




an = #Sn



a0, a1, a2, . . .



2. Find a combinatorial interpretation for the an.

In other words,find sets S0,S1, S2, . . . such that, for all n,

an = #Sn



a0, a1, a2, . . .



2. Find a combinatorial interpretation for the an. In other words,find sets S0, S1, S2, . . . such that, for all n,

an = #Sn


Outline

The theme





Let N = {0, 1, 2, . . .}.

Suppose n, k ∈ N with 0 ≤ k ≤ n. Definethe corresponding binomial coefficient by(

n

k

)=

n!

k!(n − k)!.

Ex. We have (4

2

)=

4!

2!2!=

4 · 32 · 1

= 6.

Note that it is not clear from this definition that(nk

)is an integer.

Proposition

The binomial coefficients satisfy(n0

)=(nn

)= 1 and, for 0 < k < n,(

n

k

)=

(n − 1

k

)+

(n − 1

k − 1

).

Since the sum of two integers is an integer, induction on n gives:

Corollary

For all 0 ≤ k ≤ n we have(nk

)∈ N.

Let N = {0, 1, 2, . . .}. Suppose n, k ∈ N with 0 ≤ k ≤ n.

Definethe corresponding binomial coefficient by(

n

k

)=

n!

k!(n − k)!.

Ex. We have (4

2

)=

4!

2!2!=

4 · 32 · 1

= 6.


)is an integer.

Proposition


)=(nn

)= 1 and, for 0 < k < n,(

n

k

)=

(n − 1

k

)+

(n − 1

k − 1

).


Corollary


)∈ N.

Let N = {0, 1, 2, . . .}. Suppose n, k ∈ N with 0 ≤ k ≤ n. Definethe corresponding binomial coefficient by(

n

k

)=

n!

k!(n − k)!.

Ex. We have (4

2

)=

4!

2!2!=

4 · 32 · 1

= 6.


)is an integer.

Proposition


)=(nn

)= 1 and, for 0 < k < n,(

n

k

)=

(n − 1

k

)+

(n − 1

k − 1

).


Corollary


)∈ N.


n

k

)=

n!

k!(n − k)!.

Ex. We have (4

2

)=

4!

2!2!=

4 · 32 · 1

= 6.


)is an integer.

Proposition


)=(nn

)= 1 and, for 0 < k < n,(

n

k

)=

(n − 1

k

)+

(n − 1

k − 1

).


Corollary


)∈ N.


n

k

)=

n!

k!(n − k)!.

Ex. We have (4

2

)=

4!

2!2!=

4 · 32 · 1

= 6.


)is an integer.

Proposition


)=(nn

)= 1 and, for 0 < k < n,(

n

k

)=

(n − 1

k

)+

(n − 1

k − 1

).


Corollary


)∈ N.


n

k

)=

n!

k!(n − k)!.

Ex. We have (4

2

)=

4!

2!2!=

4 · 32 · 1

= 6.


)is an integer.

Proposition


)=(nn

)= 1 and, for 0 < k < n,(

n

k

)=

(n − 1

k

)+

(n − 1

k − 1

).


Corollary


)∈ N.


n

k

)=

n!

k!(n − k)!.

Ex. We have (4

2

)=

4!

2!2!=

4 · 32 · 1

= 6.


)is an integer.

Proposition


)=(nn

)= 1 and, for 0 < k < n,(

n

k

)=

(n − 1

k

)+

(n − 1

k − 1

).


Corollary


)∈ N.

(a) Subsets.

Let

Sn,k = {S : S is a k-element subset of {1, . . . , n}}.

Ex. We have

S4,2 = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.

Note

#S4,2 = 6 =

(4

2

).

Proposition

For 0 ≤ k ≤ n we have(nk

)= #Sn,k .

(a) Subsets.Let


Ex. We have

S4,2 = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.

Note

#S4,2 = 6 =

(4

2

).

Proposition


)= #Sn,k .

(a) Subsets.Let


Ex. We have

S4,2 = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.

Note

#S4,2 = 6 =

(4

2

).

Proposition


)= #Sn,k .

(a) Subsets.Let


Ex. We have

S4,2 = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.

Note

#S4,2 = 6 =

(4

2

).

Proposition


)= #Sn,k .

(a) Subsets.Let


Ex. We have

S4,2 = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}.

Note

#S4,2 = 6 =

(4

2

).

Proposition


)= #Sn,k .

(b) Words.

A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all iEx. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.Let

Wn,k = {w = a1 . . . an : w has k zeros and n − k ones}.

Ex. We have

W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.

Note that

#W4,2 = 6 =

(4

2

).

Any w = a1 . . . an ∈ Wn,k can be obtained by choosing k of the npositions to be zeros (and the rest will be ones by default).

Proposition


)= #Wn,k .

(b) Words.A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all i

Ex. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.Let


Ex. We have

W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.

Note that

#W4,2 = 6 =

(4

2

).


Proposition


)= #Wn,k .

(b) Words.A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all iEx. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.

Let


Ex. We have

W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.

Note that

#W4,2 = 6 =

(4

2

).


Proposition


)= #Wn,k .

(b) Words.A word of length n over a set A called the alphabet is a finitesequence w = a1 . . . an where ai ∈ A for all iEx. We have that w = SEICCGTC 44 is a word of length 10 overthe alphabet A = {A, . . . ,Z , 1, . . . , 9}.Let


Ex. We have

W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.

Note that

#W4,2 = 6 =

(4

2

).


Proposition


)= #Wn,k .



Ex. We have

W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.

Note that

#W4,2 = 6 =

(4

2

).


Proposition


)= #Wn,k .



Ex. We have

W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.

Note that

#W4,2 = 6 =

(4

2

).


Proposition


)= #Wn,k .



Ex. We have

W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.

Note that

#W4,2 = 6 =

(4

2

).


Proposition


)= #Wn,k .



Ex. We have

W4,2 = {0011, 0101, 0110, 1001, 1010, 1100}.

Note that

#W4,2 = 6 =

(4

2

).


Proposition


)= #Wn,k .

(c) Lattice paths.

A NE lattice path of length n is a squence P = s1 . . . sn starting at(0, 0) and with each si being a unit step north (N) or east (E ).Ex. We have

P = ENEEN =

Let

Pn,k = {P = s1 . . . sn : P has k N-steps and n − k E -steps}

There is a bijection f : Pn,k →Wn,k where w = f (P) is obtainedby replacing each N by a 0 and each E by a 1.Ex. We have

P = ENEENf←→ w = 10110 =

1

01 1 0

Proposition


)= #Pn,k .

(c) Lattice paths.A NE lattice path of length n is a squence P = s1 . . . sn starting at(0, 0) and with each si being a unit step north (N) or east (E ).

Ex. We have

P = ENEEN =

Let



P = ENEENf←→ w = 10110 =

1

01 1 0

Proposition


)= #Pn,k .

(c) Lattice paths.A NE lattice path of length n is a squence P = s1 . . . sn starting at(0, 0) and with each si being a unit step north (N) or east (E ).Ex. We have

P = ENEEN =

Let



P = ENEENf←→ w = 10110 =

1

01 1 0

Proposition


)= #Pn,k .


P = ENEEN =

Let



P = ENEENf←→ w = 10110 =

1

01 1 0

Proposition


)= #Pn,k .


P = ENEEN =

Let


There is a bijection f : Pn,k →Wn,k where w = f (P) is obtainedby replacing each N by a 0 and each E by a 1.

Ex. We have

P = ENEENf←→ w = 10110 =

1

01 1 0

Proposition


)= #Pn,k .


P = ENEEN =

Let



P = ENEENf←→ w = 10110 =

1

01 1 0

Proposition


)= #Pn,k .


P = ENEEN =

Let



P = ENEENf←→ w = 10110 =

1

01 1 0

Proposition


)= #Pn,k .

(d) Partitions.

An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then

λ = (3, 2, 2) = ⊆ 3× 4: g←→

We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns. Let

Ln,k = {λ : λ ⊆ k × (n − k)}.

There is a bijection g : Ln,k → Pn,k : given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.

Proposition


)= #Ln,k .

(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm).

The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then

λ = (3, 2, 2) = ⊆ 3× 4: g←→


Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .

(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm).

The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row i

Ex. Suppose λ = (3, 2, 2).

Then

λ = (3, 2, 2) = ⊆ 3× 4: g←→


Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .

(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2).

Then

λ = (3, 2, 2) = ⊆ 3× 4: g←→


Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .

(d) Partitions.An (integer) partition is a weakly decreasing sequence of positiveintegers λ = (λ1, . . . , λm). The associated Ferrers diagram hasm left-justified rows of boxes with λi boxes in row iEx. Suppose λ = (3, 2, 2). Then

λ = (3, 2, 2) =

⊆ 3× 4: g←→


Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .


λ = (3, 2, 2) =

⊆ 3× 4: g←→

We say λ fits in a k × l rectangle, λ ⊆ k × l , if its Ferrers diagramhas at most k rows and at most l columns.

Let

Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .


λ = (3, 2, 2) = ⊆ 3× 4:

g←→


Let

Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .


λ = (3, 2, 2) = ⊆ 3× 4:

g←→


Let

Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .


λ = (3, 2, 2) = ⊆ 3× 4:

g←→


Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .


λ = (3, 2, 2) = ⊆ 3× 4:

g←→


Ln,k = {λ : λ ⊆ k × (n − k)}.

There is a bijection g : Ln,k → Pn,k :

given λ ⊆ k × (n − k),P = g(λ) is formed by going from the SW corner of the rectangleto the NE corner along the rectangle and the SE boundary of λ.

Proposition


)= #Ln,k .


λ = (3, 2, 2) = ⊆ 3× 4:

g←→


Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .


λ = (3, 2, 2) = ⊆ 3× 4: g←→


Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .


λ = (3, 2, 2) = ⊆ 3× 4: g←→


Ln,k = {λ : λ ⊆ k × (n − k)}.


Proposition


)= #Ln,k .

Outline

The theme





A q-analogue of a mathematical object O (number, definition,theorem) is an object O(q) with O(1) = O.

The standardq-analogue of n ∈ N is the polynomial

[n] = 1 + q + q2 + · · ·+ qn−1.

Ex. We have [4] = 1 + q + q2 + q3.Note that

[n]|q=1 =

n︷︸︸︷1 + 1 + · · ·+ 1 = n.

A q-factorial is [n]! = [1][2] · · · [n]. For 0 ≤ k ≤ n, the q-binomialcoefficients or Gaussian polynomials are[n

k

]=

[n]!

[k]![n − k]!.

Ex. We have[4

2

]=

[4]!

[2]![2]!=

[4][3]

[2][1]= 1 + q + 2q2 + q3 + q4.

Note that it is not clear from the definition that[ nk

]is always in

N[q], the set of polynomials in q with coefficients in N.

A q-analogue of a mathematical object O (number, definition,theorem) is an object O(q) with O(1) = O. The standardq-analogue of n ∈ N is the polynomial

[n] = 1 + q + q2 + · · ·+ qn−1.


[n]|q=1 =

n︷︸︸︷1 + 1 + · · ·+ 1 = n.


k

]=

[n]!

[k]![n − k]!.

Ex. We have[4

2

]=

[4]!

[2]![2]!=

[4][3]

[2][1]= 1 + q + 2q2 + q3 + q4.


]is always in



[n] = 1 + q + q2 + · · ·+ qn−1.

Ex. We have [4] = 1 + q + q2 + q3.

Note that

[n]|q=1 =

n︷︸︸︷1 + 1 + · · ·+ 1 = n.


k

]=

[n]!

[k]![n − k]!.

Ex. We have[4

2

]=

[4]!

[2]![2]!=

[4][3]

[2][1]= 1 + q + 2q2 + q3 + q4.


]is always in



[n] = 1 + q + q2 + · · ·+ qn−1.


[n]|q=1 =

n︷︸︸︷1 + 1 + · · ·+ 1 = n.


k

]=

[n]!

[k]![n − k]!.

Ex. We have[4

2

]=

[4]!

[2]![2]!=

[4][3]

[2][1]= 1 + q + 2q2 + q3 + q4.


]is always in



[n] = 1 + q + q2 + · · ·+ qn−1.


[n]|q=1 =

n︷︸︸︷1 + 1 + · · ·+ 1 = n.

A q-factorial is [n]! = [1][2] · · · [n].

For 0 ≤ k ≤ n, the q-binomialcoefficients or Gaussian polynomials are[n

k

]=

[n]!

[k]![n − k]!.

Ex. We have[4

2

]=

[4]!

[2]![2]!=

[4][3]

[2][1]= 1 + q + 2q2 + q3 + q4.


]is always in



[n] = 1 + q + q2 + · · ·+ qn−1.


[n]|q=1 =

n︷︸︸︷1 + 1 + · · ·+ 1 = n.


k

]=

[n]!

[k]![n − k]!.

Ex. We have[4

2

]=

[4]!

[2]![2]!=

[4][3]

[2][1]= 1 + q + 2q2 + q3 + q4.


]is always in



[n] = 1 + q + q2 + · · ·+ qn−1.


[n]|q=1 =

n︷︸︸︷1 + 1 + · · ·+ 1 = n.


k

]=

[n]!

[k]![n − k]!.

Ex. We have[4

2

]=

[4]!

[2]![2]!=

[4][3]

[2][1]= 1 + q + 2q2 + q3 + q4.


]is always in



[n] = 1 + q + q2 + · · ·+ qn−1.


[n]|q=1 =

n︷︸︸︷1 + 1 + · · ·+ 1 = n.


k

]=

[n]!

[k]![n − k]!.

Ex. We have[4

2

]=

[4]!

[2]![2]!=

[4][3]

[2][1]= 1 + q + 2q2 + q3 + q4.


]is always in


Here is a q-analogue for the boundary conditions and recurrencerelation for the binomial coefficients.

TheoremThe q-binomial coefficients satisfy

[n0

]=[nn

]= 1 and, for

0 < k < n, [n

k

]= qk

[n − 1

k

]+

[n − 1

k − 1

]

=

[n − 1

k

]+ qn−k

[n − 1

k − 1

].

Since sums and products of elements of N[q] are again in N[q], weimmediately get the following result.

Corollary

For all 0 ≤ k ≤ n we have[ nk

]∈ N[q].



[n0

]=[nn

]= 1 and, for

0 < k < n, [n

k

]= qk

[n − 1

k

]+

[n − 1

k − 1

]

=

[n − 1

k

]+ qn−k

[n − 1

k − 1

].


Corollary


]∈ N[q].



[n0

]=[nn

]= 1 and, for

0 < k < n, [n

k

]= qk

[n − 1

k

]+

[n − 1

k − 1

]

=

[n − 1

k

]+ qn−k

[n − 1

k − 1

].


Corollary


]∈ N[q].



[n0

]=[nn

]= 1 and, for

0 < k < n, [n

k

]= qk

[n − 1

k

]+

[n − 1

k − 1

]

=

[n − 1

k

]+ qn−k

[n − 1

k − 1

].


Corollary


]∈ N[q].

(a) Words.

If w = a1 . . . an is a word over N then the inversion set of w is

Invw = {(i , j) : i < j and ai > aj}.

The corresponding inversion number is

invw = # Invw .

Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)} and invw = 4.Consider the inversion generating function

In,k(q) =∑

w∈Wn,k

qinvw .

Ex. When n = 4 and k = 2,

W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4

=[42

].

TheoremFor all 0 ≤ k ≤ n we have

[ nk

]= In,k(q).

(a) Words.If w = a1 . . . an is a word over N then the inversion set of w is



invw = # Invw .


In,k(q) =∑

w∈Wn,k

qinvw .


W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4

=[42

].


[ nk

]= In,k(q).




invw = # Invw .

Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)}

and invw = 4.Consider the inversion generating function

In,k(q) =∑

w∈Wn,k

qinvw .


W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4

=[42

].


[ nk

]= In,k(q).




invw = # Invw .

Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)}

and invw = 4.Consider the inversion generating function

In,k(q) =∑

w∈Wn,k

qinvw .


W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4

=[42

].


[ nk

]= In,k(q).




invw = # Invw .

Ex. If w = a1a2a3a4a5 = 10110 thenInvw = {(1, 2), (1, 5), (3, 5), (4, 5)} and invw = 4.

Consider the inversion generating function

In,k(q) =∑

w∈Wn,k

qinvw .


W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4

=[42

].


[ nk

]= In,k(q).




invw = # Invw .


In,k(q) =∑

w∈Wn,k

qinvw .


W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4

=[42

].


[ nk

]= In,k(q).




invw = # Invw .


In,k(q) =∑

w∈Wn,k

qinvw .


W4,2 : 0011 0101 0110 1001 1010 1100

I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4

=[42

].


[ nk

]= In,k(q).




invw = # Invw .


In,k(q) =∑

w∈Wn,k

qinvw .


W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4

=[42

].


[ nk

]= In,k(q).




invw = # Invw .


In,k(q) =∑

w∈Wn,k

qinvw .


W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4

=[42

].


[ nk

]= In,k(q).




invw = # Invw .


In,k(q) =∑

w∈Wn,k

qinvw .


W4,2 : 0011 0101 0110 1001 1010 1100I4,2(q) = q0 + q1 + q2 + q2 + q3 + q4

=[42

].


[ nk

]= In,k(q).

(b) Even more words.

If w = a1 . . . an is a word over N then the major index of w is

majw =∑

ai>ai+1

i .

Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 somajw = 1 + 4 = 5.Consider the major index generating function

Mn,k(q) =∑

w∈Wn,k

qmajw .


W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2

=[42

].


[ nk

]= Mn,k(q).

(b) Even more words.If w = a1 . . . an is a word over N then the major index of w is

majw =∑

ai>ai+1

i .


Mn,k(q) =∑

w∈Wn,k

qmajw .


W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2

=[42

].


[ nk

]= Mn,k(q).


majw =∑

ai>ai+1

i .

Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 somajw = 1 + 4 = 5.

Consider the major index generating function

Mn,k(q) =∑

w∈Wn,k

qmajw .


W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2

=[42

].


[ nk

]= Mn,k(q).


majw =∑

ai>ai+1

i .


Mn,k(q) =∑

w∈Wn,k

qmajw .


W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2

=[42

].


[ nk

]= Mn,k(q).


majw =∑

ai>ai+1

i .


Mn,k(q) =∑

w∈Wn,k

qmajw .


W4,2 : 0011 0101 0110 1001 1010 1100

M4,2(q) = q0 + q2 + q3 + q + q4 + q2

=[42

].


[ nk

]= Mn,k(q).


majw =∑

ai>ai+1

i .


Mn,k(q) =∑

w∈Wn,k

qmajw .


W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2

=[42

].


[ nk

]= Mn,k(q).


majw =∑

ai>ai+1

i .


Mn,k(q) =∑

w∈Wn,k

qmajw .


W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2

=[42

].


[ nk

]= Mn,k(q).


majw =∑

ai>ai+1

i .


Mn,k(q) =∑

w∈Wn,k

qmajw .


W4,2 : 0011 0101 0110 1001 1010 1100M4,2(q) = q0 + q2 + q3 + q + q4 + q2

=[42

].


[ nk

]= Mn,k(q).

(c) Partitions.

If λ = (λ1, . . . , λm) is a partition then its size is

|λ| = λ1 + · · ·+ λm.

Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function

Sn,k(q) =∑λ∈Ln,k

q|λ|.

Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2

w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).

(c) Partitions.If λ = (λ1, . . . , λm) is a partition then its size is

|λ| = λ1 + · · ·+ λm.



q|λ|.


w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.

Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.

Note that |λ| is the number of squares in its Ferrers diagram.Consider the size generating function


q|λ|.


w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.

Ex. If λ = (4, 3, 3, 2) then |λ| = 4 + 3 + 3 + 2 = 12.Note that |λ| is the number of squares in its Ferrers diagram.

Consider the size generating function


q|λ|.


w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.



q|λ|.


w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.



q|λ|.

Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .

In fact, squares of λ correspond bijectively to elements of Invw .Ex. When n = 5 and k = 2

w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.



q|λ|.

Composing g : Ln,k → Pn,k and f : Pn,k →Wn,k gives a bijectionh = f ◦ g : Ln,k →Wn,k such that, if h(λ) = w then |λ| = invw .In fact, squares of λ correspond bijectively to elements of Invw .

Ex. When n = 5 and k = 2

w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.



q|λ|.


w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.



q|λ|.


w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.



q|λ|.


w = 10110h←→ λ =

10 1 1

0

1

010 1

10

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.



q|λ|.


w = 10110h←→ λ =

10 1 1

0

1

010 1

10

10 1

10

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.



q|λ|.


w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 1

10

10 1

10 1 1

0


[ nk

]= Sn,k(q).


|λ| = λ1 + · · ·+ λm.



q|λ|.


w = 10110h←→ λ =

10 1 1

0

1

010 11

0

10 11

0

10 1

10 1 1

0


[ nk

]= Sn,k(q).

(d) Subspaces.

Let q be a prime power and Fq be the Galois field with q elements.Consider the n-dimensional vector space Fn

q. Let

Vn,k(q) = {W : W is a k-dimensional subspace of Fnq}.

Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are[1 0 ∗ ∗0 1 ∗ ∗

] [1 ∗ 0 ∗0 0 1 ∗

] [1 ∗ ∗ 00 0 0 1

][

0 1 0 ∗0 0 1 ∗

] [0 1 ∗ 00 0 0 1

] [0 0 1 00 0 0 1

]where the stars are arbitrary elements of F3. Therefore

#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =

[4

2

]∣∣∣∣q=3

.

Theorem (Knuth, 1971)

For all 0 ≤ k ≤ n and q a prime power we have[ nk

]= #Vn,k(q).

(d) Subspaces.Let q be a prime power and Fq be the Galois field with q elements.

Consider the n-dimensional vector space Fnq. Let



] [1 ∗ 0 ∗0 0 1 ∗

] [1 ∗ ∗ 00 0 0 1

][

0 1 0 ∗0 0 1 ∗

] [0 1 ∗ 00 0 0 1

] [0 0 1 00 0 0 1


#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =

[4

2

]∣∣∣∣q=3

.



]= #Vn,k(q).

(d) Subspaces.Let q be a prime power and Fq be the Galois field with q elements.Consider the n-dimensional vector space Fn

q.

Let



] [1 ∗ 0 ∗0 0 1 ∗

] [1 ∗ ∗ 00 0 0 1

][

0 1 0 ∗0 0 1 ∗

] [0 1 ∗ 00 0 0 1

] [0 0 1 00 0 0 1


#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =

[4

2

]∣∣∣∣q=3

.



]= #Vn,k(q).


q. Let



] [1 ∗ 0 ∗0 0 1 ∗

] [1 ∗ ∗ 00 0 0 1

][

0 1 0 ∗0 0 1 ∗

] [0 1 ∗ 00 0 0 1

] [0 0 1 00 0 0 1


#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =

[4

2

]∣∣∣∣q=3

.



]= #Vn,k(q).


q. Let


Ex. Let q = 3.

The row echelon forms for subspaces in V4,2(3) are[1 0 ∗ ∗0 1 ∗ ∗

] [1 ∗ 0 ∗0 0 1 ∗

] [1 ∗ ∗ 00 0 0 1

][

0 1 0 ∗0 0 1 ∗

] [0 1 ∗ 00 0 0 1

] [0 0 1 00 0 0 1


#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =

[4

2

]∣∣∣∣q=3

.



]= #Vn,k(q).


q. Let



] [1 ∗ 0 ∗0 0 1 ∗

] [1 ∗ ∗ 00 0 0 1

][

0 1 0 ∗0 0 1 ∗

] [0 1 ∗ 00 0 0 1

] [0 0 1 00 0 0 1

]where the stars are arbitrary elements of F3.

Therefore

#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =

[4

2

]∣∣∣∣q=3

.



]= #Vn,k(q).


q. Let



] [1 ∗ 0 ∗0 0 1 ∗

] [1 ∗ ∗ 00 0 0 1

][

0 1 0 ∗0 0 1 ∗

] [0 1 ∗ 00 0 0 1

] [0 0 1 00 0 0 1


#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =

[4

2

]∣∣∣∣q=3

.



]= #Vn,k(q).


q. Let



] [1 ∗ 0 ∗0 0 1 ∗

] [1 ∗ ∗ 00 0 0 1

][

0 1 0 ∗0 0 1 ∗

] [0 1 ∗ 00 0 0 1

] [0 0 1 00 0 0 1


#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 =

[4

2

]∣∣∣∣q=3

.



]= #Vn,k(q).

Outline

The theme





The Fibonacci numbers are defined by F1 = F2 = 1 and, for n ≥ 2,

Fn = Fn−1 + Fn−2.

Ex. F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, . . .A fibotorial is F !

n = F1F2 · · ·Fn. For 0 ≤ k ≤ n, the correspondingfibonomial is (

n

k

)F

=F !n

F !kF !

n−k.

Ex. We have(6

3

)F

=F !6

F !3F !

3

=F6F5F4

F3F2F1=

8 · 5 · 32 · 1 · 1

= 60.

TheoremThe fibonomials satisfy

(n0

)F

=(nn

)F

= 1 and, for 0 < k < n,(n

k

)F

= Fk+1

(n − 1

k

)F

+ Fn−k−1

(n − 1

k − 1

)F

.

Corollary


)F∈ N.


Fn = Fn−1 + Fn−2.

Ex. F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, . . .

A fibotorial is F !n = F1F2 · · ·Fn. For 0 ≤ k ≤ n, the corresponding

fibonomial is (n

k

)F

=F !n

F !kF !

n−k.

Ex. We have(6

3

)F

=F !6

F !3F !

3

=F6F5F4

F3F2F1=

8 · 5 · 32 · 1 · 1

= 60.


(n0

)F

=(nn

)F

= 1 and, for 0 < k < n,(n

k

)F

= Fk+1

(n − 1

k

)F

+ Fn−k−1

(n − 1

k − 1

)F

.

Corollary


)F∈ N.


Fn = Fn−1 + Fn−2.


n = F1F2 · · ·Fn.

For 0 ≤ k ≤ n, the correspondingfibonomial is (

n

k

)F

=F !n

F !kF !

n−k.

Ex. We have(6

3

)F

=F !6

F !3F !

3

=F6F5F4

F3F2F1=

8 · 5 · 32 · 1 · 1

= 60.


(n0

)F

=(nn

)F

= 1 and, for 0 < k < n,(n

k

)F

= Fk+1

(n − 1

k

)F

+ Fn−k−1

(n − 1

k − 1

)F

.

Corollary


)F∈ N.


Fn = Fn−1 + Fn−2.



n

k

)F

=F !n

F !kF !

n−k.

Ex. We have(6

3

)F

=F !6

F !3F !

3

=F6F5F4

F3F2F1=

8 · 5 · 32 · 1 · 1

= 60.


(n0

)F

=(nn

)F

= 1 and, for 0 < k < n,(n

k

)F

= Fk+1

(n − 1

k

)F

+ Fn−k−1

(n − 1

k − 1

)F

.

Corollary


)F∈ N.


Fn = Fn−1 + Fn−2.



n

k

)F

=F !n

F !kF !

n−k.

Ex. We have(6

3

)F

=F !6

F !3F !

3

=F6F5F4

F3F2F1=

8 · 5 · 32 · 1 · 1

= 60.


(n0

)F

=(nn

)F

= 1 and, for 0 < k < n,(n

k

)F

= Fk+1

(n − 1

k

)F

+ Fn−k−1

(n − 1

k − 1

)F

.

Corollary


)F∈ N.


Fn = Fn−1 + Fn−2.



n

k

)F

=F !n

F !kF !

n−k.

Ex. We have(6

3

)F

=F !6

F !3F !

3

=F6F5F4

F3F2F1=

8 · 5 · 32 · 1 · 1

= 60.


(n0

)F

=(nn

)F

= 1 and, for 0 < k < n,(n

k

)F

= Fk+1

(n − 1

k

)F

+ Fn−k−1

(n − 1

k − 1

)F

.

Corollary


)F∈ N.


Fn = Fn−1 + Fn−2.



n

k

)F

=F !n

F !kF !

n−k.

Ex. We have(6

3

)F

=F !6

F !3F !

3

=F6F5F4

F3F2F1=

8 · 5 · 32 · 1 · 1

= 60.


(n0

)F

=(nn

)F

= 1 and, for 0 < k < n,(n

k

)F

= Fk+1

(n − 1

k

)F

+ Fn−k−1

(n − 1

k − 1

)F

.

Corollary


)F∈ N.

S. and Savage were the first to give a simple combinatorialinterpretation of

(nk

)F

.

Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot. Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes. Let

Tn = {T : T a linear tiling of a row of n squares}.

Ex. We have

T3 ={

, ,

}Note that #T3 = 3 = F4.

Proposition

For all n ≥ 1 we have Fn = #Tn−1.


(nk

)F

. Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot.

Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes. Let


Ex. We have

T3 ={

, ,


Proposition



(nk

)F

. Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot. Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes.

Let


Ex. We have

T3 ={

, ,


Proposition



(nk

)F

. Other more complicated interpretationshave been given by Benjamin-Plott, and by Gessel-Viennot. Alinear tiling, T , is a covering of a row of n squares with disjointdominoes and monominoes. Let


Ex. We have

T3 ={

, ,


Proposition



(nk

)F



Ex. We have

T3 ={

, ,

}

Note that #T3 = 3 = F4.

Proposition



(nk

)F



Ex. We have

T3 ={

, ,


Proposition



(nk

)F



Ex. We have

T3 ={

, ,


Proposition


A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi .

Let

Tλ = {T : T is a tiling of λ},Dλ = {T ∈ Tλ : every λi begins with a domino}.

Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).

If λ ⊆ k × l then there is a dual partition λ∗ = (λ∗1, . . . , λ∗r ) where

the λ∗j are the column lengths of (k × l)− λ. Let

Fn,k =⋃

λ⊆k×(n−k)

(Tλ ×Dλ∗) .

Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):

and ∈ F7,3

Proposition (S. and Savage, 2010)


)F

= #Fn,k .

A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi . Let

Tλ = {T : T is a tiling of λ},

Dλ = {T ∈ Tλ : every λi begins with a domino}.

Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).



Fn,k =⋃

λ⊆k×(n−k)

(Tλ ×Dλ∗) .

Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):

and ∈ F7,3



)F

= #Fn,k .


Tλ = {T : T is a tiling of λ},

Dλ = {T ∈ Tλ : every λi begins with a domino}.

Ex. If λ = (3, 2, 2) then∈ T(3,2,2),

∈ D(3,2,2).



Fn,k =⋃

λ⊆k×(n−k)

(Tλ ×Dλ∗) .

Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):

and ∈ F7,3



)F

= #Fn,k .



Ex. If λ = (3, 2, 2) then∈ T(3,2,2),

∈ D(3,2,2).



Fn,k =⋃

λ⊆k×(n−k)

(Tλ ×Dλ∗) .

Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):

and ∈ F7,3



)F

= #Fn,k .



Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).



Fn,k =⋃

λ⊆k×(n−k)

(Tλ ×Dλ∗) .

Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):

and ∈ F7,3



)F

= #Fn,k .



Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).


the λ∗j are the column lengths of (k × l)− λ.

Let

Fn,k =⋃

λ⊆k×(n−k)

(Tλ ×Dλ∗) .

Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):

and ∈ F7,3



)F

= #Fn,k .



Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).


the λ∗j are the column lengths of (k × l)− λ.

Let

Fn,k =⋃

λ⊆k×(n−k)

(Tλ ×Dλ∗) .

Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):

and ∈ F7,3



)F

= #Fn,k .



Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).



Fn,k =⋃

λ⊆k×(n−k)

(Tλ ×Dλ∗) .

Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):

and ∈ F7,3



)F

= #Fn,k .



Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).



Fn,k =⋃

λ⊆k×(n−k)

(Tλ ×Dλ∗) .

Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):

and ∈ F7,3



)F

= #Fn,k .



Ex. If λ = (3, 2, 2) then∈ T(3,2,2), ∈ D(3,2,2).



Fn,k =⋃

λ⊆k×(n−k)

(Tλ ×Dλ∗) .

Ex. If λ = (3, 2, 2) ⊆ 3× 4 then λ∗ = (3, 2):

and ∈ F7,3



)F

= #Fn,k .

Outline

The theme





The nth Catalan number is

Cn =1

n + 1

(2n

n

).

Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .

TheoremWe have C0 = 1 and, for n ≥ 1, Cn =

∑n−1i=0 CiCn−i−1.

Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.For example, let

Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.TheoremFor n ≥ 0 we have Cn = #Dn.

Define the fibocatalan numbers to be

Cn,F =1

Fn+1

(2n

n

)F

.

It is not hard to show Cn.F ∈ N for all n. Lou Shapiro asked: Canone find a combinatorial interpretation?


Cn =1

n + 1

(2n

n

).

Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .






Cn,F =1

Fn+1

(2n

n

)F

.



Cn =1

n + 1

(2n

n

).

Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .






Cn,F =1

Fn+1

(2n

n

)F

.



Cn =1

n + 1

(2n

n

).

Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .



Stanley’s Catalan Addendum lists almost 200 combinatorialinterpretations: http://www-math.mit.edu/˜rstan/ec/catadd.pdf.

For example, let



Cn,F =1

Fn+1

(2n

n

)F

.



Cn =1

n + 1

(2n

n

).

Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .




Dn = {P : P a NE path from (0, 0) to (n, n) not going below y = x}.

TheoremFor n ≥ 0 we have Cn = #Dn.


Cn,F =1

Fn+1

(2n

n

)F

.



Cn =1

n + 1

(2n

n

).

Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .






Cn,F =1

Fn+1

(2n

n

)F

.



Cn =1

n + 1

(2n

n

).

Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .






Cn,F =1

Fn+1

(2n

n

)F

.



Cn =1

n + 1

(2n

n

).

Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .






Cn,F =1

Fn+1

(2n

n

)F

.

It is not hard to show Cn.F ∈ N for all n.

Lou Shapiro asked: Canone find a combinatorial interpretation?


Cn =1

n + 1

(2n

n

).

Ex. C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .






Cn,F =1

Fn+1

(2n

n

)F

.


References.

1. Andrews, George E. The theory of partitions. Reprint of the1976 original. Cambridge Mathematical Library.CambridgeUniversity Press, Cambridge, (1998).

2. Benjamin, A. T., and Plott, S. S. A combinatorial approach toFibonomial coefficients, Fibonacci Quart. 46/47 (2008/09),7–9.

3. Gessel, I., and Viennot, G. Binomial determinants, paths, andhook length formulae. Adv. in Math. 58 (1985), 300–321.

4. Knuth, Donald E. Subspaces, subsets, and partitions. J.Combinatorial Theory Ser. A 10 (1971) 178–180.

5. MacMahon, Percy A. Combinatory Analysis. Volumes 1 and2. Reprint of the 1916 original. Dover, New York, (2004).

6. Sagan, Bruce E., and Savage, Carla D. Combinatorialinterpretations of binomial coefficient analogues related toLucas sequences. Integers 10 (2010), A52, 697–703.

7. Stanley, Richard P. Enumerative combinatorics. Volume 1.Second edition. Cambridge Studies in Advanced Mathematics,49. Cambridge University Press, Cambridge, (2012).

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