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a(n) = n^2*Fibonacci(n).
3

%I #31 Mar 06 2023 02:59:21

%S 0,1,4,18,48,125,288,637,1344,2754,5500,10769,20736,39377,73892,

%T 137250,252672,461533,837216,1509341,2706000,4827186,8572124,15159553,

%U 26707968,46890625,82061668,143188722,249163824,432466589,748836000,1293764509,2230588416,3838265442,6592537372,11303644625,19349736192

%N a(n) = n^2*Fibonacci(n).

%H Colin Barker, <a href="/A259451/b259451.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-5,0,3,1).

%F From _Colin Barker_, Jun 29 2015: (Start)

%F a(n) = 3*a(n-1) - 5*a(n-3) + 3*a(n-5) + a(n-6).

%F G.f.: -x*(x^4 - x^3 + 6*x^2 + x + 1)/(x^2 + x - 1)^3. (End)

%F E.g.f.: exp(x/2)*x*(sqrt(5)*(1 + x)*cosh(sqrt(5)*x/2) + (1 + 3*x)*sinh(sqrt(5)*x/2))/sqrt(5). - _Stefano Spezia_, Mar 04 2023

%p a:= n-> n^2*(<<1|1>, <1|0>>^n)[1, 2]:

%p seq(a(n), n=0..50); # _Alois P. Heinz_, Jun 30 2015

%t a[n_] := n^2 MatrixPower[{{1, 1}, {1, 0}}, n][[1, 2]]; Table[a[n], {n, 0, 50}] (* _Jean-François Alcover_, Feb 14 2016, after _Alois P. Heinz_ *)

%o (PARI) concat(0, Vec(-x*(x^4-x^3+6*x^2+x+1)/(x^2+x-1)^3 + O(x^100))) \\ _Colin Barker_, Jun 29 2015

%Y Cf. A000045, A000290, A169630.

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, Jun 27 2015