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 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. 3
 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 (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 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 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) 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 CROSSREFS Cf. A265762, A266708. Sequence in context: A133167 A133173 A307675 * A269695 A271065 A271598 Adjacent sequences: A266704 A266705 A266706 * A266708 A266709 A266710 KEYWORD nonn,easy AUTHOR Clark Kimberling, Jan 09 2016 STATUS approved

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Last modified July 19 19:50 EDT 2024. Contains 374436 sequences. (Running on oeis4.)