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.
%I A265762 #24 Sep 08 2022 08:46:15
%S A265762 -3,-5,-15,-37,-99,-257,-675,-1765,-4623,-12101,-31683,-82945,-217155,
%T A265762 -568517,-1488399,-3896677,-10201635,-26708225,-69923043,-183060901,
%U A265762 -479259663,-1254718085,-3284894595,-8599965697,-22515002499,-58945041797,-154320122895
%N A265762 Coefficient of x in minimal polynomial of the continued fraction [1^n,2,1,1,1,...], where 1^n means n ones.
%C A265762 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).
%C A265762 continued fractions d(n) e(n) f(n)
%C A265762 [1^n,2,1,1,1,...] A236428 A265762 A236428
%C A265762 [1^n,3,1,1,1,...] A236428 A265762 A236428
%C A265762 [1^n,4,1,1,1,...] A265802 A265803 A265802
%C A265762 [1^n,5,1,1,1,...] A265804 A265805 A265804
%C A265762 [1^n,1/2,1,1,1,...] A266699 A266700 A266699
%C A265762 [1^n,1/3,1,1,1,...] A266701 A266702 A266701
%C A265762 [1^n,2/3,1,1,1,...] A266703 A266704 A266703
%C A265762 [1^n,sqrt(5),1,1,1,...] A266705 A266706 A266705
%C A265762 [1^n,tau,1,1,1,...] A266707 A266708 A266707
%C A265762 [2,1^n,2,1,1,1,...] A236428 A266709 A236428
%C A265762 The following forms of continued fractions have minimal polynomials of degree 4 and, after initial terms, satisfy the following linear recurrence relation:
%C A265762 a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5).
%C A265762 [1^n,sqrt(2),1,1,1,...]: A266710, A266711, A266712, A266713, A266710
%C A265762 [1^n,sqrt(3),1,1,1,...]: A266799, A266800, A266801, A266802, A266799
%C A265762 [1^n,sqrt(6),1,1,1,...]: A266804, A266805, A266806, A266807, A277804
%C A265762 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.
%C A265762 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.
%H A265762 Colin Barker, <a href="/A265762/b265762.txt">Table of n, a(n) for n = 0..1000</a>
%H A265762 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F A265762 a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
%F A265762 G.f.: (-3 + x + x^2)/(1 - 2 x - 2 x^2 + x^3).
%F A265762 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
%e A265762 Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
%e A265762 [2,1,1,1,1,...] = (3 + sqrt(5))/2 has p(0,x) = x^2 - 3x + 1, so a(0) = -3;
%e A265762 [1,2,1,1,1,...] = (5 - sqrt(5))/2 has p(1,x) = x^2 - 5x + 5, so a(1) = -5;
%e A265762 [1,1,2,1,1,...] = (15 + sqrt(5))/10 has p(2,x) = 5x^2 - 15x + 11, so a(2) = -15.
%t A265762 Program 1:
%t A265762 u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2}, {{1}}];
%t A265762 f[n_] := FromContinuedFraction[t[n]];
%t A265762 t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
%t A265762 Coefficient[t, x, 0] (* A236428 *)
%t A265762 Coefficient[t, x, 1] (* A265762 *)
%t A265762 Coefficient[t, x, 2] (* A236428 *)
%t A265762 Program 2:
%t A265762 LinearRecurrence[{2, 2, -1}, {-3, -5, -15}, 50] (* _Vincenzo Librandi_, Jan 05 2016 *)
%o A265762 (PARI) Vec((-3+x+x^2)/(1-2*x-2*x^2+x^3) + O(x^100)) \\ _Altug Alkan_, Jan 04 2016
%o A265762 (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
%Y A265762 Cf. A236428, A265802.
%K A265762 sign,easy
%O A265762 0,1
%A A265762 _Clark Kimberling_, Jan 04 2016