Skip to main content
Added Ramsey theory tag; Rado/Schur numbers are about Ramsey theory on the integers.
Link
Added LaTeX markup.
Source Link
Stefan Kohl
  • 19.6k
  • 21
  • 75
  • 137

Consider the integer linear equation $\sum_{i=1}^{n} c_ix_i=0$, where $c_i(\ne 0) \in Z$$c_i(\ne 0) \in \mathbb{Z}$. Supposing it is given that there is a natural number N$N$ such that, if {1,2...N}$\{1,2, \dots, N\}$ is partitioned in two sets, one of these always contains a solution of the equation. The minimal such N$N$ is called the Rado number of the equation. I am looking for general bounds on such N$N$, in the cases where it exists. Where can I possibly find such results. Thanks.

Consider the integer linear equation $\sum_{i=1}^{n} c_ix_i=0$ where $c_i(\ne 0) \in Z$. Supposing it is given that there is a natural number N such that, if {1,2...N} is partitioned in two sets, one of these always contains a solution of the equation. The minimal such N is called the Rado number of the equation. I am looking for general bounds on such N, in the cases where it exists. Where can I possibly find such results. Thanks.

Consider the integer linear equation $\sum_{i=1}^{n} c_ix_i=0$, where $c_i(\ne 0) \in \mathbb{Z}$. Supposing it is given that there is a natural number $N$ such that, if $\{1,2, \dots, N\}$ is partitioned in two sets, one of these always contains a solution of the equation. The minimal such $N$ is called the Rado number of the equation. I am looking for general bounds on such $N$, in the cases where it exists. Where can I possibly find such results. Thanks.

Source Link

A generalization of Schur Numbers

Consider the integer linear equation $\sum_{i=1}^{n} c_ix_i=0$ where $c_i(\ne 0) \in Z$. Supposing it is given that there is a natural number N such that, if {1,2...N} is partitioned in two sets, one of these always contains a solution of the equation. The minimal such N is called the Rado number of the equation. I am looking for general bounds on such N, in the cases where it exists. Where can I possibly find such results. Thanks.