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$C_{n} + (\binom{2 n}{n + 1}) = (\binom{2 n}{n})$

Statement

For every $n \in N$ , in $N$ ,

$C_{n} + (\binom{2 n}{n + 1}) = (\binom{2 n}{n}),$

where $C_{n}$ is the Catalan number (The Catalan number $C_{n} : = ∣ D_{n} ∣$ ) and the coefficients are those of The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ . Equivalently $C_{n} = (\binom{2 n}{n}) - (\binom{2 n}{n + 1})$ , the subtraction being legitimate because the displayed identity has been proved.

Facts & Assumptions

Given: a natural number $n$ .

[F1]

$D_{n}$ is the set of diagonal paths of length $2 n$ from $(0, 0)$ to $(2 n, 0)$ whose height function satisfies $h (i) \geq 0$ for every $i$ with $0 \leq i \leq 2 n$ ; for $n = 1$ it has exactly one element, with step word $U D$ (Dyck paths of semilength $n$ ).

[F2]

$C_{n} = ∣ D_{n} ∣$ , and $C_{0} = 1$ (The Catalan number $C_{n} : = ∣ D_{n} ∣$ ).

[L1]

For $c, a, b \in Z$ , $ℓ \in N$ and $a > c$ , $b > c$ : if $2$ divides $ℓ + b - a$ and $b - a \geq - ℓ$ , and $u \in N$ satisfies $2 u = ℓ + b - a$ , then $u + a - c \in N$ , the set $A$ of paths in $W ((0, a), (ℓ, b))$ staying strictly above the level $c$ is finite, and $∣ A ∣ + (\binom{ℓ}{u + a - c}) = (\binom{ℓ}{u})$ (The reflection principle: paths from $(0, a)$ to $(n, b)$ staying strictly above level $c$ are counted by a difference of two binomial coefficients, clause 2).

[L2]

$(\binom{m}{k}) = 0$ for $k > m$ , $(\binom{m}{0}) = 1$ and $(\binom{m}{m}) = 1$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

Proof

technique · direct

1.1

Since heights are integers, $h (i) \geq 0$ holds exactly when $h (i) > - 1$ ; so $D_{n}$ is precisely the set of paths in $W ((0, 0), (2 n, 0))$ that stay strictly above the level $- 1$ .

2.1

Apply [L1] with $ℓ = 2 n$ , $a = 0$ , $b = 0$ and $c = - 1$ . The hypotheses hold: $a > c$ and $b > c$ because $0 > - 1$ ; and $ℓ + b - a = 2 n$ is even with $2 n \geq 0$ , so the natural number $u$ with $2 u = 2 n$ is $u = n$ and $u + a - c = n + 1$ . By step 1.1 the set $A$ is $D_{n}$ , whose cardinality is $C_{n}$ , so $C_{n} + (\binom{2 n}{n + 1}) = (\binom{2 n}{n})$ .

F2L1step 1.1algebra

3.1

Since the identity holds in $N$ , the difference form follows. At $n = 0$ it reads $1 + (\binom{0}{1}) = (\binom{0}{0})$ , that is $1 + 0 = 1$ by [L2] and $C_{0} = 1$ ; at $n = 1$ it reads $C_{1} + (\binom{2}{2}) = (\binom{2}{1})$ , that is $1 + 1 = 2$ , matching the single element of $D_{1}$ .

F1F2L2step 2.1algebra∎

Remarks

Where the reflection is spent. The level is $- 1$ and not $0$ : a Dyck path starts and ends at height $0$ , so no path stays strictly above $0$ , and it is only because heights are integers that the weak condition against $0$ is the strict condition against $- 1$ . The reflected starting height is $2 (- 1) - 0 = - 2$ , which is why the subtracted coefficient is the one attached to the endpoint pair from $- 2$ to $0$ .
The additive form is the one proved. Writing the difference first would require knowing in advance that $(\binom{2 n}{n + 1}) \leq (\binom{2 n}{n})$ , which is a consequence of the identity rather than an input to it.

Depends on

Used by

(n+1) Cₙ=C(2n, n) Corollary
The reflection bijection applied to UDDUDU Example
Why the Catalan count is proved three times, and how the three statements agree Remark

Dependency tree · two levels

20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

C. Krattenthaler, "Lattice Path Enumeration", ch. 10 of the Handbook of Enumerative Combinatorics, §10.3, Corollary 10.3.2 (standard reference, not scraped)

Cn+(2nn+1)=(2nn)Cn​+(n+12n​)=(n2n​)