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A001998 Bending a piece of wire of length n+1; walks of length n+1 on a tetrahedron; also non-branched catafusenes with n+2 condensed hexagons.
(Formerly M1211 N0468)
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 (list; graph; refs; listen; history; text; internal format)



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 self-avoiding.

Also, it appears that a(n) gives the number of equivalence classes of n-tuples of 0, 1 and 2, where two n-tuples 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)=2-x). 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 n-tuples 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 5-tuple 11111. [Emeric Deutsch, Apr 22, 2013]


A. T. Balaban, Enumeration of Cyclic Graphs, pp. 63-105 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 cata-condensed polycyclic aromatic hydrocarbons, Tetrahedron 24 (1968), 2505-2516.

L. W. Beineke and R. E. Pippert, On the enumeration of planar trees of hexagons, Glasgow Math. J., 15 (1974), 131-147.

S. J. Cyvin et al., Unbranched catacondensed polygonal systems containing hexagons and tetragons, Croatica Chem. Acta, 69 (1996), 757-774.

S. J. Cyvin et al., Enumeration of tree-like octagonal systems: catapolyoctagons, ACH Models in Chem. 134 (1997), 55-70.

R. M. Foster, Solution to Problem E185, Amer. Math. Monthly, 44 (1937), 50-51.

R. C. Read, The Enumeration of Acyclic Chemical Compounds, pp. 25-61 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 algebraic-combinatoric study of the conformational properties of n-alkanes, J. Math. Chemistry, 25, 1999, 55-64 (see p. 60).


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


a(n) = if n mod 2 = 0 then ((3^((n-2)/2)+1)/2)^2 else 3^((n-3)/2)+(1/4)*(3^(n-2)+1).

G.f.: x*(1-3*x-2*x^2+8*x^3-3*x^4)/((1-x)*(1-3*x)*(1-3*x^2)).

a(0)=1, a(1)=2, a(2)=4, a(3)=10, a(n)=4*a(n-1)-12*a(n-3)+9*a(n-4). - Harvey P. Dale, Apr 10 2013


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


A001998 := proc(n) if n = 0 then 1 elif n mod 2 = 1 then (1/4)*(3^n+4*3^((n-1)/2)+1) else (1/4)*(3^n+2*3^(n/2)+1); fi; end;

A001998:=(-1+3*z+2*z**2-8*z**3+3*z**4)/(z-1)/(3*z-1)/(3*z**2-1); [Conjectured by Simon Plouffe in his 1992 dissertation. Gives sequence with an extra leading 1.]


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}] (* Jean-François Alcover, Jan 25 2013, from formula *)

LinearRecurrence[{4, 0, -12, 9}, {1, 2, 4, 10}, 30] (* Harvey P. Dale, Apr 10 2013 *)


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




N. J. A. Sloane.


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



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Last modified October 30 11:14 EDT 2014. Contains 248797 sequences.