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%I #14 Jun 13 2015 00:52:39
%S 1,2,8,21,60,160,429,1134,2992,7865,20648,54144,141897,371722,973560,
%T 2549421,6675460,17478176,45761045,119808150,313668576,821205937,
%U 2149962768,5628704256,14736185425,38579909330,101003635304
%N a(n) = F(n) * (F(n+2)-1) = A000045(n) * A000071(n+2) = row sums of triangle A143211.
%C a(n)/a(n-1) tends to phi^2.
%C A143212(n) = Product of sum of first n Fibonacci numbers and Fibonacci number(n). - _Vladimir Joseph Stephan Orlovsky_, Oct 13 2009
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,1,-5,-1,1).
%F a(n) = F(n) * (F(n+2)-1) = A000045(n) * A000071(n+2) = row sums of triangle A143211.
%F From _R. J. Mathar_, Sep 06 2008: (Start)
%F G.f.: (1-x+x^2)/((1+x)(1-3x+x^2)(1-x-x^2)).
%F a(n) = -A000045(n+1) + 3*(-1)^n/5 + 7*A001906(n+1)/5 -3*A001906(n)/5. (End)
%F a(n) = F(n)*sum{k=0..n} F(k). - _Paul Barry_, Jan 05 2009
%e a(5) = 60 = F(5) * (F(7)-1) = 5*12.
%e a(5) = 60 = sum of row 5 terms of triangle A143211: (5 + 5 + 10 + 15 + 25).
%t Clear[lst,n,a,f]; a=0;lst={};Do[f=Fibonacci[n];a+=f;AppendTo[lst,a*Fibonacci[n]],{n,5!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Oct 13 2009 *)
%t Table[Fibonacci[n](Fibonacci[n+2]-1),{n,30}] (* _Harvey P. Dale_, Dec 14 2012 *)
%Y Cf. A000045, A000071, A143211.
%K nonn
%O 1,2
%A _Gary W. Adamson_, Jul 30 2008
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