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A266804 Coefficient of x^0 in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones. 5

%I #4 Jan 09 2016 19:59:19

%S 19,19,361,1795,14011,91489,638899,4348051,29883145,204609571,

%T 1402971259,9614651329,65903614291,451700107795,3096024736681,

%U 21220400800579,145446970016059,996907894114081,6832909585226995,46833455808339091,321001289959109449

%N Coefficient of x^0 in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones.

%C See A265762 for a guide to related sequences.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).

%F a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .

%F G.f.: (-19 + 76 x + 19 x^2 + 10 x^3 - x^4)/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).

%e Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:

%e [sqrt(6),1,1,1,...] has p(0,x)=19-14x-13x^2+2x^3+x^4, so a(0) = 19;

%e [1,sqrt(6),1,1,1,...] has p(1,x)=19-90x+143x^2-90x^3+19x^4, so a(1) = 19;

%e [1,1,sqrt(6),1,1,1...] has p(2,x)=361-722x+527x^2-166x^3+19x^4, so a(2) = 361.

%t u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[6]}, {{1}}];

%t f[n_] := FromContinuedFraction[t[n]];

%t t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];

%t Coefficient[t, x, 0] ; (* A266804 *)

%t Coefficient[t, x, 1]; (* A266805 *)

%t Coefficient[t, x, 2]; (* A266806 *)

%t Coefficient[t, x, 3]; (* A266807 *)

%t Coefficient[t, x, 4]; (* A266804 *)

%Y Cf. A265762, A266805, A266806, A266807.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Jan 09 2016

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Last modified April 25 01:06 EDT 2024. Contains 371964 sequences. (Running on oeis4.)