Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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R(s,t)R(s1,t)+R(s,t1)R(s,t)\le R(s-1,t)+R(s,t-1) for s,t2s,t\ge2, and R(k,k)(2k2k1)22k2R(k,k)\le\binom{2k-2}{k-1}\le2^{2k-2}

Statement

For s,t2s,t\ge2,

R(s,t)R(s1,t)+R(s,t1).R(s,t)\le R(s-1,t)+R(s,t-1).

For every positive kk,

R(k,k)(2k2k1)22k2.R(k,k)\le\binom{2k-2}{k-1}\le2^{2k-2}.

Here RR is The off-diagonal Ramsey number R(s,t)R(s,t) as the least NN with N(s,t)2N\to(s,t)^2, for positive s,ts,t and the binomial coefficient is The set [A]k[A]^{k} of kk-element subsets and the binomial coefficient (nk):=[n]k\binom{n}{k} := \lvert [n]^{k}\rvert; the first diagonal inequality is the specialization of Finite graph Ramsey theorem: (s+t2s1)(s,t)2\binom{s+t-2}{s-1}\to(s,t)^2 for all positive s,ts,t.

Facts & Assumptions

Given: Positive naturals s,t,ks,t,k, with s,t2s,t\ge2 for the recursion.

[L1]

If m(s1,t)2m\to(s-1,t)^2 and n(s,t1)2n\to(s,t-1)^2, then m+n(s,t)2m+n\to(s,t)^2 for s,t2s,t\ge2 (If m(s1,t)2m\to(s-1,t)^2 and n(s,t1)2n\to(s,t-1)^2, then m+n(s,t)2m+n\to(s,t)^2 for s,t2s,t\ge2).

[L2]

For all x,yRx, y \in \mathbb{R} and every nNn \in \mathbb{N}, the binomial theorem expands (x+y)n(x+y)^n as the sum of its binomial terms (The binomial theorem in R\mathbb{R}: (x+y)n=k<n+1ι ⁣(nk)xkynk(x+y)^{n} = \sum_{k<n+1} \iota\!\binom{n}{k}\, x^{k} y^{\,n-k}).

Proof

technique · direct
1.1

The numbers R(s1,t)R(s-1,t) and R(s,t1)R(s,t-1) satisfy the two hypotheses of [L1]. Hence their sum arrows to (s,t)(s,t), and leastness in the definition of R(s,t)R(s,t) gives the recursion inequality.

L1
2.1

The finite binomial theorem gives R(k,k)(2k2k1)R(k,k)\le\binom{2k-2}{k-1}. In [L2] put x=y=1x=y=1 and n=2k2n=2k-2; every summand is nonnegative, so the single central coefficient is at most their sum 22k22^{2k-2}.

L2algebra

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