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%I #2 Mar 30 2012 17:34:22
%S 1,3,6,7,1,2,8,9,1,5,9,12,1,4,6,10,2,3,4,5,1,3,6,7,3,7,9,11,3,4,8,12,
%T 1,3,6,7,1,4,6,10,2,6,8,11,5,6,11,12,1,3,6,7,2,5,7,10,3,7,9,11,7,8,10,
%U 12,1,3,6,7,1,2,8,9,1,5,9,12,1,4,6,10,2,3,4,5,1,2,8,9,2,5,7,10,2,6,8,11,1
%N A twelve vertex {3,6,3} prism (hexagon connected to two triangles) graph substitution.
%C Although designed to be similar to the {9,3} Prism in shape the {3,6,3} polyhedron sounds better to my ear.
%F 1->{2, 3, 4, 5}; 2-> {1, 3, 6, 7}; 3-> {1, 2, 8, 9}; 4-> {1, 5, 9, 12}; 5-> {1, 4, 6, 10}; 6-> {2, 5, 7, 10}; 7-> {2, 6, 8, 11}; 8-> {3, 7, 9, 11}; 9-> {3, 4, 8, 12}; 10-> {5, 6, 11, 12}; 11-> {7, 8, 10, 12}; 12-> {4, 9, 10, 11};
%t Clear[s] s[1] = {2, 3, 4, 5}; s[2] = {1, 3, 6, 7}; s[3] = {1, 2, 8, 9}; s[ 4] = {1, 5, 9, 12}; s[5] = {1, 4, 6, 10}; s[6] = {2, 5, 7, 10}; s[7] = {2, 6, 8, 11}; s[8] = {3, 7, 9, 11}; s[9] = {3, 4, 8, 12}; s[10] = {5, 6, 11, 12}; s[11] = {7, 8, 10, 12}; s[12] = {4, 9, 10, 11}; t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n - 1]]; p[4]
%K nonn,uned
%O 1,2
%A _Roger L. Bagula_, Oct 19 2007