login
A266803
Coefficient of x^0 in the minimal polynomial of the continued fraction [1^n,sqrt(2)+sqrt(3),1,1,...], where 1^n means n ones.
9
49, 49, 25281, 606409, 37676521, 1596669889, 78061422609, 3612062087761, 170677159358209, 8000461380881641, 376169445225673929, 17666248458032362369, 830040053693500377841, 38992376127586237335409, 1831844657768331755159361, 86057114020320867143580169
OFFSET
0,1
COMMENTS
See A265762 for a guide to related sequences.
LINKS
Index entries for linear recurrences with constant coefficients, signature (34, 714, -4641, -12376, 12376, 4641, -714, -34, 1).
FORMULA
a(n) = 34*a(n-1) + 714*a(n-2) - 4641*a(n-3) - 12376*a(n-4) + 12376*a(n-5) + 4641*a(n-6) - 714*a(n-7) - 34*a(n-8) + a(n-9).
G.f.: (-49 + 1617 x + 11371 x^2 + 60722 x^3 + 158186 x^4 - 21270 x^5 + 1619 x^6 + 25 x^7 - x^8)/(-1 + 34 x + 714 x^2 - 4641 x^3 - 12376 x^4 + 12376 x^5 + 4641 x^6 - 714 x^7 - 34 x^8 + x^9).
EXAMPLE
Let u = sqrt(2) and v = sqrt(3), and let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[u+v,1,1,1,...] has p(0,x) = 49 - 168 x - 50 x^2 + 212 x^3 + 47 x^4 - 68 x^5 - 18 x^6 + 4 x^7 + x^8, so that a(0) = 49.
[1,u+v,1,1,1,...] has p(1,x) = 49 - 560 x + 2498 x^2 - 5760 x^3 + 7547 x^4 - 5760 x^5 + 2498 x^6 - 560 x^7 + 49 x^8, so that a(1) = 49;
[1,1,u+v,1,1,1...] has p(2,x) = 25281 - 101124 x + 173262 x^2 - 165852 x^3 + 96847 x^4 - 35252 x^5 + 7790 x^6 - 952 x^7 + 49 x^8, so that a(2) = 25281.
MATHEMATICA
u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[2] + Sqrt[3]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0]; (* A266803 *)
Coefficient[t, x, 1]; (* A266808 *)
Coefficient[t, x, 2]; (* A267061 *)
Coefficient[t, x, 3]; (* A267062 *)
Coefficient[t, x, 4]; (* A267063 *)
Coefficient[t, x, 5]; (* A267064 *)
Coefficient[t, x, 6]; (* A267065 *)
Coefficient[t, x, 7]; (* A267066 *)
Coefficient[t, x, 8]; (* A266803 *)
KEYWORD
sign,easy
AUTHOR
Clark Kimberling, Jan 10 2016
STATUS
approved