%I #14 Dec 26 2017 03:23:44
%S 1,0,5,4,25,40,141,300,865,2064,5525,13780,35881,91000,234525,598524,
%T 1536625,3930720,10077221,25800100,66108985,169309384,433745325,
%U 1110982860,2845964161,7289895600,18673752245,47833334644,122528343625,313861682200,803975056701
%N Expansion of 1 / ((1+x)*(1-x-4*x^2)). (5,4)-Padovan sequence.
%C See A000931 (Padovan), and the W. Lang link given there.
%H Colin Barker, <a href="/A176738/b176738.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,5,4).
%F O.g.f.: 1/((1-x-4*x^2)*(1+x)) = ((3-4*x)/(1-x-4*x^2) -1/(1+x))/2.
%F a(n) = (3*b(n) - 4*b(n-1) - (-1)^n)/2, n>=0, with b(n):=A006131(n) ((1,4)-Fibonacci), b(-1):=0.
%F From _Colin Barker_, Dec 25 2017: (Start)
%F a(n) = ((-1)^(1+n) + (2^(-1-n)*((1+sqrt(17))^n*(-5+3*sqrt(17)) + (1-sqrt(17))^n*(5+3*sqrt(17)))) / sqrt(17)) / 2.
%F a(n) = 5*a(n-2) + 4*a(n-3) for n>2.
%F (End)
%t LinearRecurrence[{0,5,4},{1,0,5},40] (* _Harvey P. Dale_, May 27 2016 *)
%t f[n_] := Simplify[((-1)^(1 +n) + (2^(-1 -n)*((1 + Sqrt[17])^n*(-5 +3Sqrt[17]) + (1 -Sqrt[17])^n*(5 + 3Sqrt[17])))/Sqrt[17])/2]; Array[f, 31, 0] (* or *)
%t CoefficientList[Series[1/(1 -5x^2 -4x^3), {x, 0, 30}], x] (* or *)
%t RecurrenceTable[{a[n] == 5 a[n - 2] + 4 a[n - 3], a[0] == 1, a[1] == 0, a[2] == 5}, a, {n, 30}] (* _Robert G. Wilson v_, Dec 25 2017 *)
%o (PARI) Vec(1 / ((1 + x)*(1 - x - 4*x^2)) + O(x^40)) \\ _Colin Barker_, Dec 25 2017
%Y Cf. A176737 ((4,3)-Padovan).
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Jul 14 2010