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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.
5

%I #5 Jul 10 2011 18:22:50

%S 1,3,15,66,406,2775,19900,152076,1206681,9841266,82336528,702993756,

%T 6105180250,53822344278,480681790786,4342078862605,39621836138886,

%U 364831810979041,3386667673687950,31669036266203766

%N 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 The degree of each node is <= 4.

%C 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 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 <a href="/index/Ro#rooted">Index entries for sequences related to rooted trees</a>

%H <a href="/index/Tra#trees">Index entries for sequences related to trees</a>

%F 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 Cf. A000598, A000602, A010732, A010733, A000625, A000628, A086194.

%Y 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.

%K nonn

%O 1,2

%A Steve Strand (snstrand(AT)comcast.net), Aug 28 2003