引言

计算

烷基自由基计数

我们从卤代烃$C_nH_{2n+1}X$ 开始, X是卤素, 当然也可以是别的取代基.

Planarisomers

设 Free radicals count ,$\mathtt{Fr}(x)$

三个取代基的置换群若不考虑手性为 S_3 ;若考虑手性为 A_3

\cong{}

\mathbb{\Z{}}

$$\mathbb{Z}(\mathcal{S}_3)=\frac 16 (2s_3+3s_2s_1+s_1^3)$$

补上剩下的一个碳

$$\begin{aligned} A(x)&=1+x\mathbb{Z}[\mathcal{S}_3,A(x)]\\ &=1+\dfrac{x}{6}[A^3(x)+3A(x)A(x^2)+2A(x^3)] \end{aligned}$$

当然这个函数方程怎么展开是个大问题, 我们可以使用附录中级数的复合与反演来解决.

A[z_]:=Evaluate@Normal@Fold[
    Series[1+z/6(#^3+3# ComposeSeries[#,z^2+O[z]^#2]+2 ComposeSeries[#,z^3+O[z]^#2]),{z,0,#2}]&,
    1+O[z],Range@Floor@n
];

Stereoisomers

三个取代基的置换群若考虑手性为循环群 $\mathcal{C}_3$

$$\mathbb{Z}(\mathcal{C}_3)=\frac13(s_1^3+2 s_3)$$

A[z_]:=Evaluate@Normal@Fold[
    Series[1+z/3(#^3+2 ComposeSeries[#,z^3+O[z]^#2]),{z,0,#2}]&,
    1+O[z],Range@Floor[n/2]
];

烷烃计数

接下来考虑从烷基自由基组合生成烷烃 $C_nH_{2n+2}$

$$p^*-q^*+s=1$$

点的等价类数-边的等价类数+对称边数=1

碳的不同标记方法数-键的不同标记方法数+对称键数=1

let Alkane Count,$\mathtt{Ac}$

\mathtt{\Ac{}}

(z)=P(z)-Q(z)+S(z)

Planarisomers

$$\begin{aligned} P(x)&=x\mathbb{Z}[\mathcal{S}_4,A(x)]\\ &=\frac{1}{24}x(A^4(x)+6A^2(x)A(x^2)+3A^2(x^2)+8A(x)A(x^3)+6A(x^4))\\ Q(x)&=\mathbb{Z}[\mathcal{S}_2,A(x)-1]\\ &=\frac 12((A(x)-1)^2+A(x^2)-1)\\ S(x)&=A(z^2)-1 \end{aligned}$$

AlkaneSeries[n_Integer]:=Block[
    {A,P,Q,S,G},
    A[z_]:=Evaluate@Normal@Fold[
        Series[1+z/6(#^3+3# ComposeSeries[#,z^2+O[z]^#2]+2 ComposeSeries[#,z^3+O[z]^#2]),{z,0,#2}]&,
        1+O[z],Range@Floor[n/2]
    ];
    P[z_]=z CycleIndexPolynomial[SymmetricGroup[4],Array[A[z^#]&,4]];
    Q[z_]=CycleIndexPolynomial[SymmetricGroup[2],Array[A[z^#]-1&,2]];
    S[z_]=A[z^2];
    Series[P[z]-Q[z]+S[z]-1,{z,0,n}]
];

Stereoisomers

同样的 $p^*$ 由对称群$\mathcal{S}_4$ 变为交错群$\mathcal{A}_4$ .

$$\mathbb{Z}(\mathcal{A}_4)=\frac{1}{12}(s_1^4+8 s_3 s_1+3 s_2^2)$$

$$Q(x)=\mathbb{Z}[\mathcal{A}_4,A(x)]=\frac{1}{12}\left(8 A\left(z^3\right) A(z)+3 A\left(z^2\right)^2+A(z)^4\right)$$

AlkaneSeries[n_Integer]:=Block[
    {A,P,Q,S,G},
    A[z_]:=Evaluate@Normal@Fold[
        Series[1+z/3(#^3+2 ComposeSeries[#,z^3+O[z]^#2]),{z,0,#2}]&,
        1+O[z],Range@Floor[n/2]
    ];
    P[z_]=z CycleIndexPolynomial[AlternatingGroup[4],Array[A[z^#]&,4]];
    Q[z_]=CycleIndexPolynomial[SymmetricGroup[2],Array[A[z^#]-1&,2]];
    S[z_]=A[z^2];
    Series[P[z]-Q[z]+S[z]-1,{z,0,n}]
];

苯环

pg = GroupDirectProduct[AlternatingGroup@5, CyclicGroup@2];
A[z_, i_] := Evaluate[#^i & /@ (1 + a + b + c + d)]
CycleIndexPolynomial[pg, Array[A[z, #] &, 60]];
Coefficient[%, a b c d]

附录

旋转群与指标多项式

S,C,A,D

级数复合与反演

拉格朗日反演, 比起比较两边系数高到不知道哪里去了.

群直积与积群

GroupDirectProduct[g1_,g2_]:=With[
    {order1=GroupOrder@g1,order2=GroupOrder@g2,r,pd},
    r=Thread[Range[order2]->(order1+Range[order2])];
    pd=Outer[PermutationProduct,GroupElements[g1],GroupElements[g2]/.r];
    PermutationGroup@Flatten@pd
]

我刚开始以为Mathematica没有计算群直积的函数, 然后写了个GroupDirectProduct.

后来才发现确实没有这个函数, 因为群直积居然划分在群属性里.

GroupDirectProduct[AlternatingGroup@5,CyclicGroup@2]
FiniteGroupData[{"DirectProduct",
    {
        {"AlternatingGroup",5},
        {"CyclicGroup",2}
    }
},"PermutationGroupRepresentation"]