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A099921 a(n) = 5*Fibonacci(n)^2.

%I A099921 #35 Sep 08 2022 08:45:15
%S A099921 5,5,20,45,125,320,845,2205,5780,15125,39605,103680,271445,710645,
%T A099921 1860500,4870845,12752045,33385280,87403805,228826125,599074580,
%U A099921 1568397605,4106118245,10749957120,28143753125,73681302245,192900153620,505019158605,1322157322205
%N A099921 a(n) = 5*Fibonacci(n)^2.
%D A099921 Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 53.
%H A099921 Matthew House, <a href="/A099921/b099921.txt">Table of n, a(n) for n = 1..2380</a>
%H A099921 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F A099921 a(n) = 5 * A007598(n).
%F A099921 a(n) = Lucas(n)^2 - 4(-1)^n.
%F A099921 G.f.: x*(5-5*x) / ((1+x)*(1-3*x+x^2)).
%F A099921 a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3) for n > 3. - _Matthew House_, Jan 13 2017
%F A099921 a(n) = ((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n - 2*(-1)^n. - _Colin Barker_, Jan 14 2017
%F A099921 a(n) = 2*Fibonacci(2*n+1) - Fibonacci(2*n) - 2*(-1)^n. - _Vincenzo Librandi_, Sep 14 2017
%F A099921 a(n) = Lucas(2*n) - 2*(-1)^n. - _Amiram Eldar_, Jan 17 2022
%p A099921 A099921:=n->5*combinat[fibonacci](n)^2: seq(A099921(n), n=1..50); # _Wesley Ivan Hurt_, Sep 16 2017
%t A099921 5*Fibonacci[Range[30]]^2 (* _Harvey P. Dale_, Feb 24 2015 *)
%t A099921 LinearRecurrence[{2, 2, -1}, {5, 5, 20}, 30] (* _Vincenzo Librandi_, Sep 14 2017 *)
%o A099921 (PARI) a(n) = 5*fibonacci(n)^2; \\ _Michel Marcus_, Jan 14 2017
%o A099921 (PARI) Vec(x*(5-5*x) / ((1+x)*(1-3*x+x^2)) + O(x^40)) \\ _Colin Barker_, Jan 14 2017
%o A099921 (Magma) [5*Fibonacci(n)^2: n in [1..30]]; // _Vincenzo Librandi_, Sep 14 2017
%Y A099921 Cf. A000032, A000045 (Fibonacci numbers), A007598.
%K A099921 nonn,easy
%O A099921 1,1
%A A099921 _Ralf Stephan_, Nov 01 2004