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A265803
Coefficient of x in minimal polynomial of the continued fraction [1^n,4,1,1,1,...], where 1^n means n ones.
3
-7, -29, -67, -185, -475, -1253, -3271, -8573, -22435, -58745, -153787, -402629, -1054087, -2759645, -7224835, -18914873, -49519771, -129644453, -339413575, -888596285, -2326375267, -6090529529, -15945213307, -41745110405, -109290117895, -286125243293
OFFSET
0,1
COMMENTS
See A265762 for a guide to related sequences.
FORMULA
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (-1)*(7 + 15*x - 5*x^2)/(1 - 2*x - 2*x^2 + x^3).
a(n) = (2^(-n)*(13*(-2)^n + 12*(3-sqrt(5))^n*(-2+sqrt(5)) - 12*(2+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Oct 20 2016
a(n) = (13*(-1)^n - 12*Lucas(2*n+3))/5 = 5*(-1)^n - 12*F(n+1)*F(n+2), F=Fibonacci. - G. C. Greubel, Dec 12 2019
EXAMPLE
Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[4,1,1,1,1,...] = (7 + sqrt(5))/2 has p(0,x) = 11 - 7 x + x^2, so a(0) = 1;
[1,4,1,1,1,...] = (29 - sqrt(5))/22 has p(1,x) = 19 - 29 x + 11 x^2, so a(1) = 11;
[1,1,4,1,1,...] = (67 + sqrt(5))/38 has p(2,x) = 59 - 67 x + 19 x^2, so a(2) = 19.
MAPLE
with(combinat); f:=fibonacci; seq( 5*(-1)^n - 12*f(n+1)*f(n+2), n=0..30); # G. C. Greubel, Dec 12 2019
MATHEMATICA
u[n_]:= Table[1, {k, n}]; t[n_]:= Join[u[n], {4}, {{1}}];
f[n_]:= FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A265802 *)
Coefficient[t, x, 1] (* A265803 *)
Coefficient[t, x, 2] (* A236802 *)
LinearRecurrence[{2, 2, -1}, {-7, -29, -67}, 30] (* Vincenzo Librandi, Jan 06 2016 *)
PROG
(PARI) Vec((-7-15*x+5*x^2)/(1-2*x-2*x^2+x^3) + O(x^30)) \\ Altug Alkan, Jan 04 2016
(PARI) vector(31, n, f=fibonacci; -(5*(-1)^n + 12*f(n)*f(n+1)) ) \\ G. C. Greubel, Dec 12 2019
(Magma) I:=[-7, -29, -67]; [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 06 2016
(Sage) [(13*(-1)^n -12*lucas_number2(2*n+3, 1, -1))/5 for n in (0..30)] # G. C. Greubel, Dec 12 2019
(GAP) List([0..30], n-> (13*(-1)^n -12*Lucas(1, -1, 2*n+3)[2])/5 ); # G. C. Greubel, Dec 12 2019
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Clark Kimberling, Jan 04 2016
STATUS
approved