Search: id:A086200 Results 1-1 of 1 results found. %I A086200 %S A086200 1,3,15,66,406,2775,19900,152076,1206681,9841266,82336528, %T A086200 702993756,6105180250,53822344278,480681790786,4342078862605, %U A086200 39621836138886,364831810979041,3386667673687950,31669036266203766 %N A086200 Number of unrooted steric quartic trees with 2n (unlabeled) nodes and possessing a bicentroid; number of 2n-carbon alkanes C(2n)H(4n +2) with a bicentroid when stereoisomers are regarded as different. %C A086200 The degree of each node is <= 4. %C A086200 A bicentroid is an edge which connects two subtrees of exactly m/2 nodes, where m is the number of nodes in the tree. If a bicentroid exists it is unique. Clearly trees with an odd number of nodes cannot have a bicentroid. %C A086200 Regarding stereoisomers as different means that only the alternating group A_4 acts at each node, not the full symmetric group S_4. See A010373 for the analogous sequence when stereoisomers are not counted as different. %H A086200 Index entries for sequences related to rooted trees %H A086200 Index entries for sequences related to trees %F A086200 G.f.: replace each term x in g.f. for A000625 by x(x+1)/2. Interpretation: ways to pick 2 specific radicals (order not important) from all n carbon radicals is number of 2n carbon bicentered alkanes (join the two radicals with an edge). %Y A086200 Cf. A000598, A000602, A010732, A010733, A000625, A000628, A086194. %Y A086200 For even n A000628(n) = A086194(n) + a(n/2), for odd n A000628(n) = A086194(n), since every tree has either a centroid or a bicentroid but not both. %Y A086200 Sequence in context: A001447 A106732 A052981 this_sequence A122558 A110211 A033876 %Y A086200 Adjacent sequences: A086197 A086198 A086199 this_sequence A086201 A086202 A086203 %K A086200 nonn %O A086200 1,2 %A A086200 Steve Strand (snstrand(AT)comcast.net), Aug 28 2003 Search completed in 0.001 seconds