This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A156085 #18 Aug 16 2017 13:36:29
%S A156085 0,1,17,306,5490,98515,1767779,31721508,569219364,10214227045,
%T A156085 183286867445,3288949386966,59017802097942,1059031488375991,
%U A156085 19003548988669895,341004850307682120,6119083756549608264,109802502767585266633,1970325966059985191129
%N A156085 One fourth of the sum of the squares of the first n Fibonacci numbers with index divisible by 3.
%C A156085 Natural bilateral extension (brackets mark index 0): ..., -5490, -306, -17, -1, 0, [0], 1, 17, 306, 5490, 98515, ... This is (-A156085)-reversed followed by A156085. That is, A156085(-n) = -A156085(n-1).
%H A156085 Harvey P. Dale, <a href="/A156085/b156085.txt">Table of n, a(n) for n = 0..798</a>
%H A156085 C. Pita, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Pita/pita12.html">On s-Fibonomials</a>, J. Int. Seq. 14 (2011) # 11.3.7
%H A156085 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (17, 17, -1).
%F A156085 Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
%F A156085 a(n) = (1/4) sum_{k=1..n} F(3k)^2.
%F A156085 Closed form: a(n) = L(6n+3)/80 - (-1)^n/20.
%F A156085 Factored closed form: a(n) = (1/16) F(n) F(n+1) (L(n) - 1)(L(n) + 1)(L(2n+2) - 1) if n is even; a(n) = (1/16) F(n) F(n+1) (L(n+1) - 1)(L(n+1) + 1)(L(2n) - 1) if n is odd.
%F A156085 Recurrence: a(n) - 17 a(n-1) - 17 a(n-2) + a(n-3) = 0.
%F A156085 G.f.: A(x) = x/(1 - 17 x - 17 x^2 + x^3) = x/((1 + x)(1 - 18 x + x^2)).
%F A156085 a(n) = ((9+4*sqrt(5))^(-n)*(2-sqrt(5)-4*(-9-4*sqrt(5))^n+(2+sqrt(5))*(9+4*sqrt(5))^(2*n)))/80. - _Colin Barker_, Mar 04 2016
%t A156085 a[n_Integer] := If[ n >= 0, Sum[ (1/4) Fibonacci[3k]^2, {k, 1, n} ], -Sum[ (1/4) Fibonacci[-3k]^2, {k, 1, -n - 1} ] ]
%t A156085 Accumulate[Fibonacci[3*Range[0,20]]^2]/4 (* or *) LinearRecurrence[{17,17,-1},{0,1,17},30] (* _Harvey P. Dale_, Aug 17 2014 *)
%o A156085 (PARI) concat(0, Vec(x/((1+x)*(1-18*x+x^2)) + O(x^25))) \\ _Colin Barker_, Mar 04 2016
%o A156085 (PARI) a(n) = sum(k=1, n, fibonacci(3*k)^2)/4; \\ _Michel Marcus_, Mar 04 2016
%Y A156085 Cf. A156084, A156090, A156091.
%K A156085 nonn,easy
%O A156085 0,3
%A A156085 _Stuart Clary_, Feb 04 2009