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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A265762 Coefficient of x in minimal polynomial of the continued fraction [1^n,2,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

-3, -5, -15, -37, -99, -257, -675, -1765, -4623, -12101, -31683, -82945, -217155, -568517, -1488399, -3896677, -10201635, -26708225, -69923043, -183060901, -479259663, -1254718085, -3284894595, -8599965697, -22515002499, -58945041797, -154320122895
Offset: 0

Views

Author

Clark Kimberling, Jan 04 2016

Keywords

Comments

In the following guide to related sequences, d(n), e(n), f(n) represent the coefficients in the minimal polynomial written as d(n)*x^2 + e(n)*x + f(n), except, in some cases, for initial terms. All of these sequences (eventually) satisfy the linear recurrence relation a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
continued fractions d(n) e(n) f(n)
[1^n,2,1,1,1,...] A236428 A265762 A236428
[1^n,3,1,1,1,...] A236428 A265762 A236428
[1^n,4,1,1,1,...] A265802 A265803 A265802
[1^n,5,1,1,1,...] A265804 A265805 A265804
[1^n,1/2,1,1,1,...] A266699 A266700 A266699
[1^n,1/3,1,1,1,...] A266701 A266702 A266701
[1^n,2/3,1,1,1,...] A266703 A266704 A266703
[1^n,sqrt(5),1,1,1,...] A266705 A266706 A266705
[1^n,tau,1,1,1,...] A266707 A266708 A266707
[2,1^n,2,1,1,1,...] A236428 A266709 A236428
The following forms of continued fractions have minimal polynomials of degree 4 and, after initial terms, satisfy the following linear recurrence relation:
a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5).
[1^n,sqrt(2),1,1,1,...]: A266710, A266711, A266712, A266713, A266710
[1^n,sqrt(3),1,1,1,...]: A266799, A266800, A266801, A266802, A266799
[1^n,sqrt(6),1,1,1,...]: A266804, A266805, A266806, A266807, A277804
Continued fractions [1^n,2^(1/3),1,1,1,...] have minimal polynomials of degree 6. The coefficient sequences are linearly recurrenct with signature {13, 104, -260, -260, 104, 13, -1, 0, 0}; see A267078-A267083.
Continued fractions [1^n,sqrt(2)+sqrt(3),1,1,1,...] have minimal polynomials of degree 8. The coefficient sequences are linearly recurrenct with signature {13, 104, -260, -260, 104, 13, -1}; see A266803, A266808, A267061-A267066.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[2,1,1,1,1,...] = (3 + sqrt(5))/2 has p(0,x) = x^2 - 3x + 1, so a(0) = -3;
[1,2,1,1,1,...] = (5 - sqrt(5))/2 has p(1,x) = x^2 - 5x + 5, so a(1) = -5;
[1,1,2,1,1,...] = (15 + sqrt(5))/10 has p(2,x) = 5x^2 - 15x + 11, so a(2) = -15.

Links

Crossrefs

Cf. A236428, A265802.

Programs

  • Magma
    I:=[-3,-5,-15]; [n le 3 select I[n] else 2*Self(n-1)+2*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jan 05 2016
  • Mathematica
    Program 1:
u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A236428 *)
Coefficient[t, x, 1] (* A265762 *)
Coefficient[t, x, 2] (* A236428 *)
Program 2:
LinearRecurrence[{2, 2, -1}, {-3, -5, -15}, 50] (* Vincenzo Librandi, Jan 05 2016 *)
  • PARI
    Vec((-3+x+x^2)/(1-2*x-2*x^2+x^3) + O(x^100)) \\ Altug Alkan, Jan 04 2016

Formula

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (-3 + x + x^2)/(1 - 2 x - 2 x^2 + x^3).
a(n) = (-1)*(2^(-n)*(3*(-2)^n+2*((3-sqrt(5))^(1+n)+(3+sqrt(5))^(1+n))))/5. - Colin Barker, Sep 27 2016

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

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jan 04 2016

Keywords

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

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
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