The OEIS is supported by the many generous donors to the OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 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. 3
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS See A265762 for a guide to related sequences. LINKS Colin Barker, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (2,2,-1). FORMULA G.f.: 2*x*(-5 + x))/(1 - 2*x - 2*x^2 + x^3). 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 EXAMPLE 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. MATHEMATICA 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 *) PROG (PARI) 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 (PARI) concat(0, Vec(-2*x*(5-x)/((1+x)*(1-3*x+x^2)) + O(x^30))) \\ Colin Barker, Sep 30 2016 CROSSREFS Cf. A192914, A265762, A266708. Sequence in context: A291931 A050576 A144376 * A186235 A241053 A068642 Adjacent sequences: A266705 A266706 A266707 * A266709 A266710 A266711 KEYWORD sign,easy AUTHOR Clark Kimberling, Jan 09 2016 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified August 3 05:44 EDT 2024. Contains 374875 sequences. (Running on oeis4.)