

A001998


Bending a piece of wire of length n+1; walks of length n+1 on a tetrahedron; also nonbranched catafusenes with n+2 condensed hexagons.
(Formerly M1211 N0468)


15



1, 2, 4, 10, 25, 70, 196, 574, 1681, 5002, 14884, 44530, 133225, 399310, 1196836, 3589414, 10764961, 32291602, 96864964, 290585050, 871725625, 2615147350, 7845353476, 23535971854, 70607649841, 211822683802, 635467254244, 1906400965570
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OFFSET

0,2


COMMENTS

The wire stays in the plane, there are n bends, each is R,L or O; turning the wire over does not count as a new figure.
Equivalently, walks of n+1 steps on a tetrahedron, visiting n+2 vertices, with n "corners"; the symmetry group is S4, reversing a walk does not count as different. Simply interpret R,L,O as instructions to turn R, turn L, or retrace the last step. Walks are not selfavoiding.
Also, it appears that a(n) gives the number of equivalence classes of ntuples of 0, 1 and 2, where two ntuples are equivalent if one can be obtained from the other by a sequence of operations R and C, where R denotes reversal and C denotes taking the 2's complement (C(x)=2x). This has been verified up to a(19)=290585050. Example: for n=3 there are ten equivalence classes {000, 222}, {001, 100, 122, 221}, {002, 022, 200, 220}, {010, 212}, {011, 110, 112, 211}, {012, 210}, {020, 202}, {021, 102, 120, 201}, {101, 121}, {111}, so a(3)=10. [From John W. Layman, Oct 13 2009]
There exists a bijection between chains of n+2 hexagons and the above described equivalence classes of ntuples of 0,1, and 2. Namely, for a given chain of n+2 hexagons we take the sequence of the numbers of vertices of degree 2 (0, 1, or 2) between the consecutive contact vertices on one side of the chain; switching to the other side we obtain the 2's complement of this sequence; reversing the order of the hexagons, we obtain the reverse sequence. The inverse mapping is straightforward. For example, to a linear chain of 7 hexagons there corresponds the 5tuple 11111. [Emeric Deutsch, Apr 22, 2013]


REFERENCES

A. T. Balaban, Enumeration of Cyclic Graphs, pp. 63105 of A. T. Balaban, ed., Chemical Applications of Graph Theory, Ac. Press, 1976; see p. 75.
A. T. Balaban and F. Harary, Chemical graphs V: enumeration and proposed nomenclature of benzenoid catacondensed polycyclic aromatic hydrocarbons, Tetrahedron 24 (1968), 25052516.
L. W. Beineke and R. E. Pippert, On the enumeration of planar trees of hexagons, Glasgow Math. J., 15 (1974), 131147.
S. J. Cyvin et al., Unbranched catacondensed polygonal systems containing hexagons and tetragons, Croatica Chem. Acta, 69 (1996), 757774.
S. J. Cyvin et al., Enumeration of treelike octagonal systems: catapolyoctagons, ACH Models in Chem. 134 (1997), 5570.
R. M. Foster, Solution to Problem E185, Amer. Math. Monthly, 44 (1937), 5051.
R. C. Read, The Enumeration of Acyclic Chemical Compounds, pp. 2561 of A. T. Balaban, ed., Chemical Applications of Graph Theory, Ac. Press, 1976. [I think this reference does not mention this sequence.  N. J. A. Sloane, Aug 10 2006]
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Gy. Tasi and F. Mizukami, Quantum algebraiccombinatoric study of the conformational properties of nalkanes, J. Math. Chemistry, 25, 1999, 5564 (see p. 60).


LINKS

T. D. Noe, Table of n, a(n) for n = 0..500
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec
Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
Index entries for sequences obtained by enumerating foldings


FORMULA

a(n) = if n mod 2 = 0 then ((3^((n2)/2)+1)/2)^2 else 3^((n3)/2)+(1/4)*(3^(n2)+1).
G.f.: x*(13*x2*x^2+8*x^33*x^4)/((1x)*(13*x)*(13*x^2)).
a(0)=1, a(1)=2, a(2)=4, a(3)=10, a(n)=4*a(n1)12*a(n3)+9*a(n4).  Harvey P. Dale, Apr 10 2013


EXAMPLE

There are 2 ways to bend a piece of wire of length 2 (bend it or not).


MAPLE

A001998 := proc(n) if n = 0 then 1 elif n mod 2 = 1 then (1/4)*(3^n+4*3^((n1)/2)+1) else (1/4)*(3^n+2*3^(n/2)+1); fi; end;
A001998:=(1+3*z+2*z**28*z**3+3*z**4)/(z1)/(3*z1)/(3*z**21); [Conjectured by Simon Plouffe in his 1992 dissertation. Gives sequence with an extra leading 1.]


MATHEMATICA

a[n_?OddQ] := (1/4)*(3^n + 4*3^((n  1)/2) + 1); a[n_?EvenQ] := (1/4)*(3^n + 2*3^(n/2) + 1); Table[a[n], {n, 0, 27}] (* JeanFrançois Alcover, Jan 25 2013, from formula *)
LinearRecurrence[{4, 0, 12, 9}, {1, 2, 4, 10}, 30] (* Harvey P. Dale, Apr 10 2013 *)


CROSSREFS

Cf. A036359, A002216, A005963, A000228, A001997, A001444, A038766.
Sequence in context: A027432 A032128 A052829 * A005817 A148093 A206289
Adjacent sequences: A001995 A001996 A001997 * A001999 A002000 A002001


KEYWORD

nonn,nice,easy


AUTHOR

N. J. A. Sloane.


EXTENSIONS

Offset and Maple code corrected by Colin Mallows, Nov 12 1999.


STATUS

approved



