OFFSET
0,2
COMMENTS
See A265762 for a guide to related sequences.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (13, 104, -260, -260, 104, 13, -1).
FORMULA
a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7) for n > 8.
G.f.: (1 - 8 x - 158 x^2 - 272 x^3 + 2134 x^4 + 2168 x^5 - 1009 x^6 - 130 x^7 + 10 x^8)/(1 - 13 x - 104 x^2 + 260 x^3 + 260 x^4 - 104 x^5 - 13 x^6 + x^7).
G.f.: (1 - 8*x - 158*x^2 - 272*x^3 + 2134*x^4 + 2168*x^5 - 1009*x^6 - 130*x^7 + 10*x^8)/((1 + x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)). - Andrew Howroyd, Mar 07 2018
EXAMPLE
Let u = 2^(1/3), and let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[u,1,1,1,...] has p(0,x) = -5 - 15 x - 6 x^2 - 9 x^3 + 3 x^5 + x^6, so that a(0) = 1.
[1,u,1,1,1,...] has p(1,x) = -11 + 45 x - 66 x^2 + 35 x^3 + 6 x^4 - 15 x^5 + 5 x^6, so that a(1) = 5;
[1,1,u,1,1,1...] has p(2,x) = 131 - 633 x + 1110 x^2 - 969 x^3 + 456 x^4 - 111 x^5 + 11 x^6, so that a(2) = 11.
MATHEMATICA
u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2^(1/3)}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 30}]
Coefficient[t, x, 0]; (* A267078 *)
Coefficient[t, x, 1]; (* A267079 *)
Coefficient[t, x, 2]; (* A267080 *)
Coefficient[t, x, 3]; (* A267081 *)
Coefficient[t, x, 4]; (* A267082 *)
Coefficient[t, x, 5]; (* A267083 *)
Coefficient[t, x, 6]; (* A266527 *)
PROG
(PARI) Vec((1 - 8*x - 158*x^2 - 272*x^3 + 2134*x^4 + 2168*x^5 - 1009*x^6 - 130*x^7 + 10*x^8)/((1 + x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)) + O(x^30)) \\ Andrew Howroyd, Mar 07 2018
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Clark Kimberling, Jan 11 2016
STATUS
approved