A266707 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,tau,1,1,1,...], where 1^n means n ones and tau = golden ratio = (1 + sqrt(5))/2.

1, 5, 4, 19, 41, 116, 295, 781, 2036, 5339, 13969, 36580, 95759, 250709, 656356, 1718371, 4498745, 11777876, 30834871, 80726749, 211345364, 553309355, 1448582689, 3792438724, 9928733471, 25993761701, 68052551620, 178163893171, 466439127881, 1221153490484

Offset

0,2

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.: (1 + 3 x - 8 x^2 + 2 x^3)/(1 - 2 x - 2 x^2 + x^3).

a(n) = (2^(-1-n)*(-3*(-1)^n*2^(3+n)-(3-sqrt(5))^n*(-7+sqrt(5))+(3+sqrt(5))^n*(7+sqrt(5))))/5 for n>0. - Colin Barker, Sep 29 2016

Example

Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[tau,1,1,1,1,...] = sqrt(5) has p(0,x) = -5 + x^2, so a(0) = 1;
[1,tau,1,1,1,...] = (5 + sqrt(5))/5 has p(1,x) = 4 - 10 x + 5 x^2, so a(1) = 5;
[1,1,tau,1,1,...] = (9 - sqrt(5))/4 has p(2,x) = 19 - 18 x + 4 x^2, so a(2) = 4.

Mathematica

u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {GoldenRatio}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266707 *)
Coefficient[t, x, 1] (* A266708 *)
Coefficient[t, x, 2] (* A266707 *)

Prog

(PARI) Vec((1+3*x-8*x^2+2*x^3)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

Crossrefs

Cf. A265762, A266708.

Sequence in context: A133167 A133173 A307675 * A269695 A271065 A271598

Adjacent sequences: A266704 A266705 A266706 * A266708 A266709 A266710

Keyword

nonn,easy

Author

Clark Kimberling, Jan 09 2016

Status

approved