A266707 -id:A266707 - 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

A266708 Coefficient of x 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

0, -10, -18, -56, -138, -370, -960, -2522, -6594, -17272, -45210, -118370, -309888, -811306, -2124018, -5560760, -14558250, -38114002, -99783744, -261237242, -683927970, -1790546680, -4687712058, -12272589506, -32130056448, -84117579850, -220222683090

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) = 0;
[1,tau,1,1,1,...] = (5 + sqrt(5))/5 has p(1,x) = 4 - 10*x + 5*x^2, so a(1) = -10;
[1,1,tau,1,1,...] = (9 - sqrt(5))/4 has p(2,x) = 19 - 18*x + 4*x^2, so a(2) = -18.

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

Cf. A192914, 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 *)

a(n) = round((2^(1-n)*(3*(-1)^n*2^(1+n)+(3-sqrt(5))^n*(-3+2*sqrt(5))-(3+sqrt(5))^n*(3+2*sqrt(5))))/5) \\ Colin Barker, Sep 30 2016

concat(0, Vec(-2*x*(5-x)/((1+x)*(1-3*x+x^2)) + O(x^30))) \\ Colin Barker, Sep 30 2016

G.f.: 2*x*(-5 + x)/((1 + x)*(1 - 3*x + x^2)).
a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
a(n) = -2*A192914(n+1).
a(n) = (2^(1-n)*(3*(-1)^n*2^(1+n)+(3-sqrt(5))^n*(-3+2*sqrt(5))-(3+sqrt(5))^n*(3+2*sqrt(5))))/5. - Colin Barker, Sep 30 2016

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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.

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A266708 Coefficient of x 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.

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