A356411 - OEIS

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A356411

Sum of powers of roots of x^3 - x^2 - x - 3.

3, 1, 3, 13, 19, 41, 99, 197, 419, 913, 1923, 4093, 8755, 18617, 39651, 84533, 180035, 383521, 817155, 1740781, 3708499, 7900745, 16831587, 35857829, 76391651, 162744241, 346709379, 738628573, 1573570675, 3352327385, 7141783779

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OFFSET

0,1

COMMENTS

a(n) is the sum of the n-th powers of the three roots of x^3 - x^2 - x - 3. These roots are c1 = 2.130395..., c2 = -0.5651977... - i*1.0434274..., and c3 = -0.5651977... + i*1.0434274..., and so a(n) = c1^n + c2^n + c3^n. The real parts of c2 and c3 are A273065.

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 = 1, e2 = -1, and e3 = 3, giving us e1^3 - 3*e1*e2 + 3*e3 = 1 + 3 + 9 = 13, as expected.

LINKS

Table of n, a(n) for n=0..30.

Index entries for linear recurrences with constant coefficients, signature (1,1,3).

FORMULA

a(n) = a(n-1) + a(n-2) + 3*a(n-3) with a(0)=3, a(1)=1, a(2) = 3.

G.f.: (3 - 2*x - x^2)/(1 - x - x^2 - 3*x^3).

EXAMPLE

For n=3, a(3) = (2.130395...)^3 + (-0.5651977... - i*1.0434274...)^3 + (-0.5651977... + i*1.0434274...)^3 = 13.

MATHEMATICA

LinearRecurrence[{1, 1, 3}, {3, 1, 3}, 40]

PROG

(PARI) polsym(x^3 - x^2 - x - 3, 35) \\ Joerg Arndt, Aug 11 2022