Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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R(s,t)≤R(s−1,t)+R(s,t−1) for s,t≥2, and R(k,k)≤(2k−2k−1)≤22k−2

Statement

For s,t≥2,

R(s,t)≤R(s−1,t)+R(s,t−1).

For every positive k,

R(k,k)≤(2k−2k−1)≤22k−2.

Here R is The off-diagonal Ramsey number R(s,t) as the least N with N→(s,t)2, for positive s,t and the binomial coefficient is The set [A]k of k-element subsets and the binomial coefficient (nk):=∣[n]k∣; the first diagonal inequality is the specialization of Finite graph Ramsey theorem: (s+t−2s−1)→(s,t)2 for all positive s,t.

Facts & Assumptions

Given: Positive naturals s,t,k, with s,t≥2 for the recursion.

[L1]

If m→(s−1,t)2 and n→(s,t−1)2, then m+n→(s,t)2 for s,t≥2 (If m→(s−1,t)2 and n→(s,t−1)2, then m+n→(s,t)2 for s,t≥2).

[L2]

For all x,y∈R and every n∈N, the binomial theorem expands (x+y)n as the sum of its binomial terms (The binomial theorem in R: (x+y)n=∑k<n+1ι ⁣(nk) xky n−k).

Proof

technique · direct
1.1

The numbers R(s−1,t) and R(s,t−1) satisfy the two hypotheses of [L1]. Hence their sum arrows to (s,t), and leastness in the definition of R(s,t) gives the recursion inequality.

L1
2.1

The finite binomial theorem gives R(k,k)≤(2k−2k−1). In [L2] put x=y=1 and n=2k−2; every summand is nonnegative, so the single central coefficient is at most their sum 22k−2.

L2algebra∎

Depends on

Used by

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Dependency tree · two levels

31 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