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%I #18 Feb 26 2019 10:53:51
%S 1,1,9,1,54,225,1,243,4131,11025,1,1008,50166,457200,893025,1,4077,
%T 520218,11708154,70301925,108056025,1,16362,5020623,243313164,
%U 3274844175,14427513450,18261468225,1,65511,46789461,4535570691,119537963811,1107456067125,3821273720775,4108830350625
%N Numerator coefficients of the bivariate Maclaurin series ("inverse Kepler equation") developed as Lagrange inversion E=KeplerInv(e,M) of Kepler's equation M = Kepler(e,E) = E - e*sin(E).
%C Coefficients of the numerator polynomials of the bivariate Maclaurin series ("inverse Kepler equation") developed as Lagrange inversion E = KeplerInv(e,M) of Kepler's equation M = Kepler(e,E) = E - e*sin(E), where e=numeric eccentricity, M=mean anomaly, E=eccentric anomaly. The series is KeplerInv(e,M) = M/(1-e) + Sum_{n>=1} (-1)^n*(Sum_{j=1..n} a(n,j)*e^j)/(1-e)^(3n+1)*M^(2n+1)/(2n+1)! = M/(1-e) - (e/(1-e)^4)*M^3/3! + ((e+9*e^2)/(1-e)^7)*M^5/5! - + ... .
%C The element a(n,n) with highest index in each row (the diagonal element) has the form Product_{j=1..n} (2*j+1)^2.
%C The derivative dKepler/dE = 1 - e*cos(E) goes to zero at E = i*arccosh(1/e) in the complex plane. Thus dKeplerInv/dM goes to infinity at M = i*(arccosh(1/e) - sqrt(1-e^2)), so that the radius of convergence of KeplerInv(e,M) is arccosh(1/e) - sqrt(1-e^2). KeplerInv(e,M) converges linearly within the circle of convergence |M| < arccosh(1/e) - sqrt(1-e^2).
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Kepler%27s_equation">Kepler's equation</a>
%F While M = E - e*sin(E) = E*(1-e) - e*Sum_{n>=1} (-1)^n*E^(2n+1)/(2n+1)! the formal power series of the compositional inverse KeplerInv(e,M) is as above according to A111785 and A304462.
%e Matrix (regular triangle) lexicographically ascending in the rows:
%e 1;
%e 1, 9;
%e 1, 54, 225;
%e 1, 243, 4131, 11025;
%e 1, 1008, 50166, 457200, 893025;
%e 1, 4077, 520218, 11708154, 70301925, 108056025;
%e 1, 16362, 5020623, 243313164, 3274844175, 14427513450, 18261468225;
%e ...
%Y Generated by A111785 or A304462, diagonal elements are in A001818.
%K nonn,tabl
%O 0,3
%A _Herbert Eberle_, Feb 23 2019
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