A107767 a(n) = (1 + 3^n - 2*3^(n/2))/4 if n is even, (1 + 3^n - 4*3^((n-1)/2))/4 if n odd.
0, 1, 4, 16, 52, 169, 520, 1600, 4840, 14641, 44044, 132496, 397852, 1194649, 3585040, 10758400, 32278480, 96845281, 290545684, 871666576, 2615029252, 7845176329, 23535617560, 70607118400, 211821620920, 635465659921
Offset: 1
References
- Balaban, A. T., Brunvoll, J., Cyvin, B. N., & Cyvin, S. J. (1988). Enumeration of branched catacondensed benzenoid hydrocarbons and their numbers of Kekulé structures. Tetrahedron, 44(1), 221-228. See Eq. 5.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- 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).
- Index entries for linear recurrences with constant coefficients, signature (4,0,-12,9).
Crossrefs
Programs
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GAP
a:=[];; for n in [1..30] do if n mod 2 <> 0 then Add(a,(1+3^n-4*3^((n-1)/2))/4); else Add(a,(1+3^n-2*3^(n/2))/4); fi; od; a; # Muniru A Asiru, Oct 30 2018
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Magma
I:=[0, 1, 4, 16]; [n le 4 select I[n] else 4*Self(n-1)-12*Self(n-3)+9*Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012
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Maple
a:=proc(n) if n mod 2 = 0 then (1+3^n-2*3^(n/2))/4 else (1+3^n-4*3^((n-1)/2))/4 fi end: seq(a(n),n=1..32);
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Mathematica
CoefficientList[Series[-x/((x-1)*(3*x-1)*(3*x^2-1)),{x,0,40}],x] (* or *) LinearRecurrence[{4,0,-12,9},{0,1,4,16},50] (* Vincenzo Librandi, Jun 26 2012 *) Ach[n_, k_] := Ach[n, k] = If[n<2, Boole[n==k && n>=0], k Ach[n-2,k] + Ach[n-2,k-1] + Ach[n-2,k-2]] (* A304972 *) k=3; Table[Sum[StirlingS2[n,j]-Ach[n,j],{j,k}]/2,{n,2,40}] (* Robert A. Russell, Oct 28 2018 *) CoefficientList[Series[(1/12 E^(-Sqrt[3] x) (-3 + 2 Sqrt[3] - (3 + 2 Sqrt[3]) E^(2 Sqrt[3] x) + 3 E^((3 + Sqrt[3]) x) + 3 E^(x + Sqrt[3] x)))/x, {x, 0, 20}], x]*Table[(k+1)!, {k, 0, 20}] (* Stefano Spezia, Oct 29 2018 *) -
PARI
x='x+O('x^50); concat(0, Vec(x^2/((1-x)*(3*x-1)*(3*x^2-1)))) \\ Altug Alkan, Sep 23 2018
Formula
G.f.: -x^2 / ( (x-1)*(3*x-1)*(3*x^2-1) ). - R. J. Mathar, Dec 16 2010
a(n) = 4*a(n-1) - 12*a(n-3) + 9*a(n-4). - Vincenzo Librandi, Jun 26 2012
From Robert A. Russell, Oct 28 2018: (Start)
a(n-1) = Sum_{j=0..k} (S2(n,j) - Ach(n,j)) / 2, where k=3 is the maximum number of colors, S2 is the Stirling subset number A008277, and Ach(n,k) = [n>=0 & n<2 & n==k] + [n>1]*(k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2)).
E.g.f.: (1/12)*exp(-sqrt(3)*x)*(-3 + 2*sqrt(3) - (3 + 2*sqrt(3))*exp(2*sqrt(3)*x) + 3*exp((3 + sqrt(3))*x) + 3*exp(x + sqrt(3)*x)). - Stefano Spezia, Oct 29 2018
From Bruno Berselli, Oct 31 2018: (Start)
a(n) = (1 + 3^n - 3^((n-1)/2)*(4 + (-2 + sqrt(3))*(1 + (-1)^n)))/4. Therefore:
a(2*k) = (3^k - 1)^2/4;
a(2*k+1) = (3^k - 1)*(3^(k+1) - 1)/4. (End)
Extensions
Entry revised by N. J. A. Sloane, Jul 29 2011
Comments