Conjecture: Ramsey Number R(m,n)=(2m-1)*p(2m-6+n,m)+{1,m,m+1}, for 3<=m<=n

Question

Today(2023-11-22), I have a conjecture on Ramsey numbers.

Fence Conjecture(栅栏猜想): Ramsey Number

R(m,n)=(2m-1)*p(2m-6+n,m)+{1,m,m+1}, for 3<=m<=n.

Here p(n,k) denotes both the number of partitions of n into exactly k parts (that is, sums of k positive integers that add to n), and the number of partitions of n into parts of maximum size exactly k. see Triangle_of_partition_numbers. The last summand choose one of three values {1,m,m+1}.

    In[89]:= Column[Table[
      IntegerPartitions[n, {k}] // Length, {n, 1, 20}, {k, 1, n}]]
    T[n_, k_] := 
     T[n, k] = 
      If[n > 0 && k > 0, T[n - 1, k - 1] + T[n - k, k], 
       Boole[n == 0 && k == 0]]
    p = Table[T[n, k], {n, 1, 40}, {k, 1, 40}];
    MatrixForm@p
    
    In[90]:= Table[(2 m - 1)*p[[2*m - 6 + n, m]], {m, 3, 15}, {n, m, 15}]

Out[90]= {{5, 5, 10, 15, 20, 25, 35, 40, 50, 60, 70, 80, 95}, 
{14, 21,35, 42, 63, 77, 105, 126, 161, 189, 238, 273}, 
{45, 63, 90, 117, 162, 207, 270, 333, 423, 513, 630}, 
{121, 154, 220, 286, 385, 484, 638, 781, 990, 1210}, 
{273, 364, 494, 637, 845, 1066, 1365, 1703, 2132}, 
{600, 780, 1050, 1335, 1740, 2190, 2790, 3450}, 
{1241, 1598, 2091, 2669, 3417, 4284, 5406}, 
{2432, 3116, 4028, 5073, 6460, 8037}, 
{4599, 5838, 7455, 9345, 11760}, 
{8418, 10580, 13386, 16675}, 
{14925, 18675, 23375}, 
{25839, 32076}, 
{43732}}

If this conjecture is correct, then we are only one step away from solving Ramsey numbers R(m,n). Because this equation basically determines the value of R(m,n), it is only necessary to choose one of three remainders modulo 2m-1. Here is my Mathematica .nb file.

Do you think my Fence Conjecture is correct? Any additions or corrections? Because I haven't found a general way to build a Ramsey critical graph (I'm sure R(3,10)=41, I'm working on building a Ramsey graph of order 40 without K3 and V10, and R(5,5) of order 45 critical graph), so I’m not completely sure yet, so I’ll post it here first to see everyone’s suggestions! I know this conjecture conflicts with many bounds data from Ramsey’s Theorem,Leo Berggren, but perfectly assists the data from http://users.cecs.anu.edu.au/~bdm/data/ramsey.html

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(modified at 2023-12-02): I think what I conjectured on 2023-11-22 had made some mistake. I should be more cautious in treating math problems. I accept Misha Lavrov's answer.

(Revisioned)Fence conjecture: $$R(m,n) <= (2m-1)*A008284_T(2m-6+n,m)+m+1 \text{,for } n>=m>=3$$. $$R(3,n)≡1,3,4(mod 5) \text {,for } 1<=n$$. $$R(n,n)= 2^{n+1}-4n+2 = 2*A183155(n-1)$$, R(n,n) is double the number of dominating sets in the n-path complement graph.

It sounds rather reasonable that R(n,n) is not greater than double the number of dominating sets in the n-path complement graph.

Answer 1

This conjecture would imply that $R(3,n) \in \{5p(n,3)+1, 5p(n,3)+3, 5p(n,3)+4\}$. We know that $p(n,3)$ is equal to $\frac{n^2}{12}$ rounded to the nearest integer (see OEIS A001399), so this would tell us that $R(3,n)$ grows approximately like $\frac{5}{12}n^2$ as $n \to \infty$.

This contradicts the known asymptotic upper bound that $R(3,n) = O(\frac{n^2}{\log n})$ due to Ajtai, Komlós, and Szemerédi.