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

0,1

Comments

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.

Links

Table of n, a(n) for n=0..26.

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

Formula

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).

Example

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

Mathematica

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

Crossrefs

Cf. A077996, A348909 (c1-1).

Sequence in context: A268642 A110768 A140230 * A334588 A025240 A137602

Adjacent sequences: A356560 A356561 A356562 * A356564 A356565 A356566

Keyword

nonn,easy

Author

Greg Dresden and Hanzhang Fang, Aug 12 2022

Status

approved