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arXiv 1505.05071math.CO
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$x_1 + x_2 + \dots + x_m + c = 2x_0$的离散Rado数

The Discrete Rado Number for $x_1 + x_2 + \dots + x_m + c = 2x_0$

Tristin Lehmann, Donald L. Vestal

摘要:

对于正整数$m$和实数$c$,令$R = R(m,c,2)$表示等式$x_1 + x_2 + \dots + x_m + c = 2x_0$的离散2色Rado数。换句话说,$R$是最小的整数,使得对于整数${1, 2, \dots, R}$的任何着色,都存在具有相同颜色的数字$x_1, x_2, \dots, x_m, x_0$,使得$x_1 + x_2 + \dots + x_m + c = 2x_0$。在这篇文章中,我们表明,如果$m \geq 2$和$c > 0$,那么$$ \begin{array}{cc} R(m,c,2) = \begin{cases} \infty & \text{for $m$ even, $c$ odd} \newline \big\lceil \frac{m}{2} \big\lceil \frac{m+c}{2} \big\rceil + \frac{c}{2} \big\rceil & \text{otherwise.} \end{cases} \end{array} $$对于实数$a$和$c$,我们看看方程$x_1 + c = ax_0$的双色Rado数。我们证明了如果$a > 1$和$c > 0$,则2色连续Rado数是$$ R_\mathbb{R}(1, c, a) = \begin{cases} \infty & \text{if $a=1$ } \newline \frac{c}{a-1} & \text{otherwise.} \end{cases} $$。由此,我们将证明离散Rado数是$$ R(1, c, a) = \begin{cases} \frac{c}{a-1} & \text{if $ \left( a-1 \right) \mid c$} \newline \infty & \text{otherwise.} \end{cases} $$

英文摘要:

For a positive integer $m$ and a real number $c$, let $R = R(m,c,2)$ denote the discrete 2-color Rado number for the equation $x_1 + x_2 + \dots + x_m + c = 2x_0$. In other words, $R$ is the smallest integer such that for any coloring of the integers ${1, 2, \dots, R}$, there exist numbers $x_1, x_2, \dots, x_m, x_0$, all with the same color, such that $x_1 + x_2 + \dots + x_m + c = 2x_0$. In this article we show that if $m \geq 2$ and $c > 0$, then $$ \begin{array}{cc} R(m,c,2) = \begin{cases} \infty & \text{for $m$ even, $c$ odd} \newline \big\lceil \frac{m}{2} \big\lceil \frac{m+c}{2} \big\rceil + \frac{c}{2} \big\rceil & \text{otherwise.} \end{cases} \end{array} $$ For real numbers $a$ and $c$, we look at the 2-color Rado number for the equation $x_1 + c = ax_0$. We show that if $a > 1$ and $c > 0$, then the 2-color continuous Rado number is $$ R_\mathbb{R}(1, c, a) = \begin{cases} \infty & \text{if $a=1$ } \newline \frac{c}{a-1} & \text{otherwise.} \end{cases} $$ From this, we will show that the discrete Rado number is $$ R(1, c, a) = \begin{cases} \frac{c}{a-1} & \text{if $ \left( a-1 \right) \mid c$} \newline \infty & \text{otherwise.} \end{cases} $$

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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