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Sum of n-th powers of the roots of x^3 + 4*x^2 - 11*x - 1.
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%I #39 Dec 28 2022 15:21:36

%S 3,-4,38,-193,1186,-6829,40169,-234609,1373466,-8034394,47011093,

%T -275049240,1609284589,-9415668903,55089756851,-322322100748,

%U 1885860059450,-11033893589177,64557712909910,-377717821061137,2209972232664381,-12930227249420121

%N Sum of n-th powers of the roots of x^3 + 4*x^2 - 11*x - 1.

%C This is half of a two sided sequences.

%C The other half is A274664. - _Kai Wang_, Aug 02 2016

%C a(n) is x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of the polynomial x^3 + 4*x^2 - 11*x - 1.

%C x1 = (cos(Pi/7))^2/(cos(2*Pi/7)*cos(4*Pi/7)),

%C x2 = -(cos(2*Pi/7))^2/(cos(4*Pi/7)*cos(Pi/7)),

%C x3 = -(cos(4*Pi/7))^2/(cos(Pi/7) *cos(2*Pi/7)).

%H Colin Barker, <a href="/A274663/b274663.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 (-4,11,1).

%F a(n) = ((cos(Pi/7))^2/(cos(2*Pi/7)*cos(4*Pi/7)))^n + (-(cos(2*Pi/7))^2/(cos(4*Pi/7)*cos(Pi/7)))^n + (-(cos(4*Pi/7))^2/(cos(Pi/7)*cos(2*Pi/7)))^n.

%F a(n) = -4*a(n-1) + 11*a(n-2) + a(n-3) for n>2.

%F G.f.: (3+8*x-11*x^2)/(1+4*x-11*x^2-x^3). - _Wesley Ivan Hurt_, Jul 02 2016

%F a(n) = (-1/8)^(-n)*cos(Pi/7)^(3*n) + (-8)^n*sin(Pi/14)^(3*n) +

%F 8^n*sin(3*Pi/14)^(3*n). - _Wesley Ivan Hurt_, Jul 11 2016

%t RecurrenceTable[{a[0] == 3, a[1] == -4, a[2] == 38, a[n] == -4 a[n - 1] + 11 a[n - 2] + a[n - 3]}, a, {n, 0, 20}] (* _Michael De Vlieger_, Jul 02 2016 *)

%t LinearRecurrence[{-4,11,1},{3,-4,38},30] (* _Harvey P. Dale_, Dec 28 2022 *)

%o (PARI) polsym(x^3 + 4*x^2 - 11*x - 1, 21)

%o (PARI) Vec((3+8*x-11*x^2)/(1+4*x-11*x^2-x^3) + O(x^99)) \\ _Altug Alkan_, Jul 08 2016

%Y Cf. A215076, A248417, A274220, A274592, A274664.

%K sign,easy

%O 0,1

%A _Kai Wang_, Jul 01 2016