Exact values and lower bounds on the n-color weak Schur numbers for n=2,3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n=2,3$$\end{document}

For integers k, n with k, n⩾1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n \geqslant 1$$\end{document}, the n-color weak Schur numberWSk(n)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$W\hspace{-0.6mm}S_{k}(n)$$\end{document} is defined as the least integer N, such that for every n-coloring of the integer interval [1, N], there exists a monochromatic solution x1,⋯,xk,xk+1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x_{1},\dots , x_{k}, x_{k+1}$$\end{document} in that interval to the equation: x1+x2+⋯+xk=xk+1,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} x_{1}+x_{2}+\dots +x_{k} =x_{k+1}, \end{aligned}$$\end{document}with xi≠xj\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x_{i} \ne x_{j}$$\end{document}, when i≠j.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$i\ne j.$$\end{document} In this paper, we obtain the exact values of WS6(2)=166\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$WS_{6}(2)=166$$\end{document}, WS7(2)=253\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$WS_{7}(2)=253$$\end{document}, WS3(3)=94\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$WS_{3}(3)=94$$\end{document} and WS4(3)=259\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$WS_{4}(3)=259$$\end{document} and we show new lower bounds on n-color weak Schur number WSk(n)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$W\hspace{-0.6mm}S_{k}(n)$$\end{document} for n=2,3.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n=2,3.$$\end{document}


Introduction
For integers a ≤ b, we shall denote [a, b] the integer interval consisting of all t ∈ N + = {1, 2, . . .} such that a ≤ t ≤ b.A function where c 1 , . . ., c n ∈ N + represent different colors, is an n-coloring of the interval [1, N ].
Given an n-coloring and the equation x 1 + • • • + x k = x k+1 in k + 1 variables, we say that a solution x 1 , . . ., x k , x k+1 to the equation is monochromatic if and only if For integers k, n with k, n ≥ 1, the n-color weak Schur number WS k (n) is defined as the least integer N , such that for every n-coloring of the integer interval [1, N ], there exists a monochromatic solution x 1 , . . ., x k , x k+1 in that interval to the equation: , with x i = x j when i = j.Irwing [14] showed the existence and obtained the following general upper bound: For k = 2, we have 1 + 315 n−1 5 ≤ WS 2 (n) ≤ [n!ne] + 1, the lower bound is due to Sierpinski [20] and the upper bound to Bornsztein [5].

Schur numbers and weak Schur numbers
A set A of integers is called sum-free if it contains no elements x 1 , x 2 , x 3 ∈ A satisfying x 1 + x 2 = x 3 where x 1 , x 2 need not be distinct.
Schur [19] in 1916 proved that, given a positive integer n, there exists a greatest positive integer S 2 (n) = N with the property that the integer interval [1, N − 1] can be partitioned into n sum-free sets.The numbers S 2 (n) are called Schur numbers.The current knowledge on these numbers for 1 ≤ n ≤ 7 is given in Table 1.
Many generalizations of Schur numbers have appeared since their introduction.Now, a set A of integers is called weakly sum-free if it contains no pairwise distinct elements x 1 , x 2 , x 3 ∈ A satisfying x 1 + x 2 = x 3 .We denote by WS 2 (n), the greatest integer N , for which the integer interval [1, N − 1].The exact value of S 2 (4) was given by Baumert [2] and recently S 2 (5) has been obtained by Heule [13].Finally, the lower bounds on S 2 (6) and S 2 (7) were obtained by Fredricksen and Sweet [11] by considering symmetric sum-free partitions.A set A of integers is said to be k-sum-free [15] gave the following generalization: given two positive integers, n and k ≥ 2, there exists a greatest positive integer, S k (n) = N , such that the integer interval [1, N − 1] can be partitioned into n sets which are k-sum-free.In 1966, Znám [22] established a lower In 1982, Beutelspacher and Brestovansky [3] proved the equality for two k-sum-free sets: In 2010 [18], the last author obtained the exact value of S 3 (3) = 43.Independently, Ahmed and Schaal [1] in 2016 gave the values of S k (3) for k = 3, 4, 5.In 2019, Boza et al. [6] determined the exact formula of S k (3) = k 3 + 2k 2 − 2 for all k ≥ 3, finding an upper bound that coincides with the lower bound given by Znám [22].
The numbers WS 2 (n) are called the weak Schur numbers for the equation x 1 + x 2 = x 3 .The known weak Schur numbers are given in Table 2.
The current state of knowledge concerning WS 2 (n) is a bit confusing.The problem seems to have been first considered in [21], which is Walker's solution to Problem E985 proposed a year earlier, in 1951, by Moser.Walker considered the cases n = 3, 4 and 5 and claimed the values WS 2 (3) = 24, WS 2 (4) = 67, and WS 2 (5) = 197.Unfortunately, the short account written by Moser on Walker's solution only gives suitable partitions of [1,23] for n = 3, and no details at all for the cases n = 4 and 5. Walker's claimed values of WS 2 (3) and WS 2 (4) were later confirmed by Blanchard, Harary, and Reis using computers [4].In 2012, the two last authors et al. [9] confirmed the lower bound WS 2 (5) ≥ 197.In addition, a lower bound on WS 2 (6) was obtained in [9] and later improved to WS 2 (6) ≥ 583 in [10].The lower bounds for 7 ≤ n ≤ 9 were obtained [17] in 2015.
In terms of coloring, the WS k (n) is the least positive integer N such that for every where c 1 , . . ., c n represent n different colors, there exists a monochromatic solution to the equation where x i = x j when i = j.
In addition, for 2-coloring, the known weak Schur numbers WS k (2) are shown in Table 3.

Main results
In Section 2, we determine a general lower bound on the 2-color weak Schur numbers for the equation , with x i = x j when i = j, for k ≥ 5, improving the lower bound given in [7].
for any integer k ≥ 5.In Section 3, we determine a general lower bound on W S k (3) improving the lower bound given in [7].

A general lower bound for WS k (2)
In terms of coloring, the weak Schur number W S k (2) is the least positive integer N such that for every 2-coloring of [1, N ], where c 1 , c 2 represent 2 different colors, there exists a monochromatic solution to the equation In [7], a general lower bound of the weak Schur number W S k (2) was given, now we show a new general lower bound that improves the previous one.

Lemma 2.1 For any integer
Proof Let be a 2-coloring: where We show that the above partition of the interval [1, 1 2 (k 3 +4k 2 −5k)] has no monochromatic solution to the equation For that, it is sufficient to prove that for every i, In addition, for k ≥ 5, -If Therefore, we obtain the lower bound.
123 With this general lower bound, we improve the results shown in Table 3.In addition, in the Section 4, we will prove that these new lower bounds shown in Table 4 for k = 6 and k = 7, are exact values.

A lower bound for WS k (3)
Applying the result given in [7], the lower bounds shown in Table 5were obtained.
In the next result, we improve the general lower bound of WS k (3) obtained in [7].
Lemma 3.1 For any integer k ≥ 5, we have Proof We will show that he following partition of the interval has no monochromatic solution to the equation Consider the following 3-coloring where A 1 and A 2 are the same as used in the construction of the 2-coloring in Lemma 2.1.
Since the above 3-coloring is an extension of 2-coloring given by Lemma 2.1, we just have to try the following cases: -If Therefore, we obtain the desired lower bound.
With this general lower bound,we improve the results shown in Table 5.In addition, in Section 4, we will prove that these new lower bounds shown in Table 6 for k = 3 and = 4 are exact values.
In the next result, we improve the lower bounds of Lemma 3.1 for any integer k ≥ 8.

Lemma 3.2
For any integer k ≥ 8, we have

Proof
The following partition of the interval has no monochromatic solution to the equation This 3-coloring is an extension of 3-coloring given in Lemma 3.1, so we just have to try the following cases: We consider four cases: -If Therefore,we obtain the desired improved lower bound.
Applying Lemmas 3.1 and 3.2, the following lower bounds are shown in Table 7.

For every partition of Y into two subsets A
Proof 1.This is trivial.2. We have checked the result transforming the problem into a Boolean satisfiability problem and solving it with a SAT solver [12].Let be a 2-coloring of [1,166]: For any {y n } ∈ Y, we consider a Boolean variable φ defined on [1,42] as follows: Let S = {(y a , y b , y c , y d , y e , y f , y g ) ∈ Y 7 | y a +y b +y c +y d +y e +y f = y g , with a < b < c < d < e < f }.
For any s = (y a , y b , y c , y d , y e , y f , y g ) ∈ S, we consider two clauses: Then, p(s) is satisfiable if and only if , does not induce in s a monochromatic solution on c 1 of the equation x 1 +• • •+x 6 = x 7 .Analogously, q(s) is satisfiable if and only if does not induce in s a monochromatic solution of the equation Clearly C is satisfiable if and only if does not induce on Y a monochromatic solution of the equation The SAT-Solver shows that C is not satisfiable, hence for every 2-coloring of the set,Y has a monochromatic solution to the equation With this result, we have tested the upper bound on WS 6 (2).Therefore, we conclude with the following result:
We have to prove that the equation The following two lemmas can be proved transforming the problem into a Boolean satisfiability problem and solving it with a SAT solver [12].

Lemma 4.3 For every 2-coloring of D 1 without monochromatic solution, we have
Proof Let be a 2-coloring of D 1 : For any {u n } ∈ D 1 , we consider a Boolean variable φ defined on [1,73] as follows: For any s = (u a , u b , u c , u d , u e , u f , u g , u h ) ∈ S, we consider two clauses: Then, p(s) is satisfiable if and only if , does not induce in s a monochromatic solution on c 1 of the equation Analogously, q(s) satisfiable if and only if does not induce in s a monochromatic solution of the equation Clearly C is satisfiable if and only if does not induce on D 1 a monochromatic solution of the equation The SAT-Solver shows that C is not satisfiable, hence we have the result.
Trivially we have, Proof Let be a 2-coloring of D 1 : we consider a Boolean variable φ defined on [1,78] as follows: For any s = (u a , u b , u c , u d , u e , u f , u g , u h ) ∈ S, we consider two clauses: Then, p(s) is satisfiable if and only if

The exact value of WS 3 (3)
The weak Schur number W S 3 (3) is the least positive integer N such that for every 3-coloring of [1, N ], where c 1 , c 2 , c 3 represent 3 different colors, there exists a monochromatic solution to the equation We shall prove that W S 3 (3) = 94.Let us first show a lower bound.Proof It is easy to verify that the 3-coloring has no monochromatic solution to the equation x 1 + x 2 + x 3 = x 4 such that x i = x j when i = j.
To prove that the equation x 1 + x 2 + x 3 = x 4 has a monochromatic solution for every 3-coloring of the integer interval [1,94]

For every partition of Y into three subsets A
Proof 1.This is trivial.
2. We have checked the result transforming the problem into a Boolean satisfiability problem and solving it with a SAT solver [12].Let be a 3-coloring of [1,94]: For any {y n } ∈ Y, we consider two Boolean variables φ and ψ defined on [1,51] as follows: Thus, for any n ∈ [1,51] we have that φ For any s = (y a , y b , y c , y d ) ∈ S, we consider three clauses: does not induce in s a monochromatic solution on c 1 of the equation x 1 + x 2 + x 3 = x 4 .Analogously, q(s) or r (s) is satisfiable if and only if does not induce in s a monochromatic solution of the equation Clearly D ∧ C is satisfiable if and only if does not induce on Y a monochromatic solution of the equation x 1 + x 2 + x 3 = x 4 .The SAT-Solver shows that D ∧ C is not satisfiable, hence W S 3 (3) ≤ 94.
With this result, we have tested the upper bound on WS 3 (3).
Therefore, we conclude with the following result: To prove that the equation x 1 + x 2 + x 3 + x 4 = x 5 has a monochromatic solution for every 3-coloring of the integer interval [1,259], it is necessary to prove the following result.
Proof 1.This is trivial.2. We have checked the result transforming the problem into a Boolean satisfiability problem and solving it with a SAT solver [12].

Table 1
The first few Schur numbers S 2 (n)

Table 2
The first few weak Schur numbers WS 2 (n)

4 Computer-assisted proofs for the exact values of WS 6 (2), WS 7 (2), WS 3 (3) and WS 4 (3) 4.1 The exact value of WS 6 (2) We
for every 2-coloring of the integer interval 166], it is necessary to show the following result.
, it is necessary to prove the following result.