A266704 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,2/3,1,1,1,...], where 1^n means n ones.

-3, -19, 17, -75, -165, -463, -1181, -3123, -8145, -21355, -55877, -146319, -383037, -1002835, -2625425, -6873483, -17994981, -47111503, -123339485, -322906995, -845381457, -2213237419, -5794330757, -15169754895, -39714933885, -103975046803, -272210206481

Offset

0,1

Comments

See A265762 for a guide to related sequences.

Links

Colin Barker, Table of n, a(n) for n = 0..1000

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

Formula

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

G.f.: (-3 - 13 x + 61 x^2 - 74 x^3 - 68 x^4 + 34 x^5)/(1 - 2 x - 2 x^2 + x^3).

a(n) = 2^(-n)*(43*(-2)^n+2*(3-sqrt(5))^n*(-7+sqrt(5))-2*(3+sqrt(5))^n*(7+sqrt(5)))/5 for n>2. - Colin Barker, Sep 29 2016

Example

Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[2/3,1,1,1,...] = (1+3*sqrt(5))/6 has p(0,x) = -11 - 3 x + 9 x^2, so a(0) = 9;
[1,2/3,1,1,...] = (19+9*sqrt(5))/22 has p(1,x) = -1 - 19 x + 11 x^2, so a(1) = 11;
[1,1,2/3,1,...] = (-17+9*sqrt(5))/2 has p(2,x) = -29 + 17 x + x^2, so a(2) = 1.

Mathematica

u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2/3}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266703 *)
Coefficient[t, x, 1] (* A266704 *)
Coefficient[t, x, 2] (* A266703 *)

Prog

(PARI) Vec(-(3+13*x-61*x^2+74*x^3+68*x^4-34*x^5)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

Crossrefs

Cf. A265762, A266703.

Sequence in context: A358979 A178985 A357435 * A185446 A172032 A043073

Adjacent sequences: A266701 A266702 A266703 * A266705 A266706 A266707

Keyword

sign,easy

Author

Clark Kimberling, Jan 09 2016

Status

approved