A092387 a(n) = Fibonacci(2*n+1) + Fibonacci(2*n-1) + 2.

5, 9, 20, 49, 125, 324, 845, 2209, 5780, 15129, 39605, 103684, 271445, 710649, 1860500, 4870849, 12752045, 33385284, 87403805, 228826129, 599074580, 1568397609, 4106118245, 10749957124, 28143753125, 73681302249, 192900153620

Offset: 1

Author

Benoit Cloitre, Mar 20 2004

Keywords

• nonn
• easy

Comments

Let b(k)=sum(i=1,k,F(2*n*i)*binomial(k,i)) where F(k) denotes the k-th Fibonacci number. The (b(k)) sequence satisfies the recursion: b(k)=a(n)*(b(k-1)-b(k-2)).

Same as the number of Kekulé structures in polyphenanthrene in terms of the number of hexagons. - Parthasarathy Nambi, Aug 22 2006

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• I. Lukovits and D. Janezic, Enumeration of conjugated circuits in nanotubes, J. Chem. Inf. Comput. Sci. 44 (2004), 410-414. See Table 1 column 3 on page 411.
• Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Crossrefs

Equals A065034(n)+1.

Programs

Magma

I:=[5, 9, 20]; [n le 3 select I[n] else 4*Self(n-1)-4*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, May 06 2012

Mathematica

CoefficientList[Series[-(5-11*x+4*x^2)/((x-1)(x^2-3*x+1)),  {x,0,30}],x] (* Vincenzo Librandi, May 06 2012 *)

Formula

a(1)=5, a(2)=9, a(3)=20, a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).

a(n) = 3 + floor((1+phi)^n) where phi = (1+sqrt(5))/2.

a(n) = A005248(n) + 2.

From R. J. Mathar, Mar 18 2009: (Start)

G.f.: -x*(5 - 11*x + 4*x^2)/((x-1)(x^2 - 3*x + 1)).

a(n+1) - a(n) = A002878(n). (End)

Extensions

Better definition from Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 20 2004