A230080 - OEIS

%I #16 May 19 2017 11:02:23

%S 0,3,5,23,73,265,920,3245,11385,40018,140574,493911,1735243,6096533,

%T 21419128,75252674,264387942,928884046,3263482673,11465714843,

%U 40282921096,141527481021,497233748352,1746949771565,6137623429414

%N Sequence needed for the nonpositive powers of rho(11) = 2*cos(Pi/11) in terms of the power basis of the degree 5 number field Q(rho(11)). Coefficients of the first power.

%C The formula for the nonpositive powers of rho(11) := 2*cos(Pi/11) (the length ratio (smallest diagonal/side) in the regular 11-gon), when written in the power basis of the degree 5 algebraic number field Q(rho(11)) is: 1/rho(11)^n = A038342(n)*1 + a(n)*rho(11) - A230081(n)*rho(11)^2 - A069006(n-1)*rho(11)^3 + A038342(n-1)*rho(11)^4, n >= 0, with A069006(-1) = 0 = A038342(-1).

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,3,-4,-1,1).

%F G.f.: x*(3 - 4*x - x^2 + x^3)/(1 - 3*x - 3*x^2 + 4*x^3 + x^4 - x^5).

%F a(n) = 3*a(n-1) +3*a(n-2) -4*a(n-3) -a(n-4) +a(n-5) for n >= 0, with a(-5)=-3, a(-4)=a(-3)=a(-2)=0, a(-1)=1.

%e 1/rho(11)^4 = 146*1 + 73*rho(11) - 173*rho(11)^2 - 29*rho(11)^3  + 41*rho(11)^4 (approximately 0.07374164519).

%t LinearRecurrence[{3,3,-4,-1,1},{0,3,5,23,73},30] (* _Harvey P. Dale_, May 19 2017 *)

%Y Cf. A038342, A069006, A230081.

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, Nov 04 2013