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%I #34 Jan 14 2016 17:30:48
%S 1,1,17,961,116129,24419521,7935695921,3689986634881,2325980044877633,
%T 1909820766285110401,1980859561990210210385,2532925102150057552027201,
%U 3914882960190010585911806177,7195136206495188167173611850561,15509756436233434005672607162123889
%N E.g.f. A(x) satisfies: A(x*cos(x)) = 1/cos(x).
%H Robert Israel, <a href="/A196873/b196873.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) = (2*n)! * [x^(2*n)] (1/cos(x)^(2*n+1)) / (2*n+1).
%F a(n) = A185142(n) / (2*n+1).
%F E.g.f. satisfies:
%F (1) A(x) = 1/cos(x*A(x)).
%F (2) A(x) = Sum_{n>=0} A000364(n)*(x*A(x))^(2*n)/(2*n)!, where A000364(n) = E_{2*n} forms the Euler numbers.
%F (3) A(x) = (1/x)*Series_Reversion(x*cos(x)).
%F (4) A(x) = Sum_{n>=1} (-1)^(n-1)*4*(2*n-1)*Pi / ((2*n-1)^2*Pi^2 - 4*x^2*A(x)^2).
%F a(n) ~ s*sqrt(2/(1+r^2*s^2*(3*s^2-1))) * 2^(2*n) * n^(2*n-1) / (exp(2*n) * r^(2*n)), where s = 1.533308151311528971... is the root of the equation s*cos(1/sqrt(s^2-1)) = 1, and r = 1/(s*sqrt(s^2-1)) = 0.561096338191045067... - _Vaclav Kotesovec_, Jan 23 2014
%e E.g.f.: A(x) = 1 + x^2/2! + 17*x^4/4! + 961*x^6/6! + 116129*x^8/8! + ...
%e where A(x) = 4*Pi/(Pi^2 - 4*x^2*A(x)^2) - 4*3*Pi/(9*Pi^2 - 4*x^2*A(x)^2) + 4*5*Pi/(25*Pi^2 - 4*x^2*A(x)^2) - 4*7*Pi/(49*Pi^2 - 4*x^2*A(x)^2) +- ...
%e E.g.f. A(x) is related to the Euler numbers (A000364) by the series:
%e A(x*cos(x)) = 1/cos(x) = 1 + x^2/2! + 5*x^4/4! + 61*x^6/6! + 1385*x^8/8! + ... + A000364(n)*x^(2*n)/(2*n)! + ...
%e A(x) = 1 + x^2*A(x)^2/2! + 5*x^4*A(x)^4/4! + 61*x^6*A(x)^6/6! + 1385*x^8*A(x)^8/8! + ... + A000364(n)*x^(2*n)*A(x)^(2*n)/(2*n)! + ...
%p S:= series(RootOf(z*cos(x*z)-1,z),x,101):
%p seq(coeff(S,x,2*j)*(2*j)!, j=0..50); # _Robert Israel_, Nov 24 2015
%t Table[(CoefficientList[1/x*InverseSeries[Series[x*Cos[x],{x,0,43}],x],x]*Range[0,42]!)[[n]],{n,1,41,2}] (* _Vaclav Kotesovec_, Jan 23 2014 *)
%t Table[Sum[(-1)^(m+n) m^(2n) Binomial[2k, k-m] (2n+k)!/(k (2k-2)!! (2n+1)!), {k, 1, 2n}, {m, 1, k}] + KroneckerDelta[n], {n, 0, 20}] (* _Vladimir Reshetnikov_, Nov 24 2015 *)
%o (PARI) {a(n)=(2*n)!*polcoeff((1/x)*serreverse(x*cos(x +O(x^(2*n+2)))),2*n)}
%o (PARI) /* From _Vladimir Kruchinin_'s formula in A185142: */
%o {a(n)=if(n==0,1,2*sum(k=1,2*n,binomial(2*n+k,2*n)*sum(j=1,k,sum(i=0,floor((j-1)/2),(j-2*i)^(2*n)*binomial(j,i))*binomial(k,j)*(-1)^(n-j)/2^j)))/(2*n+1))}
%Y Cf. A000364, A185142.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Oct 06 2011
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