A121179 Related to enumeration of alkane systems - see reference for precise definition.

1, 1, 1, 4, 19, 91, 476, 2586, 14421, 82225, 476913, 2804880, 16689036, 100276894, 607588840, 3708251888, 22776251835, 140676848445, 873210347555, 5444307431052, 34080036632565, 214104150405915, 1349504948018208, 8531467913710560, 54083412667272300, 343715994386622918, 2189505804590364876

Offset: 0

Author

N. J. A. Sloane, Aug 15 2006

Keywords

• nonn

Comments

a(n) is the "number of all staggered conformers of alkyls containing n carbon atoms". It is related to sequence b(n) = A001764(n), which is the number of "space positions of conformers of alkyls related to another alkyl without C_3 symmetry" that contain n carbon atoms. The generating functions of the sequences (a(n): n >= 0) and (b(n): n >= 0), with a(0) = b(0) = 1, appear in some of the papers below. - Petros Hadjicostas, Jul 24 2019

Links

• S. J. Cyvin, Algebraic solution for the numbers of staggered conformers of alkanes, J. Math. Chem. 17 (1995), 291-293.
• S. J. Cyvin, J. Brunvoll, B. N. Cyvin, and W. Lüttke, Enumeration of the staggered conformers of alkanes, Zeitschrift für Naturforschung A 50(9) (1995), 857-863.
• S. J. Cyvin, B. N. Cyvin, J. Brunvoll, and Jianji Wang, Enumeration of staggered conformers of alkanes and monocyclic cycloalkanes, J. Molec. Struct. 445 (1998), 127-137.
• S. J. Cyvin, Jianji Wang, J. Brunvoll, Shiming Cao, Ying Li, B. N. Cyvin, and Yugang Wang, Staggered conformers of alkanes: complete solution of the enumeration problem, J. Molec. Struct. 413-414 (1997), 227-239.
• F. Hering et al., The enumeration of stack polytopes and simplicial clusters, Discrete Math., 40 (1982), 203-217.
• Jianji Wang, Shiming Cao, and Ying Li, An algebraic solution for the numbers of staggered conformers of alkanes, J. Math. Chem. 20 (1996), 211-212.

Crossrefs

Cf. A001764, A121190.

Programs

Maple

A121179 := proc(n)
    if n = 0 then
        return 1;
    elif modp(n,3) <> 1 then
        A001764(n) ;
    else
        A001764(n)+2*A001764((n-1)/3) ;
    end if;
    %/3 ;
end proc:
seq(A121179(n),n=0..30) ; # R. J. Mathar, Jul 31 2019

Mathematica

b[n_] := Binomial[3n, n]/(2n + 1);
a[n_] := If[n == 0, 1, If[Mod[n, 3] != 1, b[n], b[n] + 2 b[(n-1)/3]]/3];
a /@ Range[0, 30] (* Jean-François Alcover, Apr 01 2020 *)

Formula

From Petros Hadjicostas, Jul 24 2019: (Start)

We have a(0) = 1, while for n >= 1 we have

a(n) = (1/3) * A001764(n) = binomial(3*n, n)/(3*(2*n + 1)) if n !== 1 (mod 3), and

a(n) = (1/3) * A001764(n) + (2/3) * A001764((n-1)/3) if n == 1 (mod 3).

G.f.: 1 + (x/3) * (B(x)^3 + 2*B(x^3)), where B(x) is the g.f. of sequence A001764, which satisfies the functional equation B(x) = 1 + x*B^3(x). (It also satisfies the equation B(x) = 1/(1 - x*B^2(x)).) We have

B(x) = (2/sqrt(3*x)) * sin((1/3) * arcsin(sqrt(27*x/4))).

(End)

The g.f. for this sequence is F(3,t) on page 209 of the Hering link. - Robert A. Russell, May 12 2024

Extensions

More terms from Petros Hadjicostas, Jul 24 2019