在xy，xz，yz平面可自由变换的条件下，计算 5t16-1
结构5t16有5个点3个维度
点1，2，3共两面，点1，4共两面，点1，5共一面。点2，3共两面，点2，4共一面，点2，5不共面。点3，4共一面，点3，5不共面。点4，5共两面。所以这点2和3的对称性是一样的，点1，4，5的对称性各不相同。
所以5t16-1应该得到5个4点结构，其中有2个是一样的
5（5t16-1）=1*4t2+1*4t6+2*4t16+1*4t29
5t16减5得到4t2，减4得到4t6，减2或者3得到4t16，减1得到4t29
|----|---------------------------------------|
| 1 | 5*4t1 |
| 2 | 1*4t1+3*4t2+1*4t3 |
| 3 | 1*4t1+4*4t3 |
| 4 | 1*4t1+3*4t4+1*4t5 |
| 5 | 1*4t1+4*4t5 |
| 6 | 1*4t1+3*4t6+1*4t7 |
| 7 | 1*4t1+4*4t7 |
| 8 | 2*4t2+1*4t8+2*4t9 |
| 9 | 1*4t2+1*4t3+2*4t9+1*4t22 |
| 10 | 2*4t2+2*4t10+1*4t23 |
| 11 | 2*4t2+1*4t11+2*4t12 |
| 12 | 1*4t2+1*4t3+2*4t12+1*4t24 |
| 13 | 1*4t2+1*4t4+2*4t13+1*4t26 |
| 14 | 1*4t2+1*4t4+1*4t14+1*4t15+1*4t25 |
| 15 | 1*4t2+1*4t5+2*4t15+1*4t27 |
| 16 | 1*4t2+1*4t6+2*4t16+1*4t29 |
| 17 | 1*4t2+1*4t6+1*4t17+1*4t18+1*4t28 |
| 18 | 1*4t2+1*4t7+2*4t18+1*4t30 |
| 19 | 1*4t2+1*4t6+2*4t19+1*4t32 |
| 20 | 1*4t2+1*4t6+1*4t20+1*4t21+1*4t31 |
| 21 | 1*4t2+1*4t7+2*4t21+1*4t33 |
| 22 | 2*4t3+3*4t22 |
| 23 | 2*4t3+3*4t23 |
| 24 | 2*4t3+3*4t24 |
| 25 | 1*4t3+1*4t4+2*4t25+1*4t27 |
| 26 | 1*4t3+1*4t5+3*4t26 |
| 27 | 1*4t3+1*4t5+3*4t27 |
| 28 | 1*4t3+1*4t6+2*4t28+1*4t30 |
| 29 | 1*4t3+1*4t7+3*4t29 |
| 30 | 1*4t3+1*4t7+3*4t30 |
| 31 | 1*4t3+1*4t6+2*4t31+1*4t33 |
| 32 | 1*4t3+1*4t7+3*4t32 |
| 33 | 1*4t3+1*4t7+3*4t33 |
| 34 | 2*4t4+1*4t34+2*4t35 |
| 35 | 1*4t4+1*4t5+2*4t35+1*4t45 |
| 36 | 1*4t4+1*4t6+2*4t36+1*4t47 |
这里列出5点结构前36个减一的结果
5点的36个结构
4点的结构
