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G.f. satisfies A(x) = exp( Sum_{k>=1} A(x^k) * x^k/(k * (1 + x^k)^3) ).
2

%I #8 Jun 10 2023 11:18:30

%S 1,1,-1,1,2,-4,-1,10,-3,-20,19,38,-70,-65,221,73,-640,117,1745,-1223,

%T -4433,5770,10124,-22007,-18999,75063,19307,-235725,59665,685744,

%U -525477,-1832544,2531982,4364936,-10007555,-8468154,35302510,8542655,-114305453

%N G.f. satisfies A(x) = exp( Sum_{k>=1} A(x^k) * x^k/(k * (1 + x^k)^3) ).

%F A(x) = (1 + x)^3 * B(x) where B(x) is the g.f. of A363566.

%F a(n) = Sum_{k=0..3} binomial(3,k) * A363566(n-k).

%o (PARI) seq(n) = my(A=1); for(i=1, n, A=exp(sum(k=1, i, subst(A, x, x^k)*x^k/(k*(1+x^k)^3))+x*O(x^n))); Vec(A);

%Y Cf. A198518, A363567.

%Y Cf. A363548, A363566.

%K sign

%O 0,5

%A _Seiichi Manyama_, Jun 10 2023