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%I #15 Oct 26 2025 07:06:56
%S 1,2,6,18,-48,-2630,-52800,-824054,-8682240,54462258,7410631680,
%T 305163480578,8935815871488,167137193150954,-1440976761090048,
%U -349400091225243270,-22113174143289262080,-960586728800597050526,-26252145855684866211840,255024367557922004307442
%N E.g.f. A(x) satisfies A(x) = 1/( 1 - x * cos(x * A(x)) )^2.
%C As stated in the comment of A185951, A185951(n,0) = 0^n.
%F E.g.f.: B(x)^2, where B(x) is the e.g.f. of A381378.
%F a(n) = 2 * Sum_{k=0..n} k! * binomial(2*n-k+2,k)/(2*n-k+2) * i^(n-k) * A185951(n,k), where i is the imaginary unit.
%t terms = 20; A[_] = 0; Do[A[x_] =1/(1- x *Cos[x*A[x]])^2 + O[x]^terms // Normal, terms]; CoefficientList[A[x], x]*Range[0,terms-1]! (* _Stefano Spezia_, Oct 26 2025 *)
%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) = 2*sum(k=0, n, k!*binomial(2*n-k+2, k)/(2*n-k+2)*I^(n-k)*a185951(n, k));
%Y Cf. A185951, A381378.
%K sign
%O 0,2
%A _Seiichi Manyama_, Feb 22 2025