| …
ST （7） p1 （19）={{1,2,3},{1,4,5},{1,6,7},{1,8,9},{1,10,11},{1,12,13},{1,14,15},{1,16,17},{1,18,19},{2,4,18},{2,5,19},{2,6,16},{2,7,17},{2,8,12},{2,9,13},{2,10,14},{2,11,15},{3,4,19},{3,5,18},{3,6,17},{3,7,16},{3,8,13},{3,9,12},{3,10,15},{3,11,14},{4,6,10},{4,7,11},{4,8,16},{4,9,17},{4,12,14},{4,13,15},{5,6,11},{5,7,10},{5,8,17},{5,9,16},{5,12,15},{5,13,14},{6,8,14},{6,9,15},{6,12,18},{6,13,19},{7,8,15},{7,9,14},{7,12,19},{7,13,18},{8,10,18},{8,11,19},{9,10,19},{9,11,18},{10,12,16},{10,13,17},{11,12,17},{11,13,16},{14,16,18},{14,17,19},{15,16,19},{15,17,18}}。
类似的得P 2 类19阶Steiner三连系：
ST （1） p2 （19）={{1,2,3},{1,4,6},{1,5,7},{1,8,14},{1,9,15},{1,10,12},{1,11,13},{1,16,18},{1,17,19},{2,4,5},{2,6,7},{2,8,9},{2,10,11},{2,12,13},{2,14,15},{2,16,17},{2,18,19},{3,4,7},{3,5,6},{3,8,15},{3,9,14},{3,10,13},{3,11,12},{3,16,19},{3,17,18},{4,8,10},{4,9,11},{4,12,16},{4,13,17},{4,14,18},{4,15,19},{5,8,11},{5,9,10},{5,12,17},{5,13,16},{5,14,19},{5,15,18},{6,8,16},{6,9,17},{6,10,18},{6,11,19},{6,12,14},{6,13,15},{7,8,17},{7,9,16},{7,10,19},{7,11,18},{7,12,17},{7,13,14},{8,12,18},{8,13,19},{9,12,19},{9,13,18},{10,14,16},{10,15,17},{11,14,16},{11,15,16}}；
ST （2） p2 （19）={{1,2,3},{1,4,8},{1,5,9},{1,6,8},{1,7,19},{1,10,16},{1,11,17},{1,12,14},{1,13,15},{2,4,5},{2,6,7},{2,8,9},{2,10,11},{2,12,13},{2,14,15},{2,16,17},{2,18,19},{3,4,9},{3,5,8},{3,6,19},{3,7,18},{3,10,17},{3,11,16},{3,12,15},{3,13,14},{4,6,12},{4,7,13},{4,10,14},{4,11,15},{4,16,18},{4,17,19},{5,6,13},{5,7,12},{5,10,15},{5,11,14},{5,16,19},{5,17,18},{6,8,10},{6,9,11},{6,14,16},{6,15,17},{7,8,11},{7,9,10},{7,14,17},{7,15,16},{8,12,16},{8,13,17},{8,14,18},{8,15,19},{9,12,17},{9,13,16},{9,14,19},{9,15,18},{10,12,18},{10,13,19},{11,12,19},{11,13,18}}；
…
ST （7） p2 （19）={{1,2,3},{1,4,18},{1,5,19},{1,6,16},{1,7,17},{1,8,12},{1,9,13},{1,10,14},{1,11,15},{2,4,5},{2,6,7},{2,8,9},{2,10,11},{2,12,13},{2,14,15},{2,16,17},{2,18,19},{3,4,19},{3,5,18},{3,6,17},{3,7,16},{3,8,13},{3,9,12},{3,10,15},{3,11,14},{4,6,10},{4,7,11},{4,8,16},{4,9,17},{4,12,14},{4,13,15},{5,6,11},{5,7,10},{5,8,17},{5,9,16},{5,12,15},{5,13,14},{6,8,14},{6,9,15},{6,12,18},{6,13,19},{7,8,15},{7,9,14},{7,12,19},{7,13,18},{8,10,18},{8,11,19},{9,10,19},{9,11,18},{10,12,16},{10,13,17},{11,12,17},{11,13,16},{14,16,18},{14,17,19},{15,16,19},{15,17,18}}。
类似的得P 20 类19阶Steiner三连系：
ST （1） p20 （19）={{1,2,19},{1,3,18},{1,4,19},{1,5,15},{1,6,16},{1,7,11},{1,8,12},{1,9,13},{1,10,14},{2,3,19},{2,4,18},{2,5,16},{2,6,15},{2,7,12},{2,8,11},{2,9,14},{2,10,13},{3,4,19},{3,5,9},{3,6,10},{3,7,15},{3,8,16},{3,11,13},{3,12,14},{4,7,16},{4,8,15},{4,11,14},{4,12,13},{5,6,19},{5,7,13},{5,8,14},{5,11,18},{5,12,19},{6,7,14},{6,8,13},{6,11,19},{6,12,18},{7,8,19},{7,9,18},{7,10,19},{8,9,19},{8,10,18},{9,10,19},{9,11,15},{9,12,16},{10,11,16},{10,12,15},{11,12,19},{13,14,19},{13,15,18},{13,16,19},{14,15,19},{14,16,18},{15,16,19},{17,18,19}}；
ST （2） p20 （19）={{1,2,19},{1,3,7},{1,4,8},{1,5,17},{1,6,18},{1,9,15},{1,10,16},{1,11,13},{1,12,14},{2,3,8},{2,4,7},{2,5,18},{2,6,17},{2,9,16},{2,10,15},{2,11,14},{2,12,13},{3,4,19},{3,5,11},{3,6,12},{3,9,13},{3,10,14},{3,15,17},{3,16,18},{4,5,12},{4,6,11},{4,9,14},{4,10,13},{4,15,18},{4,16,17},{5,6,19},{5,7,9},{5,8,10},{5,13,15},{5,14,16},{6,7,10},{6,8,9},{6,13,16},{6,14,15},{7,8,19},{7,11,15},{7,12,16},{7,13,17},{7,14,18},{8,11,16},{8,12,15},{8,13,18},{8,14,17},{9,10,19},{9,11,17},{9,12,18},{10,11,18},{10,12,17},{11,12,19},{13,14,19},{15,16,19},{17,18,19}}；
…
ST （7） p20 （19）={{1,2,19},{1,3,17},{1,4,18},{1,5,15},{1,6,16},{1,7,11},{1,8,12},{1,9,13},{1,10,14},{2,3,18},{2,4,17},{2,5,16},{2,6,15},{2,7,12},{2,8,11},{2,9,14},{2,10,13},{3,4,19},{3,5,9},{3,6,10},{3,7,15},{3,8,16},{3,11,13},{3,12,14},{4,5,10},{4,6,9},{4,7,16},{4,8,15},{4,11,14},{4,12,13},{5,6,19},{5,7,13},{5,8,14},{5,11,17},{5,12,18},{6,7,14},{6,8,13},{6,11,18},{6,12,17},{7,8,19},{7,9,17},{7,10,18},{8,9,18},{8,10,17},{9,10,19},{9,11,15},{9,12,16},{10,11,16},{10,12,15},{11,12,19},{13,14,19},{13,15,17},{13,16,18},{14,15,18},{14,16,17},{15,16,19}}。
3、19阶Steiner三连系的计数
从2t+1阶Steiner三连系的实际构造发现，2t+1阶Steiner三连系的个数N决定于： t阶Steiner三连系的个数N （1） ，边矩阵K v ′的2×2子矩阵划分方案p 1 ，p 2 ，…，p v+1 的个数N （2） =v+1，2×2的三连系矩阵K  的构造方案数N （3） =2，依照乘法原则，19阶Steiner三连系的个数N= N（1）×N（2）×N（3）= N（2）×2(v+1)=105×20×2。本文仅构造出140个ST(19)。
参考文献：
[1] 杨骅飞，王朝瑞.组合数学及其应用[M].北京：北京理工大学出版社.1992
[2] 杨振生.组合数学及其算法[M].合肥：中国科学技术大学出版社.2003
[3]侴万禧.r×t阶Kirlman三连系构造的一种方法[J].数学的实践与认识.2004.(9) 144-150
[5]侴万禧.高阶Steiner三连系及其构造方法[J].安徽理工大学学报.2004. (3) 76-80
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