And what I am going to do is say start with this ion, add up the energy associated with the interactions between that ion and everybody else in the row and then multiply it by Avogadro's number, because that is the number of atoms there are in a row.

接下来我要从这一离子开始，加上相互作用的能量，也就是这一离子,和其它所有在这一行的离子之间的能量,再乘以阿伏加德罗常数，因为这是在一行的原子的数量。

麻省理工公开课 - 固态化学导论课程节选

so given that we chose an odd number of people in the row, if exactly one candidate stands and that candidate is the center candidate, then that's an equilibrium.

假若我们选中行的人数为奇数,如果确实只有一个候选人参选且,那个候选人就是在中间,则那是个均衡

耶鲁公开课 - 博弈论课程节选