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User: Hanzhang Fang

Hanzhang Fang has authored 1 sequences.

A356563 Sums of powers of roots of x^3 - 2*x^2 - x - 2.

3, 2, 6, 20, 50, 132, 354, 940, 2498, 6644, 17666, 46972, 124898, 332100, 883042, 2347980, 6243202, 16600468, 44140098, 117367068, 312075170, 829797604, 2206404514, 5866756972, 15599513666, 41478593332, 110290214274

Offset: 0

Greg Dresden and Hanzhang Fang, Aug 12 2022

The three roots of x^3 - 2*x^2 - x - 2 are c1=2.65896708... = A348909+1, c2=-0.32948354... + 0.80225455...*i, and c3=-0.32948354... - 0.80225455...*i.

a(n) can also be determined by Vieta's formulas and Newton's identities. For example, a(3) by definition is c1^3 + c2^3 + c3^3, and from Newton's identities this equals e1^3 - 3*e1*e2 + 3*e3 for e1, e2, e3 the elementary symmetric polynomials of x^3 - x^2 - x - 3. From Vieta's formulas we have e1 = 2, e2 = -1, and e3 = 2, giving us e1^3 - 3*e1*e2 + 3*e3 = 8 + 6 + 6 = 20, as expected.

			For n=3, a(3) = (2.65896708...)^3 + (-0.32948354... + 0.80225455...*i)^3 + (-0.32948354... - 0.80225455...*i)^3 = 20.

Index entries for linear recurrences with constant coefficients, signature (2,1,2).

Cf. A077996, A348909 (c1-1).

LinearRecurrence[{2, 1, 2}, {3, 2, 6}, 30]

a(n) = 2*a(n-1) + a(n-2) + 2*a(n-3).
G.f.: (3 - 4 x - x^2)/(1 - 2 x - x^2 - 2 x^3).
2*a(n) = 7*b(n) - b(n+1) for b(n) = A077996(n).

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User: Hanzhang Fang

A356563 Sums of powers of roots of x^3 - 2*x^2 - x - 2.

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