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%I #22 Nov 16 2024 11:57:38
%S -1,1,-9,31,311,1889,13599,91519,631721,4318271,29628279,202995649,
%T 1391561279,9537357311,65371447881,448058829919,3071050697399,
%U 21049268992289,144273903091551,988867867179391,6777801652728809,46455742430697599,318412398690263799
%N Coefficient of x^0 in the minimal polynomial of the continued fraction [1^n,sqrt(2),1,1,...], where 1^n means n ones.
%C See A265762 for a guide to related sequences.
%H G. C. Greubel, <a href="/A266710/b266710.txt">Table of n, a(n) for n = 0..1000</a>
%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.: (1 -6*x -x^2 -46*x^3 -301*x^4 +260*x^5 +92*x^6 -18*x^7)/(-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(2),1,1,1,...] has p(0,x) = -1 - 6 x - 5 x^2 + 2 x^3 + x^4, so a(0) = -1;
%e [1,sqrt(2),1,1,1,...] has p(1,x) = 1 + 2 x - 7 x^2 + 2 x^3 + x^4, so a(1) = 1;
%e [1,1,sqrt(2),1,1,1...] has p(2,x) = -9 + 18 x - 7 x^2 - 2 x^3 + x^4, so a(2) = -9.
%t u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[2]}, {{1}}];
%t f[n_] := FromContinuedFraction[t[n]];
%t t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
%t Coefficient[t, x, 0] ; (* A266710 *)
%t Coefficient[t, x, 1]; (* A266711 *)
%t Coefficient[t, x, 2]; (* A266712 *)
%t Coefficient[t, x, 3]; (* A266713 *)
%t Coefficient[t, x, 4]; (* A266710 *)
%t LinearRecurrence[{5,15,-15,-5,1},{-1,1,-9,31,311,1889,13599,91519},30] (* _Harvey P. Dale_, Jun 17 2016 *)
%o (PARI) x='x+O('x^30); Vec((1 -6*x -x^2 -46*x^3 -301*x^4 +260*x^5 +92*x^6 -18*x^7)/(-1 +5*x + 15*x^2 -15*x^3 -5*x^4 +x^5)) \\ _G. C. Greubel_, Jan 26 2018
%o (Magma) I:=[31,311,1889,13599,91519]; [-1,1,-9] cat [n le 5 select I[n] else 5*Self(n-1) +15*Self(n-2) -15*Self(n-3) -5*Self(n-4) + Self(n-5): n in [1..30]]; // _G. C. Greubel_, Jan 26 2018
%Y Cf. A265762, A266711, A266712, A266713.
%K sign,easy
%O 0,3
%A _Clark Kimberling_, Jan 09 2016