Two Binomial Coefficient Analogues Bruce Sagan Department of - - PowerPoint PPT Presentation

Two Binomial Coefficient Analogues

Bruce Sagan Department of Mathematics Michigan State University East Lansing, MI 48824-1027 sagan@math.msu.edu www.math.msu.edu/˜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 Variation 1: binomial coefficients Variation 2: q-binomial coefficients Variation 3: fibonomial coefficients Coda: an open question and bibliography

Suppose we are given a sequence a0, a1, a2, . . . defined so that we only know that the an are rational numbers.

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.

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?

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.

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.

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.

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.

Outline

The theme Variation 1: binomial coefficients Variation 2: q-binomial coefficients Variation 3: fibonomial coefficients Coda: an open question and bibliography

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

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

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

• =

n! k!(n − k)!.

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

• =

n! k!(n − k)!.

• Ex. We have

4 2

• = 4!

2!2! = 4 · 3 2 · 1 = 6.

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

• =

n! k!(n − k)!.

• Ex. We have

4 2

• = 4!

2!2! = 4 · 3 2 · 1 = 6. Note that it is not clear from this definition that n

k

• is an integer.

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

• =

n! k!(n − k)!.

• Ex. We have

4 2

• = 4!

2!2! = 4 · 3 2 · 1 = 6. Note that it is not clear from this definition that n

k

• is an integer.

Proposition

The binomial coefficients satisfy n

• =

n

n

• = 1 and, for 0 < k < n,

n k

• =

n − 1 k

• +

n − 1 k − 1

• .

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

• =

n! k!(n − k)!.

• Ex. We have

4 2

• = 4!

2!2! = 4 · 3 2 · 1 = 6. Note that it is not clear from this definition that n

k

• is an integer.

Proposition

The binomial coefficients satisfy n

• =

n

n

• = 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 n

k

• ∈ N.

(a) Subsets.

(a) Subsets. Let Sn,k = {S : S is a k-element subset of {1, . . . , 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}}.

(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

• .

(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 n

k

• = #Sn,k.

(b) Words.

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

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

• Ex. We have that w = SEICCGTC44 is a word of length 10 over

the alphabet A = {A, . . . , Z, 1, . . . , 9}.

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

• Ex. We have that w = SEICCGTC44 is a word of length 10 over

the alphabet A = {A, . . . , Z, 1, . . . , 9}. Let Wn,k = {w = a1 . . . an : w has k zeros and n − k ones}.

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

• Ex. We have that w = SEICCGTC44 is a word of length 10 over

the 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}.

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

• Ex. We have that w = SEICCGTC44 is a word of length 10 over

the 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

• .

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

• Ex. We have that w = SEICCGTC44 is a word of length 10 over

the 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 n positions to be zeros (and the rest will be ones by default).

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

• Ex. We have that w = SEICCGTC44 is a word of length 10 over

the 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 n positions to be zeros (and the rest will be ones by default).

Proposition

For 0 ≤ k ≤ n we have n

k

• = #Wn,k.

(c) Lattice paths.

(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).

(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 =

(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}

(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 obtained by replacing each N by a 0 and each E by a 1.

(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 obtained by replacing each N by a 0 and each E by a 1.

• Ex. We have

P = ENEEN

f

← → w = 10110 = 1 1 1 0

(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 obtained by replacing each N by a 0 and each E by a 1.

• Ex. We have

P = ENEEN

f

← → w = 10110 = 1 1 1 0

Proposition

For 0 ≤ k ≤ n we have n

k

• = #Pn,k.

(d) Partitions.

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

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

• Ex. Suppose λ = (3, 2, 2).

(d) Partitions. An (integer) partition is a weakly decreasing sequence of positive integers λ = (λ1, . . . , λm). The associated Ferrers diagram has m left-justified rows of boxes with λi boxes in row i

• Ex. Suppose λ = (3, 2, 2).

(d) Partitions. An (integer) partition is a weakly decreasing sequence of positive integers λ = (λ1, . . . , λm). The associated Ferrers diagram has m left-justified rows of boxes with λi boxes in row i

• Ex. Suppose λ = (3, 2, 2).

Then λ = (3, 2, 2) =

(d) Partitions. An (integer) partition is a weakly decreasing sequence of positive integers λ = (λ1, . . . , λm). The associated Ferrers diagram has m left-justified rows of boxes with λi boxes in row i

• Ex. Suppose λ = (3, 2, 2).

Then λ = (3, 2, 2) = We say λ fits in a k × l rectangle, λ ⊆ k × l, if its Ferrers diagram has at most k rows and at most l columns.

(d) Partitions. An (integer) partition is a weakly decreasing sequence of positive integers λ = (λ1, . . . , λm). The associated Ferrers diagram has m left-justified rows of boxes with λi boxes in row i

• Ex. Suppose λ = (3, 2, 2).

Then λ = (3, 2, 2) = ⊆ 3 × 4: We say λ fits in a k × l rectangle, λ ⊆ k × l, if its Ferrers diagram has at most k rows and at most l columns.

(d) Partitions. An (integer) partition is a weakly decreasing sequence of positive integers λ = (λ1, . . . , λm). The associated Ferrers diagram has m left-justified rows of boxes with λi boxes in row i

• Ex. Suppose λ = (3, 2, 2).

Then λ = (3, 2, 2) = ⊆ 3 × 4: We say λ fits in a k × l rectangle, λ ⊆ k × l, if its Ferrers diagram has at most k rows and at most l columns.

(d) Partitions. An (integer) partition is a weakly decreasing sequence of positive integers λ = (λ1, . . . , λm). The associated Ferrers diagram has m left-justified rows of boxes with λi boxes in row i

• Ex. Suppose λ = (3, 2, 2).

Then λ = (3, 2, 2) = ⊆ 3 × 4: We say λ fits in a k × l rectangle, λ ⊆ k × l, if its Ferrers diagram has at most k rows and at most l columns. Let Ln,k = {λ : λ ⊆ k × (n − k)}.

(d) Partitions. An (integer) partition is a weakly decreasing sequence of positive integers λ = (λ1, . . . , λm). The associated Ferrers diagram has m left-justified rows of boxes with λi boxes in row i

• Ex. Suppose λ = (3, 2, 2).

Then λ = (3, 2, 2) = ⊆ 3 × 4: We say λ fits in a k × l rectangle, λ ⊆ k × l, if its Ferrers diagram has 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:

(d) Partitions. An (integer) partition is a weakly decreasing sequence of positive integers λ = (λ1, . . . , λm). The associated Ferrers diagram has m left-justified rows of boxes with λi boxes in row i

• Ex. Suppose λ = (3, 2, 2).

Then λ = (3, 2, 2) = ⊆ 3 × 4: We say λ fits in a k × l rectangle, λ ⊆ k × l, if its Ferrers diagram has 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 rectangle to the NE corner along the rectangle and the SE boundary of λ.

(d) Partitions. An (integer) partition is a weakly decreasing sequence of positive integers λ = (λ1, . . . , λm). The associated Ferrers diagram has m 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

← → We say λ fits in a k × l rectangle, λ ⊆ k × l, if its Ferrers diagram has 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 rectangle to the NE corner along the rectangle and the SE boundary of λ.

(d) Partitions. An (integer) partition is a weakly decreasing sequence of positive integers λ = (λ1, . . . , λm). The associated Ferrers diagram has m 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

← → We say λ fits in a k × l rectangle, λ ⊆ k × l, if its Ferrers diagram has 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 rectangle to the NE corner along the rectangle and the SE boundary of λ.

Proposition

For 0 ≤ k ≤ n we have n

k

• = #Ln,k.

Outline

The theme Variation 1: binomial coefficients Variation 2: q-binomial coefficients Variation 3: fibonomial coefficients Coda: an open question and bibliography

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

A q-analogue of a mathematical object O (number, definition, theorem) is an object O(q) with O(1) = O. The standard q-analogue of n ∈ N is the polynomial [n] = 1 + q + q2 + · · · + qn−1.

A q-analogue of a mathematical object O (number, definition, theorem) is an object O(q) with O(1) = O. The standard q-analogue of n ∈ N is the polynomial [n] = 1 + q + q2 + · · · + qn−1.

• Ex. We have [4] = 1 + q + q2 + q3.

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

A q-analogue of a mathematical object O (number, definition, theorem) is an object O(q) with O(1) = O. The standard q-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-binomial coefficients or Gaussian polynomials are n k

• =

[n]! [k]![n − k]!.

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

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

k

• is always in

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

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

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

Theorem

The q-binomial coefficients satisfy n

• =

n

n

• = 1 and, for

0 < k < n, n k

• =

qk n − 1 k

• +

n − 1 k − 1

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

Theorem

The q-binomial coefficients satisfy n

• =

n

n

• = 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

• .

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

Theorem

The q-binomial coefficients satisfy n

• =

n

n

• = 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], we immediately get the following result.

Corollary

For all 0 ≤ k ≤ n we have n

k

• ∈ N[q].

(a) Words.

(a) Words. If w = a1 . . . an is a word over N then the inversion set of w is Inv w = {(i, j) : i < j and ai > aj}.

(a) Words. If w = a1 . . . an is a word over N then the inversion set of w is Inv w = {(i, j) : i < j and ai > aj}.

• Ex. If w = a1a2a3a4a5 = 10110 then

Inv w = {(1, 2), (1, 5), (3, 5), (4, 5)}

(a) Words. If w = a1 . . . an is a word over N then the inversion set of w is Inv w = {(i, j) : i < j and ai > aj}. The corresponding inversion number is inv w = # Inv w.

• Ex. If w = a1a2a3a4a5 = 10110 then

Inv w = {(1, 2), (1, 5), (3, 5), (4, 5)}

(a) Words. If w = a1 . . . an is a word over N then the inversion set of w is Inv w = {(i, j) : i < j and ai > aj}. The corresponding inversion number is inv w = # Inv w.

• Ex. If w = a1a2a3a4a5 = 10110 then

Inv w = {(1, 2), (1, 5), (3, 5), (4, 5)} and inv w = 4.

(a) Words. If w = a1 . . . an is a word over N then the inversion set of w is Inv w = {(i, j) : i < j and ai > aj}. The corresponding inversion number is inv w = # Inv w.

• Ex. If w = a1a2a3a4a5 = 10110 then

Inv w = {(1, 2), (1, 5), (3, 5), (4, 5)} and inv w = 4. Consider the inversion generating function In,k(q) =

• w∈Wn,k

qinv w.

(a) Words. If w = a1 . . . an is a word over N then the inversion set of w is Inv w = {(i, j) : i < j and ai > aj}. The corresponding inversion number is inv w = # Inv w.

• Ex. If w = a1a2a3a4a5 = 10110 then

Inv w = {(1, 2), (1, 5), (3, 5), (4, 5)} and inv w = 4. Consider the inversion generating function In,k(q) =

• w∈Wn,k

qinv w.

• Ex. When n = 4 and k = 2,

W4,2 : 0011 0101 0110 1001 1010 1100

(a) Words. If w = a1 . . . an is a word over N then the inversion set of w is Inv w = {(i, j) : i < j and ai > aj}. The corresponding inversion number is inv w = # Inv w.

• Ex. If w = a1a2a3a4a5 = 10110 then

Inv w = {(1, 2), (1, 5), (3, 5), (4, 5)} and inv w = 4. Consider the inversion generating function In,k(q) =

• w∈Wn,k

qinv w.

• Ex. When n = 4 and k = 2,

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

(a) Words. If w = a1 . . . an is a word over N then the inversion set of w is Inv w = {(i, j) : i < j and ai > aj}. The corresponding inversion number is inv w = # Inv w.

• Ex. If w = a1a2a3a4a5 = 10110 then

Inv w = {(1, 2), (1, 5), (3, 5), (4, 5)} and inv w = 4. Consider the inversion generating function In,k(q) =

• w∈Wn,k

qinv w.

• Ex. When n = 4 and k = 2,

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

• 4

2

• .

(a) Words. If w = a1 . . . an is a word over N then the inversion set of w is Inv w = {(i, j) : i < j and ai > aj}. The corresponding inversion number is inv w = # Inv w.

• Ex. If w = a1a2a3a4a5 = 10110 then

Inv w = {(1, 2), (1, 5), (3, 5), (4, 5)} and inv w = 4. Consider the inversion generating function In,k(q) =

• w∈Wn,k

qinv w.

• Ex. When n = 4 and k = 2,

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

• 4

2

• .

Theorem

For all 0 ≤ k ≤ n we have n

k

• = In,k(q).

(b) Even more words.

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

• ai>ai+1

i.

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

• ai>ai+1

i.

• Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 so

maj w = 1 + 4 = 5.

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

• ai>ai+1

i.

• Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 so

maj w = 1 + 4 = 5. Consider the major index generating function Mn,k(q) =

• w∈Wn,k

qmaj w.

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

• ai>ai+1

i.

• Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 so

maj w = 1 + 4 = 5. Consider the major index generating function Mn,k(q) =

• w∈Wn,k

qmaj w.

• Ex. When n = 4 and k = 2,

W4,2 : 0011 0101 0110 1001 1010 1100

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

• ai>ai+1

i.

• Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 so

maj w = 1 + 4 = 5. Consider the major index generating function Mn,k(q) =

• w∈Wn,k

qmaj w.

• Ex. When n = 4 and k = 2,

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

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

• ai>ai+1

i.

• Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 so

maj w = 1 + 4 = 5. Consider the major index generating function Mn,k(q) =

• w∈Wn,k

qmaj w.

• Ex. When n = 4 and k = 2,

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

• 4

2

• .

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

• ai>ai+1

i.

• Ex. If w = a1a2a3a4a5 = 10110 then a1 > a2 and a4 > a5 so

maj w = 1 + 4 = 5. Consider the major index generating function Mn,k(q) =

• w∈Wn,k

qmaj w.

• Ex. When n = 4 and k = 2,

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

• 4

2

• .

Theorem

For all 0 ≤ k ≤ n we have n

k

• = Mn,k(q).

(c) Partitions.

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

(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.

(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.

(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|.

(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 bijection h = f ◦ g : Ln,k → Wn,k such that, if h(λ) = w then |λ| = inv w.

(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 bijection h = f ◦ g : Ln,k → Wn,k such that, if h(λ) = w then |λ| = inv w. In fact, squares of λ correspond bijectively to elements of Inv w.

(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 bijection h = f ◦ g : Ln,k → Wn,k such that, if h(λ) = w then |λ| = inv w. In fact, squares of λ correspond bijectively to elements of Inv w.

• Ex. When n = 5 and k = 2

w = 10110

h

← → λ = 1 01 1

(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 bijection h = f ◦ g : Ln,k → Wn,k such that, if h(λ) = w then |λ| = inv w. In fact, squares of λ correspond bijectively to elements of Inv w.

• Ex. When n = 5 and k = 2

w = 10110

h

← → λ = 1 01 1

(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 bijection h = f ◦ g : Ln,k → Wn,k such that, if h(λ) = w then |λ| = inv w. In fact, squares of λ correspond bijectively to elements of Inv w.

• Ex. When n = 5 and k = 2

w = 10110

h

← → λ = 1 1 1

(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 bijection h = f ◦ g : Ln,k → Wn,k such that, if h(λ) = w then |λ| = inv w. In fact, squares of λ correspond bijectively to elements of Inv w.

• Ex. When n = 5 and k = 2

w = 10110

h

← → λ = 1 1 1

(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 bijection h = f ◦ g : Ln,k → Wn,k such that, if h(λ) = w then |λ| = inv w. In fact, squares of λ correspond bijectively to elements of Inv w.

• Ex. When n = 5 and k = 2

w = 10110

h

← → λ = 1 1 01

(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 bijection h = f ◦ g : Ln,k → Wn,k such that, if h(λ) = w then |λ| = inv w. In fact, squares of λ correspond bijectively to elements of Inv w.

• Ex. When n = 5 and k = 2

w = 10110

h

← → λ = 1 01 1

Theorem

For all 0 ≤ k ≤ n we have n

k

• = Sn,k(q).

(d) Subspaces.

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

(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.

(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 Fn

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 Fn

q}.

• Ex. Let q = 3.

(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 Fn

q}.

• Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are

1 ∗ ∗ 1 ∗ ∗

• 1

∗ ∗ 1 ∗

• 1

∗ ∗ 1

• 1

∗ 1 ∗

• 1

∗ 1

• 1

1

• where the stars are arbitrary elements of F3.

(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 Fn

q}.

• Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are

1 ∗ ∗ 1 ∗ ∗

• 1

∗ ∗ 1 ∗

• 1

∗ ∗ 1

• 1

∗ 1 ∗

• 1

∗ 1

• 1

1

• where the stars are arbitrary elements of F3. Therefore

#V4,2(3) = 34 + 33 + 32 + 32 + 3 + 1 = 4 2

• q=3

.

(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 Fn

q}.

• Ex. Let q = 3. The row echelon forms for subspaces in V4,2(3) are

1 ∗ ∗ 1 ∗ ∗

• 1

∗ ∗ 1 ∗

• 1

∗ ∗ 1

• 1

∗ 1 ∗

• 1

∗ 1

• 1

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 n

k

• = #Vn,k(q).

Outline

The theme Variation 1: binomial coefficients Variation 2: q-binomial coefficients Variation 3: fibonomial coefficients Coda: an open question and bibliography

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

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, . . .

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.

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 corresponding

fibonomial is n k

• F

= F !

n

F !

kF ! n−k

.

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 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 · 3 2 · 1 · 1 = 60.

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 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 · 3 2 · 1 · 1 = 60.

Theorem

The fibonomials satisfy n

• F =

n

n

• F = 1 and, for 0 < k < n,

n k

• F

= Fk+1 n − 1 k

• F

+ Fn−k−1 n − 1 k − 1

• F

.

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 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 · 3 2 · 1 · 1 = 60.

Theorem

The fibonomials satisfy n

• F =

n

n

• 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

For all 0 ≤ k ≤ n we have n

k

• F ∈ N.

• S. and Savage were the first to give a simple combinatorial

interpretation of n

k

• F.

• S. and Savage were the first to give a simple combinatorial

interpretation of n

k

• F. Other more complicated interpretations

have been given by Benjamin-Plott, and by Gessel-Viennot.

• S. and Savage were the first to give a simple combinatorial

interpretation of n

k

• F. Other more complicated interpretations

have been given by Benjamin-Plott, and by Gessel-Viennot. A linear tiling, T, is a covering of a row of n squares with disjoint dominoes and monominoes.

• S. and Savage were the first to give a simple combinatorial

interpretation of n

k

• F. Other more complicated interpretations

have been given by Benjamin-Plott, and by Gessel-Viennot. A linear tiling, T, is a covering of a row of n squares with disjoint dominoes and monominoes. Let Tn = {T : T a linear tiling of a row of n squares}.

• S. and Savage were the first to give a simple combinatorial

interpretation of n

k

• F. Other more complicated interpretations

have been given by Benjamin-Plott, and by Gessel-Viennot. A linear tiling, T, is a covering of a row of n squares with disjoint dominoes and monominoes. Let Tn = {T : T a linear tiling of a row of n squares}.

• Ex. We have

T3 =

• ,

,

• S. and Savage were the first to give a simple combinatorial

interpretation of n

k

• F. Other more complicated interpretations

have been given by Benjamin-Plott, and by Gessel-Viennot. A linear tiling, T, is a covering of a row of n squares with disjoint dominoes and monominoes. Let Tn = {T : T a linear tiling of a row of n squares}.

• Ex. We have

T3 =

• ,

,

• Note that #T3 = 3 = F4.

• S. and Savage were the first to give a simple combinatorial

interpretation of n

k

• F. Other more complicated interpretations

have been given by Benjamin-Plott, and by Gessel-Viennot. A linear tiling, T, is a covering of a row of n squares with disjoint dominoes 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.

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

A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi. Let Tλ = {T : T is a tiling of λ},

A tiling of λ = (λ1, . . . , λm) is a union of tilings of each λi. Let Tλ = {T : T is a tiling of λ},

• Ex. If λ = (3, 2, 2) then

∈ T(3,2,2),

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),

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).

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.

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.

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

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):

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

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)

For 0 ≤ k ≤ n we have n

k

• F = #Fn,k.

Outline

The theme Variation 1: binomial coefficients Variation 2: q-binomial coefficients Variation 3: fibonomial coefficients Coda: an open question and bibliography

The nth Catalan number is Cn = 1 n + 1 2n n

• .

The nth Catalan number is Cn = 1 n + 1 2n n

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

The nth Catalan number is Cn = 1 n + 1 2n n

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

Theorem

We have C0 = 1 and, for n ≥ 1, Cn = n−1

i=0 CiCn−i−1.

The nth Catalan number is Cn = 1 n + 1 2n n

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

Theorem

We have C0 = 1 and, for n ≥ 1, Cn = n−1

i=0 CiCn−i−1.

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

The nth Catalan number is Cn = 1 n + 1 2n n

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

Theorem

We have C0 = 1 and, for n ≥ 1, Cn = n−1

i=0 CiCn−i−1.

Stanley’s Catalan Addendum lists almost 200 combinatorial interpretations: 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}.

The nth Catalan number is Cn = 1 n + 1 2n n

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

Theorem

We have C0 = 1 and, for n ≥ 1, Cn = n−1

i=0 CiCn−i−1.

Stanley’s Catalan Addendum lists almost 200 combinatorial interpretations: 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}.

Theorem

For n ≥ 0 we have Cn = #Dn.

The nth Catalan number is Cn = 1 n + 1 2n n

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

Theorem

We have C0 = 1 and, for n ≥ 1, Cn = n−1

i=0 CiCn−i−1.

Stanley’s Catalan Addendum lists almost 200 combinatorial interpretations: 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}.

Theorem

For n ≥ 0 we have Cn = #Dn. Define the fibocatalan numbers to be Cn,F = 1 Fn+1 2n n

• F

.

The nth Catalan number is Cn = 1 n + 1 2n n

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

Theorem

We have C0 = 1 and, for n ≥ 1, Cn = n−1

i=0 CiCn−i−1.

Stanley’s Catalan Addendum lists almost 200 combinatorial interpretations: 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}.

Theorem

For 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.

The nth Catalan number is Cn = 1 n + 1 2n n

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

Theorem

We have C0 = 1 and, for n ≥ 1, Cn = n−1

i=0 CiCn−i−1.

Stanley’s Catalan Addendum lists almost 200 combinatorial interpretations: 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}.

Theorem

For 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: Can

• ne find a combinatorial interpretation?

References.

• 1. Andrews, George E. The theory of partitions. Reprint of the

1976 original. Cambridge Mathematical Library.Cambridge University Press, Cambridge, (1998).

• 2. Benjamin, A. T., and Plott, S. S. A combinatorial approach to

Fibonomial coefficients, Fibonacci Quart. 46/47 (2008/09), 7–9.

• 3. Gessel, I., and Viennot, G. Binomial determinants, paths, and

hook 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 and
• 2. Reprint of the 1916 original. Dover, New York, (2004).
• 6. Sagan, Bruce E., and Savage, Carla D. Combinatorial

interpretations of binomial coefficient analogues related to Lucas 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).

THANKS FOR LISTENING!