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%I #49 Mar 03 2024 20:10:02
%S 1,3,15,315,36855,4833675,711485775,133449190875,33399969978375,
%T 10845524928112875,4368604540935009375,2121018409773134746875,
%U 1222083076784378918484375,826013017674132244878796875,647724113841936142199672859375,583169643524919352829283528046875
%N Expansion of e.g.f. arctan(1 - sqrt(1 - x^2)) (even powers only).
%H Deping Huang, <a href="/A333691/a333691.pdf">The bowl sequence</a>
%F a(1) = 1, a(n+1) = n*(2*n+1)*a(n) + (1/2^(2*n))*Sum_{k=1..n-1} A(n,k)*a(k),
%F A(n,k) = 2^(2*k)*(2*n-2*k-2)!*(4*n-4*k-5)*(2*n+1)!/((2*k-1)*(n-k+1)*(n-k-1)).
%e a(1) = 1.
%e a(2) = 1*(2*1+1)*a(1) = 3.
%e a(3) = -15*a(1) + 2*(2*2+1)*a(2) = 15.
%t m = 32; (Range[0, m]! * CoefficientList[Series[ArcTan[1 - Sqrt[1 - x^2]], {x, 0, m}], x])[[3 ;; -1 ;; 2]] (* _Amiram Eldar_, Apr 04 2020 *)
%o (PARI) a(n)={(2*n)!*polcoef(atan(1 - sqrt(1 - x^2 + O(x*x^(2*n)))), 2*n)} \\ _Andrew Howroyd_, Apr 04 2020
%K nonn
%O 1,2
%A _Deping Huang_, Apr 02 2020
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