A265802 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,4,1,1,1,...], where 1^n means n ones.

1, 11, 19, 59, 145, 389, 1009, 2651, 6931, 18155, 47521, 124421, 325729, 852779, 2232595, 5845019, 15302449, 40062341, 104884561, 274591355, 718889491, 1882077131, 4927341889, 12899948549, 33772503745, 88417562699, 231480184339, 606022990331, 1586588786641

Offset: 0

Author

Clark Kimberling, Jan 04 2016

Keywords

• nonn
• easy

Comments

See A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[4,1,1,1,1,...] = (7 + sqrt(5))/2 has p(0,x) = 11 - 7 x + x^2, so a(0) = 1;
[1,4,1,1,1,...] = (29 - sqrt(5))/22 has p(1,x) = 19 - 29 x + 11 x^2, so a(1) = 11;
[1,1,4,1,1,...] = (67 + sqrt(5))/38 has p(2,x) = 59 - 67 x + 19 x^2, so a(2) = 19.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Crossrefs

Cf. A000032, A000045, A265762, A265803.

Programs

GAP

List([0..30], n-> 6*Fibonacci(n+1)^2 - 5*(-1)^n); # G. C. Greubel, Dec 11 2019

Magma

I:=[1,11,19,59]; [n le 4 select I[n] else 2*Self(n-1)+2*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jan 06 2016

Maple

with(combinat); seq(6*fibonacci(n+1)^2 - 5*(-1)^n, n=0..30); # G. C. Greubel, Dec 11 2019

Mathematica

u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {4}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A265802 *)
Coefficient[t, x, 1] (* A265803 *)
Coefficient[t, x, 2] (* A236802 *)
Join[{1}, LinearRecurrence[{2, 2, -1}, {11, 19, 59}, 30]] (* Vincenzo Librandi, Jan 06 2016 *)
Table[6*Fibonacci[n+1]^2 - 5*(-1)^n, {n,0,30}] (* G. C. Greubel, Dec 11 2019 *)

PARI

Vec((1+9*x-5*x^2)/(1-2*x-2*x^2+x^3) + O(x^30)) \\ Altug Alkan, Jan 04 2016

PARI

vector(31, n, 6*fibonacci(n)^2 + 5*(-1)^n) \\ G. C. Greubel, Dec 11 2019

Sage

[6*fibonacci(n+1)^2 - 5*(-1)^n for n in (0..30)] # G. C. Greubel, Dec 11 2019

Formula

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

G.f.: (1 + 9*x - 5*x^2)/(1 - 2*x - 2*x^2 + x^3).

a(n) = (2^(-n)*(-13*(-2)^n + 3*(3-sqrt(5))^(1+n) + 3*(3+sqrt(5))^(1+n)))/5. - Colin Barker, Oct 20 2016

From Klaus Purath, Oct 28 2019: (Start)

(a(n-3) - a(n-2) - a(n-1) + a(n))/6 = Fibonacci(2*n-1).

(a(n-5) + a(n))/30 = Fibonacci(2*n-3).

(a(n) - a(n-4))/18 = Fibonacci(2*n-2). (End)

E.g.f.: (1/5)*exp(-x)*(-13 + exp(-(1/2)*(-5 + sqrt(5))*x)*(9 - 3*sqrt(5) + 3*(3 + sqrt(5))*exp(sqrt(5)*x))). - Stefano Spezia, Dec 09 2019

a(n) = 6*Fibonacci(n+1)^2 - 5*(-1)^n = (6*Lucas(2*n+2) - 13*(-1)^n)/5. - G. C. Greubel, Dec 11 2019