%I #9 Feb 20 2025 08:39:31
%S 1,1,2,3,-12,-235,-2400,-18067,-51520,1701009,49829760,872355319,
%T 9861874176,-8805084275,-4518287900672,-159719520182055,
%U -3608706518138880,-44358720138978463,748112236681789440,72503399560668659531,2875934090148742430720,73418478070342765464741
%N E.g.f. A(x) satisfies A(x) = 1/( 1 - x * cos(x * A(x)^(1/2)) ).
%C As stated in the comment of A185951, A185951(n,0) = 0^n.
%F a(n) = Sum_{k=0..n} k! * binomial(n/2+k/2+1,k)/(n/2+k/2+1) * i^(n-k) * A185951(n,k), where i is the imaginary unit.
%o (PARI) a185951(n, k) = binomial(n, k)/2^k*sum(j=0, k, (2*j-k)^(n-k)*binomial(k, j));
%o a(n) = sum(k=0, n, k!*binomial(n/2+k/2+1, k)/(n/2+k/2+1)*I^(n-k)*a185951(n, k));
%Y Cf. A185951.
%K sign
%O 0,3
%A _Seiichi Manyama_, Feb 19 2025