A266708 -id:A266708 - cp's OEIS frontend

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

-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

Clark Kimberling, Jan 04 2016

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.

			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.

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

Cf. A236428, A265802.

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

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 *)

Vec((-3+x+x^2)/(1-2*x-2*x^2+x^3) + O(x^100)) \\ Altug Alkan, Jan 04 2016

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

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.

Original entry on oeis.org

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

Clark Kimberling, Jan 09 2016

See A265762 for a guide to related sequences.

			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.

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

Cf. A265762, A266708.

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 *)

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

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

cp's OEIS Frontend

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

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