%I #49 Oct 13 2017 05:06:25

%S 1,1,2,8,56,536,6336,89984,1502080,28816000,624183040,15068741632,

%T 401279717376,11686228873216,369481496674304,12604143513731072,

%U 461450753988067328,18047560103055097856,750979001470906073088,33127742823550018125824,1544257684260700601712640

%N E.g.f. satisfies: A(x) = 1 + tan(x*A(x)).

%C Radius of convergence of e.g.f. A(x) is r = 2*atan(Pi/2-1)/Pi = 0.3301951... where A(r) = Pi/2.

%C Radius of convergence of e.g.f. A(x) is not 2*atan(Pi/2-1)/Pi, but r = 0.39828426936099395926..., where r is the root of the equation cos(r + sqrt(r*(1-r))) = sqrt(r), and A(r) = 1 + sqrt((1-r)/r) = 2.22913364222177489... - _Vaclav Kotesovec_, Jan 11 2014

%H Vaclav Kotesovec, <a href="/A201128/b201128.txt">Table of n, a(n) for n = 0..380</a>

%H Vaclav Kotesovec, <a href="/A201128/a201128.jpg">Graph - asymptotics for 1000 terms</a>

%F a(n) = [x^n/n!] (1 + tan(x))^(n+1)/(n+1).

%F E.g.f. satisfies:

%F (1) A(x/(1 + tan(x))) = 1 + tan(x).

%F (2) A(x) = (1/x)*Series_Reversion(x/(1 + tan(x))).

%F (3) A(x) = 1 + Sum_{n>=1} 8*x*A(x) / ((2*n-1)^2*Pi^2 - 4*x^2*A(x)^2).

%F (4) A(x) = 1 + x*A(x)/(1 - x^2*A(x)^2/(3 - x^2*A(x)^2/(5 - x^2*A(x)^2/(7 - x^2*A(x)^2/(9 - ...))))), a recursive continued fraction.

%F a(n) = (Sum_{k=0..n/2} binomial(n+1, 2*k+1)*(Sum_{j=0..2*k} binomial(j+n-2*k-1, n-2*k-1)*(j+n-2*k)!*2^(-j+2*k)*(-1)^(k+j)*Stirling2(n, j+n-2*k)))/(n+1), n > 0, a(0)=1. - _Vladimir Kruchinin_, Feb 03 2012

%F a(n) ~ n^(n-1) * sqrt((1 + 1/sqrt(1/r-1))/(2*r))/ (exp(n) * r^n), where r = 0.39828426936099395926... is the root of the equation cos(r + sqrt(r*(1-r))) = sqrt(r). - _Vaclav Kotesovec_, Jan 11 2014

%e E.g.f.: A(x) = 1 + x + 2*x^2/2! + 8*x^3/3! + 56*x^4/4! + 536*x^5/5! + ...

%e where A(x) = 1 + 8*x*A(x)/(Pi^2 - 4*x^2*A(x)^2) + 8*x*A(x)/(9*Pi^2 - 4*x^2*A(x)^2) + 8*x*A(x)/(25*Pi^2 - 4*x^2*A(x)^2) + 8*x*A(x)/(49*Pi^2 - 4*x^2*A(x)^2) + ...

%e The coefficients of x^n/n! in initial powers of G(x) = 1 + tan(x) begin:

%e G^1: [(1), 1, 0, 2, 0, 16, 0, 272, 0, 7936, ...];

%e G^2: [1,(2), 2, 4, 16, 32, 272, 544, 7936, ...];

%e G^3: [1, 3, (6), 12, 48, 168, 816, 4512, 23808, ...];

%e G^4: [1, 4, 12,(32), 120, 544, 2592, 15872, 96000, ...];

%e G^5: [1, 5, 20, 70, (280), 1400, 7520, 46720, 321280, ...];

%e G^6: [1, 6, 30, 132, 600, (3216), 19200, 125952, 925440, ...];

%e G^7: [1, 7, 42, 224, 1176, 6832, (44352), 312704, 2424576, ...];

%e G^8: [1, 8, 56, 352, 2128, 13568, 94976, (719872), 5907328, ...];

%e G^9: [1, 9, 72, 522, 3600, 25344, 191232, 1552752, (13518720), ...]; ...

%e where coefficients in parenthesis form initial terms of this sequence:

%e [1/1, 2/2, 6/3, 32/4, 280/5, 3216/6, 44352/7, 719872/8, 13518720/9, ...].

%t CoefficientList[1/x*InverseSeries[Series[x/(1 + Tan[x]), {x, 0, 21}], x],x] * Range[0, 20]! (* _Vaclav Kotesovec_, Jan 11 2014 *)

%t Table[Sum[Binomial[n+1,2*k+1]*Sum[Binomial[j+n-2*k-1,n-2*k-1]*(j+n-2*k)!*2^(-j+2*k)*(-1)^(k+j)*StirlingS2[n,j+n-2*k]/(n+1),{j,0,2*k}],{k,0,n/2}],{n,0,20}] (* _Vaclav Kotesovec_ after _Vladimir Kruchinin_, Jan 11 2014 *)

%o (PARI) {a(n)=n!*polcoeff(1/x*serreverse(x/(1+tan(x+x^2*O(x^n)))),n)}

%o (Maxima) a(n):=if n=0 then 1 else sum(binomial(n+1,2*k+1)*sum(binomial(j+n-2*k-1,n-2*k-1)*(j+n-2*k)!*2^(-j+2*k)*(-1)^(k+j)*stirling2(n,j+n-2*k),j,0,2*k),k,0,(n)/2)/(n+1); /* _Vladimir Kruchinin_, Feb 03 2012 */

%Y Cf. A009006.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Dec 01 2011