%I #7 Sep 18 2022 12:36:49
%S 1,1,0,0,-7,-3,-17,52,51,384,-227,-52,-6311,-2722,-18733,79229,67453,
%T 620385,-619315,85796,-13137380,-595833,-43282243,205480697,66895157,
%U 1551910768,-2300631561,1546386060,-36481481081,15982958026,-135266506195,652843485153
%N Coefficients in the power series A(x) such that: 1 = Sum_{n=-oo..+oo} x^n * (1 - x^(n+1))^(n+1) * A(x)^n.
%C Compare to A356783 and A357160.
%C Related identity: 0 = Sum_{n=-oo..+oo} x^n * (1 - x^(n+1))^(n+1).
%C Related identity: 0 = Sum_{n=-oo..+oo} x^(k*n) * (y - x^(n+1-k))^n, which holds for any positive integer k and real y.
%H Paul D. Hanna, <a href="/A357200/b357200.txt">Table of n, a(n) for n = 0..300</a>
%F G.f. A(x) = Sum_{n>=0} a(n) * x^n satisfies the following relations.
%F (1) 1 = Sum_{n=-oo..+oo} x^n * (1 - x^(n+1))^(n+1) * A(x)^n.
%F (2) x*A(x) = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n-1)) / ( (1 - x^n)^n * A(x)^n ).
%F (3) -x*A(x)^2 = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n-1)) * A(x)^n / (1 - x^n*A(x))^n.
%F (4) -A(x)^3 = Sum_{n=-oo..+oo} x^n * (A(x) - x^(n+1))^(n+1) / A(x)^n.
%F (5) 0 = Sum_{n=-oo..+oo} x^n * (1 - x^(n+1)*A(x))^(n+1) / A(x)^n.
%F (6) 0 = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n-1)) * A(x)^n / (A(x) - x^n)^n.
%e G.f.: A(x) = 1 + x - 7*x^4 - 3*x^5 - 17*x^6 + 52*x^7 + 51*x^8 + 384*x^9 - 227*x^10 - 52*x^11 - 6311*x^12 - 2722*x^13 - 18733*x^14 + ...
%e such that
%e 1 = ... + x^(-2)*(1 - 1/x)^(-1)/A(x)^2 + x^(-1)/A(x) + (1 - x) + x*(1 - x^2)*A(x) + x^2*(1 - x^3)^3*A(x)^2 + x^3*(1 - x^4)^4*A(x)^3 + ... + x^n*(1 - x^(n+1))^(n+1)*A(x)^n + ...
%e also
%e -A(x)^3 = ... + x^(-2)*(A(x) - 1/x)^(-1)*A(x)^2 + x^(-1)*A(x) + (A(x) - x) + x*(A(x) - x^2)^2/A(x) + x^2*(A(x) - x^3)^3/A(x)^2 + x^3*(A(x) - x^4)^4/A(x)^3 + ... + x^n*(A(x) - x^(n+1))^(n+1)/A(x)^n + ...
%o (PARI) {a(n) = my(A=[1]); for(i=0, n, A = concat(A, 0);
%o A[#A] = polcoeff(1 - sum(n=-#A-2, #A+2, x^n * (1 - x^(n+1) +x*O(x^#A))^(n+1) * Ser(A)^n ), #A-2); ); A[n+1]}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A356783, A357160, A357201, A357202, A357203, A357204, A357205.
%K sign
%O 0,5
%A _Paul D. Hanna_, Sep 17 2022