A010373 Number of unrooted quartic trees with 2n (unlabeled) nodes and possessing a bicentroid; number of 2n-carbon alkanes C(2n)H(4n+2) with a bicentroid, ignoring stereoisomers.

1, 1, 3, 10, 36, 153, 780, 4005, 22366, 128778, 766941, 4674153, 29180980, 185117661, 1193918545, 7800816871, 51584238201, 344632209090, 2324190638055, 15804057614995, 108277583483391, 746878494484128, 5183852459907628

Offset

1,3

Comments

The degree of each node is <= 4.

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.

Ignoring stereoisomers means that the children of a node are unordered. They can be permuted in any way and it is still the same tree. See A086200 for the analogous sequence with stereoisomers counted.

References

F. Harary, Graph Theory, p. 36, for definition of bicentroid.

Links

Alois P. Heinz, Table of n, a(n) for n = 1..500 (terms n = 1..100 from Vincenzo Librandi)

A. Cayley, Über die analytischen Figuren, welche in der Mathematik Bäume genannt werden und ihre Anwendung auf die Theorie chemischer Verbindungen, Chem. Ber. 8 (1875), 1056-1059. (Annotated scanned copy)

E. M. Rains and N. J. A. Sloane, On Cayley's Enumeration of Alkanes (or 4-Valent Trees), J. Integer Sequences, Vol. 2 (1999), Article 99.1.1.

Index entries for sequences related to trees

Formula

a(n) = b(n)*(b(n)+1)/2, where b(n) = A000598[ n ].

Maple

M[1146] := [ T, {T=Union(Epsilon, U), U=Prod(Z, Set(U, card<=3))}, unlabeled ]:
bicenteredHC := proc(n) option remember; if n mod 2<>0 then 0 else binomial(count(M[ 1146 ], size=n/2)+1, 2) fi end:

Mathematica

m = 24; a[x_] = Sum[c[k]*x^k, {k, 0, m}]; s[x_] = Series[ 1 + (1/6)*x*(a[x]^3 + 3*a[x]*a[x^2] + 2*a[x^3]) - a[x], {x, 0, m}]; eq = Thread[ CoefficientList[s[x], x] == 0];
Do[so[k] = Solve[eq[[1]], c[k-1]][[1]]; eq = Rest[eq] /. so[k], {k, 1, m+1}]; b = Array[c, m, 0] /. Flatten[ Array[so, m+1] ]; Rest[b*(b+1)/2] (* Jean-François Alcover, Jul 25 2011, after A000598 *)

Crossrefs

A000602(n) = A010372(n) + a(n/2) for n even, A000602(n) = A010372(n) for n odd.

Cf. A000200, A000598.

Sequence in context: A081921 A353262 A165792 * A322726 A104603 A337341

Adjacent sequences: A010370 A010371 A010372 * A010374 A010375 A010376

Keyword

nonn,easy

Author

Paul Zimmermann, N. J. A. Sloane

Extensions

Description revised by Steve Strand (snstrand(AT)comcast.net), Aug 20 2003

Status

approved