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A220673 Coefficients of formal series in powers of (tan(x))^2 for tan(5*x)/tan(x).

%I A220673 #29 May 23 2025 11:08:22
%S A220673 5,40,376,3560,33720,319400,3025400,28657000,271443000,2571145000,
%T A220673 24354235000,230686625000,2185095075000,20697517625000,
%U A220673 196049700875000,1857009420625000,17589845701875000,166613409915625000,1578184870646875000
%N A220673 Coefficients of formal series in powers of (tan(x))^2 for tan(5*x)/tan(x).
%C A220673 Formally Sum_{n>=0} a(n)*(tan(x))^(2*n) = tan(5*x)/ tan(x).
%C A220673 Convergence holds for x from the two open intervals (1-sqrt(6/5), 1-2/sqrt(5)) and (1+2/sqrt(5), 1+sqrt(6/5)), namely (-0.095445115, 0.105572809) and (1.894427191, 2.095445115) (10 digits).
%C A220673 These intervals follow from the denominator of the o.g.f. G(5,x) = (5 - 10*x + x^2)/(1 - 10*x + 5*x^2).
%C A220673 If one replaces x by (tan(x))^2 in this o.g.f. one obtains the formula for tan(5*x)/tan(x) in terms of (tan(x))^2. This formula is the special (n=5) solution of a general recurrence derivable from the addition theorem for tan(n*x) = tan(x + (n-1)*x), namely, with Q(n,x) := tan(n,x)/tan(x), Q(n,x) = (1 + Q(n-1,x))/(1 - v*Q(n-1,x)), where v = v(x) =(tan(x))^2, and the input is Q(1,x) = 1. Read as function of v the solution for Q(5,x) is just G(5,v) with replaced v=v(x).
%C A220673 See the irregular triangles A034867 and A034839 whose row polynomials N(n,x) and D(n,x), respectively, give for n >= 1 the solution to the recurrences N(n,x) = D(n-1,x) + N(n-1,x), D(n,x) = D(n-1,x) + x*N(n-1,x), with inputs N(1,x) = 1 and D(1,x) = 1. The proof by the Pascal triangle A007318 recurrence is trivial. Therefore, Q(n,x) from the preceding comment is given by Q(n,x) = N(n,-v)/D(n,-v) with v=v(x) = (tan(x))^2.
%C A220673 One also has, with Chebyshev's S polynomials (see A049310) Q(n,x) = tan(n*x)/tan(x) = (S(n,y) + S(n-2,y))/(S(n,y) - S(n-2,y)) = y*S(n-1,y)/(S(n,y) - S(n-2,y)) = 1/(1 - (2/y)*S(n-2,y)/S(n-1,y)), where y = y(x) = 2/sqrt(1 + (tan(x))^2). This derives from sin(n*x)/cos(n*x) in terms of Chebyshev S polynomials with argument 2*cos(x) = y(x). Note that S(n-2,y)/S(n-1,y) has the continued fraction representation 1/(y-1/(y- ... -1/(y )..(n-1)brackets..), i.e. (n-1) y's.
%C A220673 These calculations have been motivated by e-mails from Thomas Olsen.
%H A220673 G. C. Greubel, <a href="/A220673/b220673.txt">Table of n, a(n) for n = 0..1000</a>
%F A220673 O.g.f.: G(5,x) = (5 - 10*x + x^2)/(1 - 10*x + 5*x^2).
%F A220673 a(n) = delta(n,0)/5 - 8*b(n) + 24*b(n+1)/5, n>=0, with Kronecker's delta and b(n):= A190987(n).
%F A220673 E.g.f.: (1 + 8*exp(5*x)*(3*cosh(2*sqrt(5)*x) + sqrt(5)*sinh(2*sqrt(5)*x)))/5. - _Stefano Spezia_, May 23 2025
%e A220673 Q(5,x=0.1) = tan(0.5)/tan(0.1) = 5.444802663 (Maple 10 digits);
%e A220673 G(5,tan(0.1)^2) = 5.444802664;
%e A220673 Sum_{n>=0} a(n)*(tan(0.1))^(2*n) = 5.444802664.
%t A220673 CoefficientList[Series[(5-10*x+x^2)/(1-10*x+5*x^2), {x,0,50}], x] (* _G. C. Greubel_, Mar 06 2018 *)
%o A220673 (PARI) my(x='x+O('x^30)); Vec((5-10*x+x^2)/(1-10*x+5*x^2)) \\ _G. C. Greubel_, Mar 06 2018
%o A220673 (Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((5-10*x+x^2)/(1-10*x+5*x^2))); // _G. C. Greubel_, Mar 06 2018
%Y A220673 Cf. 2*A000012 (case n=2), A080923(n+1), n>=0 (case n=3), A077445(n+1), n>=0 (case n=4), A034867, A034839.
%K A220673 nonn,easy
%O A220673 0,1
%A A220673 _Wolfdieter Lang_, Jan 16 2013