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[不等式] n元多项式可能有多少种对称

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hbghlyj posted 2023-8-5 14:25 |Read mode
$x_1+x_2+x_3$在全部3!=6种置换下不变
$x_1^2x_2+x_2^2x_3+x_3^2x_1$在3种轮换下不变
$x_1^2x_2x_3$有2个对称(恒等、交换$x_2x_3$)
$x_1^3x_2^2x_3$只有1个对称(恒等)
看来3元多项式可能有1,2,3,6种对称
那么4元多项式可能有多少种对称猜测是Divisors[4!]={1, 2, 3, 4, 6, 8, 12, 24}

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original poster hbghlyj posted 2023-8-5 15:20
四元多项式 1, 2, 3, 4, 6, 12, 24都容易构造出来。8怎么构造呢

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