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A103433 a(n) = Sum_{i=1..n} Fibonacci(2i-1)^2.

%I A103433 #43 Sep 08 2022 08:45:16
%S A103433 0,1,5,30,199,1355,9276,63565,435665,2986074,20466835,140281751,
%T A103433 961505400,6590256025,45170286749,309601751190,2122041971551,
%U A103433 14544692049635,99690802375860,683290924581349,4683345669693545
%N A103433 a(n) = Sum_{i=1..n} Fibonacci(2i-1)^2.
%D A103433 A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 234.
%H A103433 Belgacem Bouras, <a href="http://www.emis.de/journals/JIS/VOL16/Bouras/bouras4.html">A New Characterization of Catalan Numbers Related to Hankel Transforms and Fibonacci Numbers</a>, Journal of Integer Sequences, 16 (2013), #13.3.3.
%H A103433 M. Dougherty, C. French, B. Saderholm, W. Qian,, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/French/french2.html">Hankel Transforms of Linear Combinations of Catalan Numbers</a>, J. Int. Seq. 14 (2011) # 11.5.1.
%H A103433 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (9,-16,9,-1).
%F A103433 G.f.: x*(1-4*x+x^2) / ((1-7*x+x^2)(1-x)^2).
%F A103433 a(n) = (1/5)*(Fibonacci(4n) + 2n).
%F A103433 a(n) = (floor(5*n*phi) + 4*Fibonacci(4*n))/20, where phi =(1+sqrt(5))/2. - _Gary Detlefs_, Mar 10 2011
%F A103433 a(n) = (8*(n+2)*(Sum_{k=1..n} 1/(2*k^2 + 6*k + 4)) + Fibonacci(4*n))/5. - _Gary Detlefs_, Dec 07 2011
%F A103433 a(n) = | Sum_{i=0..2n-1} (-1)^i*F(i)*F(i+1) |, where F(n) = Fibonacci numbers (A000045). - _Rigoberto Florez_, May 04 2019
%t A103433 Table[(Fibonacci[4n]+2n)/5, {n,0,20}] (* _Rigoberto Florez_, May 04 2019 *)
%o A103433 (Magma) [(1/5)*(Fibonacci(4*n)+2*n): n in [0..50]]; // _Vincenzo Librandi_, Apr 20 2011
%o A103433 (PARI) a(n)=(fibonacci(4*n)+2*n)/5 \\ _Charles R Greathouse IV_, Oct 07 2015
%Y A103433 Partial sums of A081068. Bisection of A077916.
%K A103433 nonn,easy
%O A103433 0,3
%A A103433 _Ralf Stephan_, Feb 08 2005