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a(n) = F(n)^3 - F(n)^2, where F(n) is the n-th Fibonacci number (A000045).
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%I #36 Sep 08 2022 08:46:08

%S 0,0,0,4,18,100,448,2028,8820,38148,163350,697048,2965248,12595048,

%T 53440504,226608900,960530634,4070452764,17246835648,73069580980,

%U 309555981900,1311374255620,5555264316910,23532984885744,99688652356608,422291386890000

%N a(n) = F(n)^3 - F(n)^2, where F(n) is the n-th Fibonacci number (A000045).

%H Colin Barker, <a href="/A244309/b244309.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (5,2,-22,-4,14,-1,-1).

%F G.f.: 2*x^3*(x^2-x+2) / ((x+1)*(x^2-3*x+1)*(x^2-x-1)*(x^2+4*x-1)).

%F a(n) = A045991(A000045(n)). - _Michel Marcus_, Jun 25 2014

%F a(n) = (F(3*n) - 3*(-1)^n*F(n))/5 - (L(2*n) - 2*(-1)^n)/5, where F=A000045 and L=A000032. - _Ehren Metcalfe_, Mar 26 2016

%e a(4) is 18 because F(4)^3 - F(4)^2 = 3^3 - 3^2 = 18.

%t CoefficientList[Series[2 x^3 (x^2 - x + 2)/((x + 1) (x^2 - 3 x + 1) (x^2 - x - 1) (x^2 + 4 x - 1)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Jun 26 2014 *)

%t Table[#^3 - #^2 &@ Fibonacci@ n, {n, 0, 25}] (* _Michael De Vlieger_, Mar 27 2016 *)

%t LinearRecurrence[{5,2,-22,-4,14,-1,-1},{0,0,0,4,18,100,448},30] (* _Harvey P. Dale_, Aug 22 2020 *)

%o (PARI) vector(50, n, fibonacci(n-1)^3-fibonacci(n-1)^2)

%o (Magma) [Fibonacci(n)^3 - Fibonacci(n)^2: n in [0..30]]; // _Vincenzo Librandi_, Jun 26 2014

%Y Cf. A000045, A045991, A244310, A056570, A007598.

%K nonn,easy

%O 0,4

%A _Colin Barker_, Jun 25 2014