TheoremBase

Binomial Coefficients: Existence and Uniqueness by Pascal's Recursion, Vanishing above the Diagonal, Diagonal and First Values, the Factorial Formula and Symmetry

There is exactly one map on pairs of natural numbers with zero satisfying Pascal's recursion with the boundary values 1 and 0, and it vanishes above the diagonal, is 1 on the diagonal and n at k=1, satisfies the factorial formula and is symmetric.

Statement

In the setting of The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion, let factorials be as in The Factorial of a Natural Number with Zero §factorial, and for m≤nm\le n in N0\mathbb{N}_{0} let n−mn-m be the difference.

There is exactly one map C:N0×N0→N0C:\mathbb{N}_{0}\times\mathbb{N}_{0}\to\mathbb{N}_{0} with C(n,0)=1C(n,0)=1, C(0,k)=0C(0,k)=0 for k≥1k\ge1, and C(n+1,k+1)=C(n,k)+C(n,k+1)C(n+1,k+1)=C(n,k)+C(n,k+1), for all n,k∈N0n,k\in\mathbb{N}_{0}.

Let CC be this map, and let n,k∈N0n,k\in\mathbb{N}_{0}.

If k>nk>n, then C(n,k)=0C(n,k)=0.

C(n,n)=1C(n,n)=1.

C(n,1)=nC(n,1)=n.

If k≤nk\le n, then k! (n−k)! C(n,k)=n!k!\,(n-k)!\,C(n,k)=n!.

If k≤nk\le n, then C(n,k)=C(n,n−k)C(n,k)=C(n,n-k).

Proofs

Log in to submit a proof.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…