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

%I A092387 #21 Sep 08 2022 08:45:13
%S A092387 5,9,20,49,125,324,845,2209,5780,15129,39605,103684,271445,710649,
%T A092387 1860500,4870849,12752045,33385284,87403805,228826129,599074580,
%U A092387 1568397609,4106118245,10749957124,28143753125,73681302249,192900153620
%N A092387 a(n) = Fibonacci(2*n+1) + Fibonacci(2*n-1) + 2.
%C A092387 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)).
%C A092387 Same as the number of Kekulé structures in polyphenanthrene in terms of the number of hexagons. - _Parthasarathy Nambi_, Aug 22 2006
%H A092387 Vincenzo Librandi, <a href="/A092387/b092387.txt">Table of n, a(n) for n = 1..1000</a>
%H A092387 I. Lukovits and D. Janezic, <a href="http://dx.doi.org/10.1021/ci034240k">Enumeration of conjugated circuits in nanotubes</a>, J. Chem. Inf. Comput. Sci. 44 (2004), 410-414. See Table 1 column 3 on page 411.
%H A092387 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4,1).
%F A092387 a(1)=5, a(2)=9, a(3)=20, a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
%F A092387 a(n) = 3 + floor((1+phi)^n) where phi = (1+sqrt(5))/2.
%F A092387 a(n) = A005248(n) + 2.
%F A092387 From _R. J. Mathar_, Mar 18 2009: (Start)
%F A092387 G.f.: -x*(5 - 11*x + 4*x^2)/((x-1)(x^2 - 3*x + 1)).
%F A092387 a(n+1) - a(n) = A002878(n). (End)
%t A092387 CoefficientList[Series[-(5-11*x+4*x^2)/((x-1)(x^2-3*x+1)),  {x,0,30}],x] (* _Vincenzo Librandi_, May 06 2012 *)
%o A092387 (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
%Y A092387 Equals A065034(n)+1.
%K A092387 nonn,easy
%O A092387 1,1
%A A092387 _Benoit Cloitre_, Mar 20 2004
%E A092387 Better definition from Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 20 2004